PAS MATEMATIKA BLOG

Lanjutan Materi (5) Turunan fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\color{blue}\textrm{D. Aturan Rantai Turunan Fungsi Trigonometri}$

Jika fungsi $y=\left ( f\circ g \right )(x)=f\left ( g(x) \right )=f(u)$ dengan $u=g(x)$, maka turunan dari fungsi komposisi tersebut adalah:

$\color{blue}\begin{matrix} y'=\left ( f\circ g \right )'(x)=f'\left ( g(x) \right )\times g'(x)\\\\ \color{black}\textbf{atau}\\\\ \displaystyle \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx} \end{matrix}$

Perluasan dari teorema di atas, adalah berikut:

Diberikan $y=\left ( f\circ g\circ h \right )(x)=h\left (f\left ( g(x) \right ) \right )=f(u)$ dengan $u=g(v)$ dan $v=h(x)$, maka turunan pertama dari fungsi komposisi tersebut adalah:

$\color{blue}\begin{matrix} y'=\left ( f\circ g\circ h \right )'(x)=f'\left ( g\left ( h(x) \right ) \right )\times g'\left ( h(x) \right )\times h'(x)\\\\ \color{black}\textbf{atau}\\\\ \displaystyle \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx} \end{matrix}$

$\LARGE\color{magenta}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukan turunan pertama dari}\\ &f(x)=\sin ^{20}\left ( 8x^{5}+\pi \right )\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}f(x)&=\sin ^{20}\left ( 8x^{5}+\pi \right )=\left ( \sin \left ( 8x^{5}+\pi \right ) \right )^{20}\\ \textrm{Dim}&\textrm{isalkan}\\ y& =u^{20},\: \: \textrm{dengan}\\ u&=\sin \left ( 8x^{5}+\pi \right )\: \: \textrm{serta}\: \: u=\sin v\\ &\textrm{dan}\: \: v=\left ( 8x^{5}+\pi \right ),\\ \color{black}\textrm{mak}&\color{black}\textrm{a}\\ \displaystyle \frac{dy}{du}&=20u^{19}=20\sin ^{19}\left ( 8x^{5}+\pi \right ),\\ \displaystyle \frac{du}{dv}&=\cos v=\cos \left ( 8x^{5}+\pi \right ),\\ \displaystyle \frac{dv}{dx}&=40x^{4}\\ \color{black}\textrm{Seh}&\color{black}\textrm{ingga}\\ f'(x)&=\displaystyle \frac{dy}{dx}\\ &=\displaystyle \frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}\\ &=20\sin ^{19}\left ( 8x^{5}+\pi \right )\times \cos \left ( 8x^{5}+\pi \right )\times 40x^{4}\\ &=800x^{4}\sin ^{19}\left ( 8x^{5}+\pi \right )\cos \left ( 8x^{5}+\pi \right ) \end{aligned} \\\\ &\color{red}\textbf{atau kalau ingin langsungan saja}\\ &\color{red}\textrm{Tentunya jika Anda sudah lancar adalah}\\\\ &\color{purple}\begin{aligned}f(x)&=\sin ^{20}\left ( 8x^{5}+\pi \right )\\ f'(x)&=20\left ( \sin ^{19}\left ( 8x^{5}+\pi \right ) \right )\times \cos \left ( 8x^{5}+\pi \right )\times \left ( 40x^{4} \right )\\ &=800x^{4}\sin ^{19}\left ( 8x^{5}+\pi \right )\cos \left ( 8x^{5}+\pi \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukan turunan pertama dari}\\ &g(x)=\sqrt[5]{\cos ^{3}\left ( x^{2}-\pi \right )}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}g(x)&=\sqrt[5]{\cos ^{3}\left ( x^{2}-\pi \right )}=\cos ^{.^{\frac{3}{5}}}\left ( x^{2}-\pi \right )\\ \textrm{Dim}&\textrm{isalkan}\\ y& =u^{.^{\frac{3}{5}}},\: \: \textrm{dengan}\\ u&=\cos \left ( x^{2}-\pi \right )\: \: \textrm{serta}\: \: u=\cos v\\ &\textrm{dan}\: \: v=\left ( x^{2}-\pi \right ),\\ \color{black}\textrm{mak}&\color{black}\textrm{a}\\ \displaystyle \frac{dy}{du}&=\displaystyle \frac{3}{5}u^{.^{-\frac{2}{5}}}=\frac{3}{5}\cos ^{.^{-\frac{2}{5}}}\left ( x^{2}-\pi \right ),\\ \displaystyle \frac{du}{dv}&=-\sin v=-\sin \left ( x^{2}-\pi \right ),\\ \displaystyle \frac{dv}{dx}&=2x\\ \color{black}\textrm{Seh}&\color{black}\textrm{ingga}\\ g'(x)&=\displaystyle \frac{dy}{dx}\\ &=\displaystyle \frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}\\ &=\displaystyle \frac{3}{5}\cos ^{.^{-\frac{2}{5}}}\left ( x^{2}-\pi \right )\times \left ( -\sin \left ( x^{2}-\pi \right ) \right )\times (2x)\\ &=-\displaystyle \frac{6x\sin \left ( x^{2}-\pi \right )}{5\cos ^{.^{\frac{2}{5}}}\left ( x^{2}-\pi \right )}\\ &=-\displaystyle \frac{6x\sin \left ( x^{2}-\pi \right )}{5\sqrt[5]{\cos^{2} \left ( x^{2}-\pi \right )}} \end{aligned} \\\\ &\color{red}\textbf{atau kalau ingin langsungan saja}\\\\ &\color{purple}\begin{aligned}g(x)&=\sqrt[5]{\cos ^{3}\left ( x^{2}-\pi \right )}=\cos ^{.^{\frac{3}{5}}}\left ( x^{2}-\pi \right )\\ g'(x)&=\displaystyle \frac{3}{5}\cos ^{.^{-\frac{2}{5}}}\left ( x^{2}-\pi \right )\times \left ( -\sin \left ( x^{2}-\pi \right ) \right )\times (2x)\\ &=-\displaystyle \frac{6x\sin \left ( x^{2}-\pi \right )}{5\cos ^{.^{\frac{2}{5}}}\left ( x^{2}-\pi \right )}\\ &=-\displaystyle \frac{6x\sin \left ( x^{2}-\pi \right )}{5\sqrt[5]{\cos^{2} \left ( x^{2}-\pi \right )}} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Tentukan turunan pertama dari}\\ &h(x)=\cos \left ( \sin x^{2020} \right )\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}h(x)&=\cos \left ( \sin x^{2020} \right )\\ \textrm{Dim}&\textrm{isalkan}\\ y& =\cos \left ( \sin x^{2020} \right )=\cos u,\: \: \textrm{dengan}\\ u&=\sin x^{2020}=\sin v\: ,\: \textrm{serta}\: \: v=x^{2020}\\ \color{black}\textrm{mak}&\color{black}\textrm{a}\\ \displaystyle \frac{dy}{du}&=-\sin u=-\sin \left ( \sin x^{2020} \right ),\\ &\: \: \color{red}\textrm{atau}\: \: \color{black}dy=-\sin u\: \: du\\ \displaystyle \frac{du}{dv}&=\cos v\: \: \color{red}\textrm{atau}\: \: \color{black}du=\cos v\: \: dv\\ \displaystyle \frac{dv}{dx}&=2020x^{2019}\: \: \color{red}\textrm{atau}\: \: \color{black}dv=2020x^{2019}\: \: dx\\ \color{black}\textrm{Seh}&\color{black}\textrm{ingga}\\ h'(x)&=\displaystyle \frac{dy}{dx}\\ &=\displaystyle \frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}\\ &=-\sin \left ( \sin x^{2020} \right )\times \cos x^{2020}\times \left ( 2020x^{2019} \right )\\ &=-2020x^{2019}\sin \left ( \sin x^{2020} \right )\cos x^{2020} \end{aligned} \\\\ &\color{black}\textbf{atau}\\ &\color{purple}\begin{aligned}dy&=-\sin u\: \: du\\ &=-\sin u\times \cos v\: \: dv\\ &=-\sin u\times \cos v\times \left ( 2020x^{2019} \right )\: \: dx\\ \displaystyle \frac{dy}{dx}&=-\sin u\times \cos v\times \left ( 2020x^{2019} \right )\\ &=.......(\color{black}\textrm{tinggal dimasukkan}) \end{aligned} \end{array}$

DAFTAR PUSTAKA

Wirodikromo, S. 2007. Matematika Jilid 2 IPA untuk Kelas XI Berdasarkan Standar Isi 2006. Jakarta: ERLANGGA.

Lanjutan Materi (4) Turunan fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\color{blue}\textrm{C. Sifat-Sifat Turunan Fungsi Trigonometri}$

Sebelumnya silahkan ingat kembali pada dalil-dalil yang berlaku pada materi turunan fungsi aljabar di kelas XI, maka turunan fungsi trigonometri pun serupa, yaitu:

$\begin{array}{|c|l|l|}\hline \textrm{No}&\color{red}\textrm{Fungsi}&\color{blue}\textrm{Turunan Pertama}\\\hline 1.&y=k.u&y'=k.u'\\\hline 2.&y=u\pm v&y'=u'\pm v'\\\hline 3.&y=u.v&y'=v.u'+u.v'\\\hline 4.&y=k.u^{n}&=n.k.u^{(n-1)}.u'\\\hline 5.&y=\displaystyle \frac{u}{v}&y'=\displaystyle \frac{u'.v-u.v'}{v^{2}}\\\hline \end{array}$

Selanjutnya untuk turunan pertama fungsi di atas semisal fungsi $y=f(x)$ diturunkan terhadap $x$, maka turunan pertamnya dapat dituliskan dengan

$\color{blue}y'=\displaystyle \frac{dy}{dx}=f'(x)=\underset{h\rightarrow 0}{\textrm{lim}}\: \frac{f(x+h)-f(x)}{h}$

dan untuk turunan keduanya dari fungsi di atas adalah:

$\color{blue}y''=\displaystyle \frac{d^{2}y}{dx^{2}}=f''(x)\: \: \color{black}\textrm{atau kadang dituliskan}\: \: \displaystyle \color{purple}\frac{df'(x)}{dx}=\frac{d^{2}f}{dx^{2}}$

Selanjutnya perhatikanlah tabel berikut

$\color{blue}\begin{array}{|l|l|}\hline \textrm{Turunan}&\qquad\quad\quad\textrm{Notasi}\\\hline \textrm{Pertama}&y'=f'(x)=\displaystyle \frac{dy}{dx}=\frac{df}{dx}\\ &\\ \textrm{Kedua}&y''=f''(x)=\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{d^{2}f}{dx^{2}}\\ &\\ \textrm{Ketiga}&y'''=f'''(x)=\displaystyle \frac{d^{3}y}{dx^{3}}=\frac{d^{3}f}{dx^{3}}\\ &\\ \cdots &\cdots \qquad\cdots \qquad\cdots \qquad\cdots \\ \textrm{Ke-n}&y^{n}=f^{n}(x)=\displaystyle \frac{d^{n}y}{dx^{n}}=\frac{d^{n}f}{dx^{n}}\\\hline \end{array}$

$\LARGE\color{magenta}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah turunan pertama dari}\\ &\begin{array}{ll}\\ \textrm{a}.&y=\sin 2x\\ \textrm{b}.&y=\cos 4x\\ \textrm{c}.&y=\sin^{2} x\\ \textrm{d}.&y=\cos^{4} x\\ \textrm{e}.&y=-\sin x\\ \textrm{f}.&y=-5\cos 2x\\ \textrm{g}.&y=-7\tan x\\ \textrm{h}.&y=3\sin^{3} x\\ \textrm{i}.&y=-10\cos^{5} x\\ \textrm{j}.&y=-4\tan^{2} x\\ \textrm{k}.&y=\sqrt{\cos x}\\ \textrm{l}.&y=2\sin x+5x\\ \textrm{m}.&y=3\cos^{2} x+2x^{2}\\ \textrm{n}.&y=\csc x-2\tan ^{2}x+4x\\ \end{array} \end{array}$

$.\: \: \quad\color{blue}\begin{aligned}\textbf{Jawab}&\\ \textrm{Turun}&\textrm{an pertamanya masing}\\ \textrm{fungsi}&\: \textrm{di atas adalah berikut}:\\ (\textrm{a}).\: \: y&=\sin 2x\\ y'&=2\cos 2x\\ (\textrm{b}).\: \: y&=\cos 4x\\ y'&=-4\sin 4x\\ (\textrm{c}).\: \: y&=\sin^{2} x\\ y'&=2\sin x\cos x,\: \: \color{red}\textrm{atau boleh juga}\\ &=\sin 2x\\ (\textrm{d}).\: \: y&=\cos^{4} x\\ y'&=4\cos ^{3}(-\sin x)=-4\cos ^{3}x.\sin x\\ (\textrm{e}).\: \: y&=-2\sin x\\ y'&=-2\cos x\\ (\textrm{f}).\: \: y&=-5\cos 2x\\ y'&=-5(-\sin 2x.(2))=10x\sin 2x\\ (\textrm{g}).\: \: y&=-7\tan x\\ y'&=-7\sec ^{2}x\\ (\textrm{h}).\: \: y&=3\sin^{3} x\\ y'&=3.\left ( 3\sin ^{2}x \right ).(\cos x)=9\sin ^{2}x\cos x\\ (\textrm{i}).\: \: y&=-10\cos^{5} x\\ y'&=5\left ( -10\cos ^{4}x \right ).(-\sin x)\\ &=50\cos ^{4}x\sin x\\ (\textrm{j}).\: \: y&=-4\tan^{2} x\\ &=2\left ( -4\tan x \right ).\left ( \sec ^{2}x \right )\\ &=-8\tan x\sec ^{2}x\\ (\textrm{k}).\: \: y&=\sqrt{\cos x}=\cos ^{.^{\frac{1}{2}}}x\\ y'&=\displaystyle \frac{1}{2}\left ( \cos ^{.^{-\frac{1}{2}}}x \right ).\left ( -\sin x \right )\\ &=-\displaystyle \frac{1}{2}\cos ^{.^{-\frac{1}{2}}}x\sin x\\ &=-\displaystyle \frac{\sin x}{2\sqrt{\cos x}}\\ (\textrm{l}).\: \: y&=2\sin x+5x\\ y'&=2\cos x+5\\ (\textrm{m}).\: \: y&=3\cos^{2} x+2x^{2}\\ y'&=2\left ( 3\cos x \right ).(-\sin x)+4x\\ &=-6\cos x\sin x+4x,\: \: \color{red}\textrm{atau}\\ &=-3\sin 2x+4x=4x-3\sin 2x\\ (\textrm{n}).\: \: y&=\csc x-2\tan ^{2}x+4x\\ &=-\csc x\cot x-2\left ( 2\tan x \right ).\left ( \sec ^{2}x \right )+4\\ &=-\csc x\cot x-4\tan x\sec ^{2}x+4 \end{aligned}$

$\begin{array}{ll}\\ 2.&\textrm{Jika diketahui}\\ &\begin{array}{ll}\\ \textrm{a}.&f(x)=\displaystyle \frac{1+\sin x}{\cos x}.\: \: \textrm{Tentukanlah}\: \: f'(x)\\ \textrm{b}.&g(x)=\displaystyle \frac{\sin x+\cos x}{\cos x}.\: \: \textrm{Tentukanlah nilai}\\ &\textrm{saat}\: \: x=\displaystyle \frac{\pi }{6}\\ \textrm{c}.&h(x)=\sin x\tan x.\: \: \textrm{Tentukanlah nilai}\\ &\textrm{saat}\: \: x=45^{\circ}\\ \textrm{d}.&k(x)=\sin x+n\cos x\: \: \textrm{dan}\: \: k'\left ( \displaystyle \frac{\pi }{3} \right ) =0.\\ &\textrm{Tentukanlah nilai}\: \: n \end{array}\\\\ &\color{blue}\textbf{Jawab}: \end{array}$

$.\: \: \quad\color{blue}\begin{aligned}2.(\textrm{a})\: \: \textrm{dike}&\textrm{tahui}\: \: f(x)=\displaystyle \frac{1+\sin x}{\cos x}\\ \textrm{gun}&\textrm{akan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-uv'}{v^{2}}\\ f'(x)&=\displaystyle \frac{(\cos x)(\cos x)-(1+\sin x)(-\sin x)}{\cos ^{2}x}\\ &=\displaystyle \frac{\cos ^{2}x+\sin x+\sin ^{2}x}{\cos ^{2}x}\\ &=\displaystyle \frac{\color{red}\cos ^{2}x+\sin ^{2}x+\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\color{red}1+\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}+\frac{\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}+\frac{1}{\cos x}.\frac{\sin x}{\cos x}\\ &=\sec ^{2}x+\sec x\tan x \end{aligned}$

$.\: \: \quad\color{blue}\begin{aligned}2.(\textrm{b})\: \: \textrm{dike}&\textrm{tahui}\: \: g(x)=\displaystyle \frac{\sin x+\cos x}{\cos x}\\ \textrm{gun}&\textrm{akan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-uv'}{v^{2}}\\ g'(x)&=\displaystyle \frac{(\cos x-\sin x)(\cos x)-(\sin x+\cos x)(-\sin x)}{\cos ^{2}x}\\ &=\displaystyle \frac{\cos ^{2}x-\sin x\cos x+\sin x+\sin ^{2}x+\sin x\cos x}{\cos ^{2}x}\\ &=\displaystyle \frac{\color{red}\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}\\ &=\displaystyle \frac{\color{red}1}{\cos ^{2}x}\\ g\left ( \displaystyle \frac{\pi }{6} \right )&=\displaystyle \frac{1}{\cos ^{2}\left ( \displaystyle \frac{\pi }{6} \right )}\\ &=\left (\displaystyle \frac{1}{\cos \left ( \displaystyle \frac{\pi }{6} \right )} \right )^{2}\\ &=\left (\displaystyle \frac{1}{\cos 30^{\circ}} \right )^{2}\\ &=\left (\displaystyle \frac{1}{\displaystyle \frac{1}{2}\sqrt{3}} \right )^{2}\\ &=\left ( \displaystyle \frac{2}{\sqrt{3}} \right )^{2}\\ &=\frac{4}{3}\: \: \color{red}\textrm{Jika Anda tidak terganggu dengan nilai}\\ &\color{red}\textrm{perbandingan trigonometri, Anda bisa langsung saja}\\ &\color{red}\textrm{ke jawabannya, yaitu}\: \: \color{blue}\displaystyle \frac{4}{3} \end{aligned}$

$.\: \: \quad\color{blue}\begin{aligned}2.(\textrm{c})\: \: \textrm{dike}&\textrm{tahui}\: \: h(x)=\sin x\tan x\\ \textrm{gun}&\textrm{akan formula}\: \: \color{red}y=u.v\Rightarrow y'=u'.v+u.v'\\ h'(x)&=\cos x.(\tan x)+\sin x.\left ( \sec ^{2}x \right )\\ h\left ( 45^{\circ} \right )&=\cos \left ( 45^{\circ} \right )\tan \left ( 45^{\circ} \right )+\sin \left ( 45^{\circ} \right )\sec ^{2}\left ( 45^{\circ} \right )\\ &=\left ( \displaystyle \frac{1}{2}\sqrt{2} \right ).1+\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\left ( \sqrt{2} \right )^{2}\\ &=\displaystyle \frac{1}{2}\sqrt{2}+\sqrt{2}\\ &=\displaystyle \frac{3}{2}\sqrt{2} \end{aligned}$

$.\: \: \quad\color{blue}\begin{aligned}2.(\textrm{d})\: \: \textrm{dike}&\textrm{tahui}\: \: k(x)=\sin x+n\cos x\\ k'(x)&=\cos x-n\sin x,\: \: \color{red}\textrm{dengan}\: \: k'\left ( \displaystyle \frac{\pi }{3} \right ) =0\\ k'\left ( \displaystyle \frac{\pi }{6} \right )&=\cos \left ( \displaystyle \frac{\pi }{3} \right )-n\sin \left ( \displaystyle \frac{\pi }{3} \right )\\ 0&=\cos 60^{\circ}-n\sin 60^{\circ}=\displaystyle \frac{1}{2}-n\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ \Leftrightarrow \quad&n\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )=\displaystyle \frac{1}{2}\\ \Leftrightarrow \quad&n\: \: \, \,\quad\quad\quad =\displaystyle \frac{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{2}\sqrt{3}}\\ \Leftrightarrow \quad&n\: \: \, \,\quad\quad\quad =\displaystyle \frac{1}{\sqrt{3}}\times \color{black}\frac{\sqrt{3}}{\sqrt{3}}\\ \Leftrightarrow \quad&n\: \: \, \,\quad\quad\quad =\displaystyle \frac{1}{3}\sqrt{3} \end{aligned}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui fungsi}\: \: y=\displaystyle \frac{1}{2}\sin ^{2}x.\: \: \textrm{Tentukanlah}\\ &\textrm{Turunan pertama, kedua, ketiga, keempat},\\ &\textrm{dan kelima dari fungsi tersebut di atas}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\qquad y&=\displaystyle \frac{1}{2}\sin ^{2}x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan pertama}\\ \displaystyle \frac{dy}{dx}&=2\left ( \displaystyle \frac{1}{2}\sin ^{1}x \right )\times \cos x\\ &=\sin x\cos x=\displaystyle \frac{1}{2}\left ( 2\sin x\cos x \right )=\frac{1}{2}\sin 2x\\ \color{black} \textrm{Turu}&\color{black}\textrm{nan keduanya}\\ \displaystyle \frac{d^{2}y}{dx^{2}}&= \displaystyle \frac{1}{2}(\cos 2x).(2)=\color{red}\cos 2x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan ketiganya}\\ \displaystyle \frac{d^{3}y}{dx^{3}}&=-\sin 2x.(2)=\color{red}-2\sin 2x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan keempatnya}\\ \displaystyle \frac{d^{4}y}{dx^{4}}&=-2\cos 2x.(2)=\color{red}-4\cos 2x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan kelimanya}\\ \displaystyle \frac{d^{5}y}{dx^{5}}&=-4(-\sin 2x),(2)=\color{red}8\sin 2x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan keenamnya}\\ \displaystyle \frac{d^{6}y}{dx^{6}}&=8\cos 2x.(2)=\color{red}16\cos 2x \end{aligned} \end{array}$

DAFTAR PUSTAKA

Kurnia, N. 2018. Jelajah Matematika 3 SMA Kelas XII Peminatan MIPA. Bogor: Yudhistira
Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
Sembiring, S., Zulkifli, M., Marsito & Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT.
Tasari, Aksin, N., Miyanto & Muklis. 2016. Matematika untuk SMA/MA Kelas XII Peminatan Matematika dan Ilmu-Ilmu Alam. Klaten: INTAN PARIWARA.
Wirodikromo, S. 2007. Matematika Jilid 2 IPA untuk Kelas XI Berdasarkan Standar Isi 2006. Jakarta: ERLANGGA.

Lanjutan Materi (3) Turunan fungsi Trigonometri (Matematika Peminatan Kelas XII)

Selanjutnya saat kita masih kukuh menggu nakan rumus semual, maka saat menentukan turunan pertama fungsi $\tan x$, kita akan ketemu bentuk $\color{blue}\sin (x+h)\cos x$ dan $\color{blue}\cos (x+h)\sin x$, maka saat ketemu bentuk itu kita gunakan rumus:

$\color{purple}\begin{cases} \sin A\cos B&=\displaystyle \frac{1}{2}\left ( \sin (A+B)+\sin (A-B) \right ) \\\\ \cos A\sin B&=\displaystyle \frac{1}{2}\left ( \sin (A+B)-\sin (A-B) \right ) \end{cases}$

Coba perhatikanlah uraian turunan fungsi tangen berikut:

$\color{purple}\begin{aligned}f'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan (x+h)-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{ \sin (x+h)}{\cos (x+h)}-\displaystyle \frac{\sin x}{\cos x}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{ \sin (x+h)\cos x}{\cos (x+h)\cos x}-\displaystyle \frac{\cos (x+h)\sin x}{\cos (x+h)\cos x}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h).\cos x.h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{...+\displaystyle \frac{1}{2}\sin h-...+\displaystyle \frac{1}{2}\sin h}{\cos (x+h).\cos x.h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin h}{h.\cos (x+h).\cos x}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \left ( \displaystyle \frac{\sin h}{h} \right )\left ( \displaystyle \frac{1}{\cos (x+h)\cos x} \right )\\ &=1\times \displaystyle \frac{1}{\cos (x+0).\cos x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}\\ &=\sec ^{2}x \end{aligned}$

Atau kita juga dapat menggunakan rumus $\color{red}\sin (A-B)=\sin A\cos B-\cos A\sin B$ sebagaimana berikut ini (perhatikanlah proses langkah 5 ke langkah 6):

$\color{blue}\begin{aligned}f'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan (x+h)-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{ \sin (x+h)}{\cos (x+h)}-\displaystyle \frac{\sin x}{\cos x}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{ \sin (x+h)\cos x}{\cos (x+h)\cos x}-\displaystyle \frac{\cos (x+h)\sin x}{\cos (x+h)\cos x}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h).\cos x.h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ((x+h)-x \right )}{\cos (x+h).\cos x.h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin h}{h.\cos (x+h).\cos x}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \left ( \displaystyle \frac{\sin h}{h} \right )\left ( \displaystyle \frac{1}{\cos (x+h)\cos x} \right )\\ &=1\times \displaystyle \frac{1}{\cos (x+0).\cos x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}\\ &=\sec ^{2}x \end{aligned}$

Berikut hasil turunan pertama untuk fungsi trigonometri yang perlu diingat:

$\color{purple}\begin{aligned}1.\quad &f(x)=\sin x\Rightarrow f'(x)=\cos x\\ 2.\quad &f(x)=\cos x\Rightarrow f'(x)=-\sin x\\ 3.\quad &f(x)=\tan x\Rightarrow f'(x)=\sec ^{2}x\\ 4.\quad &f(x)=\cot x\Rightarrow f'(x)=-\csc ^{2}x\\ 5.\quad &f(x)=\sec x\Rightarrow f'(x)=\sec x\tan x\\ 6.\quad &f(x)=\csc x\Rightarrow f'(x)=-\csc x\cot x \end{aligned}$

Lanjutan Materi (2) Turunan fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\color{blue}\textrm{B. Turunan Fungsi Trigonometri}$

Fungsi trigonometri di sini adalah suatu fungsi yang mengandung perbandingan trigonometri serta perbandingan trigonometri tersebut bukan merupakan ekponen

Kita ingat sebelumnya untuk menentukan turunan pertama suatu fungsi $f(x)$ yang selanjutnya di dinotasikan dengan $f'(x)$ adalah:

$\color{blue}f'(x)=\underset{h\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{f(x+h)-f(x)}{h}$

Selanjutnya dalam menentukan turunan formula di atas dapat digunakan untuk menentukan turunan pertama fungsi trigonometri, sebagai mana contoh berikut:

Ambil contoh $\color{purple}f(x)=\sin x$, maka kita akan menentukan turuan pertamanya, yaitu:

Pada salah satu langkah di antara langkah di atas ada beberapa rumus yang perlu diingat saat Anda duduk di kelas XI, yaitu penggunaan rumus

$\color{blue}\sin A-\sin B=2\cos \displaystyle \frac{1}{2}(A+B)\sin \displaystyle \frac{1}{2}(A-B)$.

Anda boleh juga menggunakan rumus yang lain. Karena di dalamnya ada $\sin (x+h)$, Anda dapat menggunakan rumus berikut:

$\color{blue}\sin (A+B)=\sin A\cos B+\cos A\sin B$

Coba perhatikan penggunaanya berikut, tapi malah agak panjang dikit jadinya

$\color{purple}\begin{aligned}f'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin (x+h)-\sin x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin x\left ( \cos h-1 \right )+\cos x\sin h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin x\left ( \cos h-1 \right )}{h}+\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\cos x\sin h}{h}\\ &=\sin x.\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\cos h-1}{h}+\cos x.\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin h}{h}\\ &=\sin x.\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{-2\sin ^{2}\displaystyle \frac{1}{2}h}{h}+\cos x.1\\ &=\sin x.0+\cos x\\ &=\cos x \end{aligned}$

Sampai di sini kita akan bisa coba lagi menentukan turunan pertama fungsi kosinus, sebagaimana uraian berikut:

Contoh Soal 6 Fungsi Logaritma (Uraian)

$\begin{array}{l}\\ 26.&\textrm{Diketahui bahwa}\\ & ^{^{2}}\log 3=p \: \: \textrm{dan}\: \: ^{^{3}}\log 11=q,\\ &\textrm{maka nilai}\: \: ^{^{44}}\log 66=....\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}^{^{44}}\log 66&=\displaystyle \frac{^{^{^{...}}}\log 66}{^{^{^{...}}}\log 44}\\ &=\displaystyle \frac{^{^{^{...}}}\log \left (2\times 3\times 11 \right )}{^{^{^{...}}}\log \left (2^{2}\times 11 \right )}\\ &=\displaystyle \frac{^{^{^{3}}}\log 2+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{^{^{^{3}}}\log 2^{2}+\: ^{^{^{3}}}\log 11}\\ &=\displaystyle \frac{\frac{1}{^{^{^{2}}}\log 3}+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{\frac{2}{^{^{^{3}}}\log 2^{2}}+\: ^{^{^{3}}}\log 11}\\ &=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\\ &=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\times \displaystyle \frac{p}{p}\\ &=\frac{1+p+pq}{2+pq} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textbf{(AIME 1984)}\\ &\textrm{Diketahui bahwa}\: \: x\: \: \textrm{dan}\: \: y\\ & \textrm{adalah bilangan real yang memenuhi}\\\\ &\left\{\begin{matrix} ^{^{8}}\log x+\: ^{^{4}}\log y^{2}=5\\ \\ ^{^{8}}\log y+\: ^{^{4}}\log x^{2}=7 \end{matrix}\right.\\\\ &\textrm{Tentukanlah nilai dari}\: \: xy\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&^{^{^{8}}}\log x+\: ^{^{4}}\log y^{2}=5\\ &\Leftrightarrow \: ^{^{^{2^{3}}}}\log x+\: ^{^{^{2^{2}}}}\log y^{2}=5....(1)\\ &^{^{^{8}}}\log y+\: ^{^{4}}\log x^{2}=7\\ &\Leftrightarrow \: ^{^{^{2^{3}}}}\log y+\: ^{^{^{2^{2}}}}\log x^{2}=7....(2)\\ &\textrm{selanjutnya},&\\ &\frac{1}{3}\:. ^{^{^{2}}}\log x+\: ^{^{^{2}}}\log y=5\: \: |\times \frac{1}{3}|\\ &\Rightarrow \frac{1}{9}\: .^{^{^{2}}}\log x+\: \frac{1}{3}.\: ^{^{^{2}}}\log y=\frac{5}{3}....(3)\\ &^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7\: \: |\times 1|\\ &\Rightarrow ^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7....(4)\\ &\textrm{saat persamaan}\: \: (3)-(4)\\ &=-\frac{8}{9}.\: ^{^{^{2}}}\log x=\frac{5}{3}-7=-\frac{16}{3}\\ &\color{purple}\textrm{maka}\\ &^{^{^{2}}}\log x=\left ( -\frac{16}{3} \right )\left ( -\frac{9}{8} \right )\\ &^{^{^{2}}}\log x=6\Leftrightarrow x=2^{6}\Leftrightarrow x=64\\ &\color{purple}\textrm{Pada persamaan 1 selanjutnya}\\ &\frac{1}{3}.\: ^{^{^{2}}}\log x+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: \frac{1}{3}.\: ^{^{^{2}}}\log 2^{6}+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: \frac{1}{3}.6+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: 2+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: ^{^{^{2}}}\log y=5-2=3\Leftrightarrow y=2^{3}=8\\ &\textrm{Jadi},\: \: x.y=64.8=512 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&\textrm{Tentukanlah nilai dari}\\ &\textrm{a}.\quad \left ( 2^{\: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &\textrm{b}.\quad \left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &\textrm{c}.\quad \left ( ^{625}\log 19 \right )\left ( ^{7}\log \displaystyle \frac{1}{25} \right )\left ( ^{19}\log 7 \right )\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{array}{|l|}\hline \begin{aligned}\textrm{a}.\quad &\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3^{2}}}}\log 5} \right )\left ( 5^{\: \: ^{^{^{5^{-1}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3}}}\log 5^{^{\frac{1}{2}}}} \right )\left ( 5^{\: \: ^{^{^{5}}}\log 2^{-1}} \right )\\ &=\left ( 2^{\: ^{^{2}}\log 6} \right )\left ( 3^{\: ^{^{3}}\log \sqrt{5}} \right )\left ( 5^{\: ^{^{5}}\log \frac{1}{2}} \right )\\ &=6\times \sqrt{5}\times \frac{1}{2}\\ &=3\sqrt{5} \end{aligned}\\\hline \end{array}\\ &\color{blue}\begin{array}{|l|}\hline \begin{aligned}\textrm{b}.\quad &\left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &=\left ( ^{^{3^{3}}}\log 5^{3} \right )\left ( ^{^{5^{2}}}\log 4^{-3} \right )\left ( ^{^{4^{3}}}\log 3^{-2} \right )\\ &=\displaystyle \frac{3}{3}.\left ( -\frac{3}{2} \right ).\left ( -\frac{2}{3} \right ).\: ^{^{3}}\log 5.\: ^{^{5}}\log 4.\: ^{^{4}}\log 3\\ &=1\end{aligned}\\\hline \end{array}\\ &\color{purple}\textrm{Pembahasan diserahkan kepada}\\ &\color{purple}\textrm{Pembaca yang budiman untuk poin c} \end{array}$

$\begin{array}{ll}\\ 29.&\textrm{Tentukanlah nilai}\: \: a+b\: \: \textrm{dimana}\: \: a\: \: \textrm{dan}\: \: b\\ &\textrm{adalah bilangan riil positif}.\\ &^{7}\log \left ( 1+a^{2} \right )-\: ^{7}\log 25=\: ^{7}\log \left ( 2ab-15 \right )-\: ^{7}\log \left ( 25+b^{2} \right )\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&^{7}\log \displaystyle \frac{\left ( 1+a^{2} \right )}{25}=\: ^{7}\log \displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ &\textrm{diambil}\: \textrm{persamaannya, maka}\\ &\displaystyle \frac{\left ( 1+a^{2} \right )}{25}=\displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ &\displaystyle \left ( 1+a^{2} \right )\left ( 25+b^{2} \right )=25\left ( 2ab-15 \right )\\ &\begin{cases} \left ( 1+a^{2} \right ) & \text{ faktor dari} \: \: 25,\: \: a> 0,\: a\in \mathbb{R} \\ &\textrm{atau}\\ \left ( 25+b^{2} \right ) & \text{ faktor dari }\: \: 25,\: \: b> 0,\: b\in \mathbb{R}\: \: \: \textrm{juga} \end{cases}\\ \end{aligned}\\\\ &\color{purple}\begin{array}{|c|l|c|l|c|c|c|}\hline \textrm{No}&a&\left ( 1+a^{2} \right )&b&\left ( 25+b^{2} \right )&\textrm{Keterangan}&a+b\\\hline 1&2&5&10&125&\textrm{Memenuhi}&12\\\hline 2&7&50&\cdots &\cdots &\textrm{tidak}&\cdots \\\hline 3&\cdots &\cdots &5&50&\textrm{tidak}&\cdots \\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 30.&\textrm{Jika}\: \: 60^{a}=3\: \: \textrm{dan}\: \: 60^{b}=5\\ &\textrm{maka hasil dari}\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}\\\\ &\textrm{Jawab}:\\ &\color{blue}\textrm{Perhatikanlah bahwa}\\ &\color{blue}\begin{aligned}&\begin{cases} 60^{a}=3 & \Rightarrow \: ^{60}\log 3=a \\ 60^{b}=5 & \Rightarrow \: ^{60}\log 5=b \end{cases} \end{aligned} \\ &\color{blue}\textrm{Selanjutnya}\\ &\color{blue}\textrm{Untuk}\: \: (1-a-b),\: \textrm{maka}\\ &\color{blue}\begin{aligned}1-a-b&=1-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log 60-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log \displaystyle \frac{60}{3\times 5}\\ &=\: ^{60}\log 4\\ &=\: ^{60}\log 2^{2} \end{aligned} \\ &\color{blue}\textrm{Untuk}\: \: 2(1-b),\: \textrm{maka}\\ &\color{blue}\begin{aligned}2(1-b)&=2\left ( 1-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log 60-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log \displaystyle \frac{60}{5} \right )\\ &=2\left ( ^{60}\log 12 \right )\\ &=\: ^{60}\log 12^{2} \end{aligned} \\ &\color{blue}\textrm{Untuk}\: \: \left ( \displaystyle \frac{1-x-y}{2(1-b)} \right ),\: \textrm{maka}\\ &\color{blue}\begin{aligned}\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )&=\displaystyle \frac{\: ^{60}\log 2^{2}}{\: ^{60}\log 12^{2}}\\ &=\displaystyle \frac{2\: ^{60}\log 2}{2\: ^{60}\log 12}\\ &=\displaystyle \frac{\: ^{60}\log 2}{\: ^{60}\log 12}\\ &=\: ^{12}\log 2 \end{aligned}\\ &\color{blue}\textrm{Jadi},\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}=12^{^{\: ^{12}\log 2}}=2 \end{array}$

$\begin{array}{ll}\\ 31.&\textrm{Diberikan bilangan riil positif}\: \: x,\: y,\: \textrm{dan}\: z\\ & \textrm{yang memenuhi persamaan}\\ &2\: ^{x}\log (2y)=2\: ^{2x}\log (4z)=2\: ^{4x}\log (8yz)\neq 0.\\ &\textrm{Jika nilai}\: \: xy^{5}z\: \: \textrm{dapat dinyatakan dengan}\: \: \displaystyle \frac{1}{2^{\displaystyle \frac{p}{q}}}\\ & \textrm{dengan}\: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{bilangan asli yang saling prima},\\ &\textrm{maka nilai dari}\: \: p+q=....\\\\ &\textrm{Jawab}:\\\\ &\color{blue}\begin{aligned}&2\: ^{x}\log (2y)=2\: ^{2x}\log (4z)=2\: ^{4x}\log (8yz)\neq 0\\ &\textrm{maka}\\ &^{x}\log (2y)=\: ^{2x}\log (4y)\\ &\Rightarrow \quad \log (2y)\times \log (2x)=\log x\times \log (4y)...(1)\\ &^{x}\log (2y)=\: ^{4x}\log (8yz)\\ &\Rightarrow \quad \log (2y)\times \log (4x)=\log x\times \log (8yz)....(2)\\ &^{2x}\log (4y)=\: ^{4x}\log (8yz)\\ &\Rightarrow \quad \log (4y)\times \log (4x)=\log (2x)\times \log (8yz)....(3) \end{aligned}\\\\ &\color{blue}\begin{aligned}&\textrm{Perhatikan persamaan}\: \: (2),\: \textrm{yaitu}:\\ &\log (2y)\times \log (4x)=\log x\times \log (8yz)\\ &\log (2y)\times \left (\log (2x)+\log 2 \right )=\log x\times \log (8yz)\\ &\log (2y)\times \log (2x)+\log (2y)\times \log 2=\log x\times \log (8yz)\\ &\log x\times \log (4y)+\log (2y)\times \log 2=\log x\times \log (8yz)\\ &\quad\textrm{persamaan di atas, persamaan}\: \: (1)\: \: \textrm{disubstitusikan}\\ &\log (2y)=\displaystyle \frac{\log x\times \log (8yz)-\log x\times \log (4y)}{\log 2}\\ &\log (2y)=\displaystyle \frac{\log x\times \left ( \log \displaystyle \frac{8yz}{4y} \right )}{\log 2}\\ &\log (2y)=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\: ......(4) \end{aligned}\\\\ &\color{blue}\begin{aligned}&\textrm{Perhatikan juga persamaan}\: \: (3),\: \textrm{yaitu}:\\ &\log (4y)\times \log (4x)=\log (2x)\times \log (8yz)\\ &\left (\log (2y)+\log 2 \right )\times \log (4x)=\log (2x)\times \log (8yz)\\ &\log (2y)\times \log (4x)+\log 2\times \log (4x)=\log (2x)\times \log (8yz)\\ &\log x\times \log (8yz)+\log 2\times \log (4x)=\log (2x)\times \log (8yz)\\ &\quad\textrm{di atas, persamaan}\: \: (2)\: \: \textrm{disubstitusikan}\\ &\log 2\times \log (4x)=\log (2x)\times \log (8yz)-\log x\times \log (8yz)\\ &\log 2\times \log (4x)=\log (8yz)\times \left ( \log (2x)-\log x \right )\\ &\log 2\times \log (4x)=\log (8yz)\times \left ( \log \displaystyle \frac{2x}{x} \right )\\ &\log 2\times \log (4x)=\log (8yz)\times \log 2\\ &\log 4x=\log (8yz)\\ &4x=8yz\\ &\displaystyle \frac{x}{z}=2y\: ....(5) \end{aligned} \end{array}$

$.\: \: \qquad\color{purple}\begin{aligned}&\textrm{dari persamaan}\: \: (4)\: \: \textrm{dan}\: \: (5)\\ &\log (2y)=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ &\log \left ( \displaystyle \frac{x}{z} \right )=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ &\log 2\left ( \log x-\log z \right )=\log x\times \log (2z)\\ &\log 2\times \log x-\log 2\times \log z=\log x\times \left ( \log 2+\log z \right )\\ &\log 2\times \log x-\log 2\times \log z=\log x\times \log 2+\log x\times \log z\\ &-\log 2\times \log z=\log x\times \log z\\ &\log 2^{-1}=\log x\\ &\displaystyle \frac{1}{2}=x\: .....(6) \end{aligned}$

$.\: \: \qquad\color{blue}\begin{array}{|c|c|}\hline \textrm{persamaan}\: \: (2)&\textrm{Menentukan nilai}\: \: z\\\hline \begin{aligned} \log 2y\times \log (4x)&=\log x\times \log (8yz)\\ \log 2y\times \log (4(2yz))&=\log x\times \log (8yz)\\ \log 2y\times \log (8yz)&=\log x\times \log (8yz)\\ \log (2y)&=\log x\\ 2y&=x\\ y&=\displaystyle \frac{1}{2}x\\ &=\displaystyle \frac{1}{2}\times \frac{1}{2}\\ &=\displaystyle \frac{1}{4}\: .....(7)\\ & \end{aligned}&\begin{aligned}&\\ x&=2yz\\ \displaystyle \frac{1}{2}&=2\left ( \displaystyle \frac{1}{4} \right )z\\ 1&=z\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}$

$.\: \: \qquad\color{purple}\begin{aligned}&\textrm{maka nilai untuk}\: \: xy^{5}z\: \: \textrm{adalah}\\ &xy^{5}z=\left ( \displaystyle \frac{1}{2} \right ).\left ( \displaystyle \frac{1}{4} \right )^{5}.1\\ &=\displaystyle \frac{1}{2\times 4^{5}}\\ &=\displaystyle \frac{1}{2\times \left ( 2^{2} \right )^{5}}=\displaystyle \frac{1}{2^{1+10}}\\ &=\displaystyle \frac{1}{2^{11}}=\displaystyle \frac{1}{2^{^{\frac{11}{1}}}}=\displaystyle \frac{1}{2^{^{\frac{p}{q}}}}\\ &\begin{cases} p & =11 \\ q & =1 \end{cases}\quad \textrm{dan jelas bahwa} \: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{saling prima}\\ &\textrm{Jadi},\\ &p+q=11+1=12 \end{aligned}$

DAFTAR PUSTAKA

Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
Sembiring, S. 2002. Olimpiade Matematika untuk SMU. Bandung: YRAMA WIDYA.

Contoh Soal 5 Fungsi Logaritma

$\begin{array}{ll}\\ 21.&\textbf{(SPMB '04)}\\ &\textrm{Jika}\: \: a>1\: ,\: \textrm{maka penyelesaian untuk}\\ &\left (^{a}\log (2x+1) \right )\left ( ^{3}\log \sqrt{a} \right )=1\: \: \textrm{adalah}\: ....\\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 1\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 3\\ \color{red}\textrm{d}.&\displaystyle 4\\ \textrm{e}.&\displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\left (^{a}\log (2x+1) \right )\left ( ^{3}\log \sqrt{a} \right )&=1\\ \left ( ^{3}\log \sqrt{a} \right )\left (^{a}\log (2x+1) \right )&=1\\ \left ( ^{3}\log a^{.^{^{\frac{1}{2}}}} \right )\left (^{a}\log (2x+1) \right )&=1\\ \displaystyle \frac{1}{2}\left ( ^{3}\log a \right )\left (^{a}\log (2x+1) \right )&=1\\ ^{3}\log (2x+1)&=2\\ 2x+1&=3^{2}\\ 2x&=9-1\\ 2x&=8\\ x&=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textbf{(SPMB '04)}\\ &\textrm{Nilai}\: \: \displaystyle \frac{\left ( ^{5}\log 10 \right )^{2}-\left ( ^{5}\log 2 \right )^{2}}{^{5}\log \sqrt{20}}\: \: \textrm{adalah}\: ....\\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\ \textrm{b}.&\displaystyle 1\\ \color{red}\textrm{c}.&\displaystyle 2\\ \textrm{d}.&\displaystyle 4\\ \textrm{e}.&\displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\displaystyle \frac{\left ( ^{5}\log 10 \right )^{2}-\left ( ^{5}\log 2 \right )^{2}}{^{5}\log \sqrt{20}}\\ &=\displaystyle \frac{\left ( ^{5}\log 10+\: ^{5}\log 2 \right )\left ( ^{5}\log 10-\: ^{5}\log 2 \right )}{^{5}\log (20)^{.^{\frac{1}{2}}}}\\ &=\displaystyle \frac{^{5}\log (10.2)\times ^{5}\log \left (\frac{10}{2} \right )}{\displaystyle \frac{1}{2}\times \: ^{5}\log 20}\\ &=2\times \left ( \displaystyle \frac{^{5}\log 20}{^{5}\log 20} \right )\times \: ^{5}\log 5\\ &=2.1.1\\ &=2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 23.&\textbf{(SPMB '03)}\\ &\textrm{Jika diketahui bahwa}\\ &^{4}\log ^{4}\log x-\: ^{4}\log ^{4}\log ^{4}\log 16=2\\ & \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ^{2}\log x=8\\ \textrm{b}.&\displaystyle ^{2}\log x=4\\ \color{red}\textrm{c}.&\displaystyle ^{4}\log x=8\\ \textrm{d}.&\displaystyle ^{4}\log x=16\\ \textrm{e}.&\displaystyle ^{16}\log x=8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 16=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 4^{2}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log 2=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{2^{2}}\log 2^{1}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \left ( \displaystyle \frac{1}{2} \right )=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{2^{2}}\log 2^{-1}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\left ( -\displaystyle \frac{1}{2} \right )=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x+\frac{1}{2}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x=2-\displaystyle \frac{1}{2}=\frac{3}{2}\\ &\Leftrightarrow \: ^{4}\log x=4^{.\frac{3}{2}}\\ &\Leftrightarrow \: ^{4}\log x=\left (2^{2} \right )^{.^{\frac{3}{2}}}\\ &\Leftrightarrow \: ^{4}\log x=2^{3}\\ &\Leftrightarrow \: ^{4}\log x=8\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 24.&\textbf{(UMPTN '92)}\\ &\textrm{Jika}\: \: x\: \: \textrm{memenuhi persamaan}\\ &^{4}\log ^{4}\log x-\: ^{4}\log ^{4}\log ^{4}\log 16=2\\ & \textrm{maka nilai}\: \: ^{16}\log x=\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle 4\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle -2\\ \textrm{e}.&\displaystyle -4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 16=2\\ &\textrm{menyebabkan}\\ & ^{4}\log x=8\Rightarrow x=4^{8}\\ &(\color{purple}\textrm{lihat pembahasan no.23})\\ &\textrm{maka},\\ &\: ^{16}\log x=\: ^{16}\log 4^{8}=\: ^{4^{2}}\log 4^{8}\\ &=\displaystyle \frac{8}{2}\\ &=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textbf{(UMPTN '94)}\\ &\textrm{Hasil kali akar-akar persamaan}\\ &^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{9}\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle 3\\ \textrm{e}.&\displaystyle 9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ &\Leftrightarrow \: \left ({2+\: ^{3}\log x} \right )^{3}\log x-15=0\\ &\Leftrightarrow 2\: ^{3}\log x+\: \left ( ^{3}\log x \right )^{2}-15=0\\ &\Leftrightarrow \: \left ( ^{3}\log x \right )^{2}+2\: ^{3}\log x-15=0\\ &\Leftrightarrow \: \left (^{3}\log x_{1}+5 \right )\left ( ^{3}\log x_{2}-3 \right )=0\\ &\Leftrightarrow \: ^{3}\log x_{1}+5=0\: \: \textrm{atau}\: \: ^{3}\log x_{1}-3=0\\ &\Leftrightarrow \: ^{3}\log x_{1}=-5\: \: \textrm{atau}\: \: ^{3}\log x_{2}=3\\ &\Leftrightarrow \: x_{1}=3^{-5}\: \: \textrm{atau}\: \: x_{2}=3^{3}\\ &\qquad \textrm{maka}\\ &\Leftrightarrow \: x_{1}\times x_{2}=3^{-5}\times 3^{3}=3^{-5+3}=3^{-2}\\ &\Leftrightarrow \qquad =\displaystyle \frac{1}{3^{2}}\\ &\Leftrightarrow \qquad =\frac{1}{9} \end{aligned} \end{array}$

Contoh Soal 4 Fungsi Logaritma

$\begin{array}{ll}\\ 16.&\textbf{(UMPTN '01)}\\ &\textrm{Jika}\: \: ^{10}\log x=b\: ,\: \textrm{maka}\: \: ^{10x}\log 100=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{b+1}\\ \color{red}\textrm{b}.&\displaystyle \frac{2}{b+1}\\ \textrm{c}.&\displaystyle \frac{1}{b}\\ \textrm{d}.&\displaystyle \frac{2}{b}\\ \textrm{e}.&\displaystyle \frac{2}{10b} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&^{10x}\log 100\\ &=\displaystyle \frac{\log 100}{\log 10x}\\ &=\displaystyle \frac{^{10}\log 100}{^{10}\log 10x},\quad \color{magenta}\textrm{pilih basis 10}\\ &\color{purple}\textrm{alasannya: supaya sama dengan soal}\\ &=\displaystyle \frac{^{10}\log 10^{2}}{^{10}\log 10+\: ^{10}\log x}\\ &=\displaystyle \frac{2}{1+b}\: \: \textrm{atau}\\ &=\frac{2}{b+1} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textbf{(UM UGM '03)}\\ &\textrm{Jika}\: \: ^{4}\log 6=m+1\: ,\: \textrm{maka}\: \: ^{9}\log 8=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{3}{4m-2}\\ \color{red}\textrm{b}.&\displaystyle \frac{3}{4m+2}\\ \textrm{c}.&\displaystyle \frac{3}{2m+4}\\ \textrm{d}.&\displaystyle \frac{3}{2m-4}\\ \textrm{e}.&\displaystyle \frac{3}{2m+2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\color{purple}\textrm{Sebelumnya perhatikanlah}\\ &^{4}\log 6=m+1\\ &\Leftrightarrow \: ^{2^{2}}\log (2.3)^{1}=m+1\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\times \: ^{2}\log (2.3)=m+1\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\times \: \left (^{2}\log 2 +\: ^{2}\log 3 \right )=m+1\\ &\Leftrightarrow \: \displaystyle \frac{1}{2}\times \: \left (1 +\: ^{2}\log 3 \right )=m+1\\ &\Leftrightarrow \: 1 +\: ^{2}\log 3=2m+2\\ &\Leftrightarrow \: ^{2}\log 3=2m+1\\ &\color{purple}\textrm{Selanjutnya adalah}:\\ &^{9}\log 8=\: \displaystyle \frac{1}{^{8}\log 9}\\ &=\: \displaystyle \frac{1}{^{2^{3}}\log 3^{2}}\\ &=\: \displaystyle \frac{1}{\displaystyle \frac{2}{3}\: ^{2}\log 3}\\ &=\: \displaystyle \frac{3}{2\: ^{2}\log 3}\\ &=\: \displaystyle \frac{3}{2(2m+1)}\\ &=\: \displaystyle \frac{3}{4m+2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textbf{(UMPTN '00)}\\ &\textrm{Jika}\: \: ^{3}\log 5=p\: \: \textrm{dan}\: \: ^{3}\log 4=q,\\ &\textrm{maka}\: \: ^{4}\log 15=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{pq}{1+p}\\ \textrm{b}.&\displaystyle \frac{p+q}{pq}\\ \color{red}\textrm{c}.&\displaystyle \frac{p+1}{pq}\\ \textrm{d}.&\displaystyle \frac{p+1}{q+1}\\ \textrm{e}.&\displaystyle \frac{pq}{1-p} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{4}\log 15\\ &=\displaystyle \frac{^{...}\log 15}{^{...}\log 4},\: \: \color{purple}\textrm{pilih basis 5}\\ &\color{magenta}\textrm{mengapa tidak pilih basis selain 5}\\ &\color{magenta}\textrm{lihat penyebut, di sana terdapat numerus 4}\\ &\color{magenta}\textrm{pada soal, pasangan numerus 4 adalah 5},\\ &\color{black}\textrm{makanya basis 5 dipilih, bukan yang lain}\\ &=\displaystyle \frac{^{5}\log 15}{^{5}\log 4}=\displaystyle \frac{^{5}\log (3.5)}{^{5}\log 4}\\ &=\displaystyle \frac{^{5}\log 3+\: ^{5}\log 5}{^{5}\log 4}=\displaystyle \frac{\displaystyle \frac{1}{^{3}\log 5}+\: ^{5}\log 5}{^{5}\log 4}\\ &=\displaystyle \frac{\displaystyle \frac{1}{p}+1}{q}=\displaystyle \frac{1+p}{pq},\: \: \textrm{atau}\\ &=\displaystyle \frac{p+1}{pq} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textbf{(UMPTN '94)}\\ &\textrm{Jika}\: \: ^{6}\log 5=a\: \: \textrm{dan}\: \: ^{5}\log 4=b,\\ &\textrm{maka}\: \: ^{4}\log 0,24=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{a+2}{ab}\\ \textrm{b}.&\displaystyle \frac{2a+1}{ab}\\ \textrm{c}.&\displaystyle \frac{a-2}{ab}\\ \textrm{d}.&\displaystyle \frac{2a+1}{2ab}\\ \color{red}\textrm{e}.&\displaystyle \frac{1-2a}{ab} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&^{4}\log 0,24\\ &=\displaystyle \frac{^{...}\log 0,24}{^{...}\log 4}=\frac{^{...}\log \displaystyle \frac{6}{25}}{^{...}\log 4},\: \: \color{purple}\textrm{pilih basis 5}\\ &\color{magenta}\textrm{mengapa tidak pilih basis selain 5}\\ &\color{magenta}\textrm{lihat penyebut, di sana terdapat numerus 4}\\ &\color{magenta}\textrm{pada soal, pasangan numerus 4 adalah 5},\\ &\color{black}\textrm{makanya basis 5 dipilih, bukan yang lain}\\ &=\frac{^{5}\log \displaystyle \frac{6}{25}}{^{5}\log 4}=\displaystyle \frac{^{5}\log 6-\: ^{5}\log 25}{^{5}\log 4}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{6}\log 5}-\: ^{5}\log 5^{2}}{^{5}\log 4}=\displaystyle \frac{\displaystyle \frac{1}{a}-2}{b}=\frac{1-2a}{ab} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textbf{(SPMB '05)}\\ &\textrm{Jika}\: \: ^{3}\log 2=p\: \: \textrm{dan}\: \: ^{2}\log 7=q,\\ &\textrm{maka}\: \: ^{14}\log 54=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{p+3}{p+q}\\ \textrm{b}.&\displaystyle \frac{2p}{p+q}\\ \color{red}\textrm{c}.&\displaystyle \frac{p+3}{p(q+1)}\\ \textrm{d}.&\displaystyle \frac{p+q}{p(q+1)}\\ \textrm{e}.&\displaystyle \frac{p(q+1)}{p+q} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{14}\log 54\\ &=\displaystyle \frac{^{...}\log 54}{^{...}\log 14}=\frac{^{...}\log (2.27)}{^{...}\log (2.7)},\: \: \color{purple}\textrm{pilih basis 2}\\ &\color{magenta}\textrm{mengapa tidak pilih basis selain 2}\\ &\color{magenta}\textrm{lihat penyebut, di sana terdapat numerus 7}\\ &\color{magenta}\textrm{pada soal, pasangan numerus 7 adalah 2},\\ &\color{black}\textrm{makanya basis 2 dipilih, bukan yang lain}\\ &=\frac{^{2}\log (2.27)}{^{2}\log (2.7)}=\displaystyle \frac{^{2}\log 2+\: ^{2}\log 27}{^{2}\log 2+\: ^{2}\log 7}\\ &=\displaystyle \frac{^{2}\log 2+\: ^{2}\log 3^{3}}{^{2}\log 2+\: ^{2}\log 7}=\displaystyle \frac{^{2}\log 2+\left (3\times \: \displaystyle \frac{1}{^{3}\log 2} \right )}{^{2}\log 2+\: ^{2}\log 7}\\ &=\displaystyle \frac{1+\displaystyle \frac{3}{p}}{1+q}\\ &=\displaystyle \frac{p+3}{p(q+1)} \end{aligned} \end{array}$

Contoh Soal 3 Fungsi Logaritma

$\begin{array}{ll}\\ 11.&\textrm{Nilai dari}\\ & \displaystyle \frac{1}{6}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -3\\ \textrm{b}.&\displaystyle -2\\ \textrm{c}.&-2 \displaystyle \frac{1}{2}\\ \textrm{d}.& -\displaystyle \frac{1}{2}\\ \color{red}\textrm{e}.&- \displaystyle \frac{1}{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&=\: \displaystyle \frac{1}{6}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\\ &=\displaystyle \frac{1}{3}.\frac{1}{2}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\\ &=\frac{1}{3}\left ( ^{2}\log 25^{\frac{1}{2}} -\: ^{2}\log 10\right )\\ &=\frac{1}{3}\left ( ^{2}\log 5-\: ^{2}\log 10 \right )\\ &=\frac{1}{3}\left ( ^{2}\log \frac{5}{10} \right )\\ &=\frac{1}{3}\left ( ^{2}\log \frac{1}{2} \right )\\ &=\frac{1}{3}\left ( ^{2}\log 2^{-1} \right )\\ &=-\frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Nilai dari}\\ & ^{2}\log \left ( ^{2}\log \sqrt{\sqrt{2}} \right )\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -4\\ \color{red}\textrm{b}.&\displaystyle -2\\ \textrm{c}.&-1 \displaystyle \frac{1}{2}\\ \textrm{d}.& -\displaystyle \frac{1}{2}\\ \textrm{e}.&- \displaystyle \frac{1}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&=\: ^{2}\log \left ( ^{2}\log \sqrt{\sqrt{2}} \right )\\ &=\: ^{2}\log \left ( ^{2}\log 2^{\frac{1}{4}} \right )\\ &=\: ^{2}\log \frac{1}{4}\\ &=\: ^{2}\log \left (2 \right )^{-2}\\ &=-2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textbf{(UMPTN '99)}\\ &\textrm{Diketahui}\: \: \log 2=0,3010\: \: \textrm{dan}\: \: \log 3=0,4771\\ &\textrm{maka}\: \: \log \left ( \sqrt[3]{2}\times \sqrt{3} \right )\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 0,1505\\ \textrm{b}.&\displaystyle 0,1590\\ \textrm{c}.&\displaystyle 0,2007\\ \color{red}\textrm{d}.&\displaystyle 0,3389\\ \textrm{e}.&\displaystyle 0,3891 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\log \left ( \sqrt[3]{2}\times \sqrt{3} \right )\\ &=\log \sqrt[3]{2}+\log \sqrt{3}\\ &=\log 2^{\frac{1}{3}}+\log 3^{\frac{1}{2}}\\ &=\displaystyle \frac{1}{3}\log 2+\displaystyle \frac{1}{2}\log 3\\ &=\displaystyle \frac{1}{3}(0,3010)+\displaystyle \frac{1}{2}(0,4771)\\ &=0,1003+0,2386\\ &=0,3389 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textbf{(UMPTN '98)}\\ &\textrm{Nilai}\: \: ^{a}\log \displaystyle \frac{1}{b}\times \: ^{b}\log \displaystyle \frac{1}{c^{2}}\times \: ^{c}\log \displaystyle \frac{1}{a^{3}}\: =\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle -6\\ \textrm{b}.&\displaystyle 6\\ \textrm{c}.&\displaystyle \frac{b}{a^{2}c}\\ \textrm{d}.&\displaystyle \frac{a^{2}c}{b}\\ \textrm{e}.&\displaystyle -\frac{1}{6} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{a}\log \displaystyle \frac{1}{b}\times \: ^{b}\log \displaystyle \frac{1}{c^{2}}\times \: ^{c}\log \displaystyle \frac{1}{a^{3}}\\ &=\: ^{a}\log \displaystyle b^{-1}\times \: ^{b}\log \displaystyle c^{-2}\times \: ^{c}\log \displaystyle a^{-3}\\ &=(-1).(-2).(-3)\times \: ^{a}\log \displaystyle a\times \: ^{b}\log \displaystyle c\times \: ^{c}\log \displaystyle a\\ &=-6\times \: ^{a}\log a\\ &=-6\times 1\\ &=-6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textbf{(UMPTN '01)}\\ &\textrm{Jika}\: \: \displaystyle \frac{^{2}\log a}{^{3}\log b}=m\: \: \textrm{dan}\: \: \displaystyle \frac{^{3}\log a}{^{2}\log b}=n\\ &\textrm{dengan}\: \: a> 1,\: b> 1,\: \textrm{maka}\: \: \displaystyle \frac{m}{n}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ^{2}\log 3\\ \textrm{b}.&\displaystyle ^{3}\log 2\\ \textrm{c}.&\displaystyle ^{4}\log 9\\ \textrm{d}.&\displaystyle \left ( ^{3}\log 2 \right )^{2}\\ \color{red}\textrm{e}.&\displaystyle \left ( ^{2}\log 3 \right )^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}\displaystyle \frac{m}{n}&=\displaystyle \frac{\displaystyle \frac{^{2}\log a}{^{3}\log b}}{\displaystyle \frac{^{3}\log a}{^{2}\log b}}\\ &=\displaystyle \frac{^{2}\log a\times \: ^{2}\log b}{^{3}\log b\times \: ^{3}\log a}\\ &=\displaystyle \frac{^{2}\log a\times \: \displaystyle \frac{1}{^{b}\log 2}}{^{3}\log b\times \: \displaystyle \frac{1}{^{a}\log 3}}\\ &=\displaystyle \frac{^{2}\log a\times \: ^{a}\log 3}{^{3}\log b\times \: ^{b}\log 2}\\ &=\displaystyle \frac{^{2}\log 3}{^{3}\log 2}=\displaystyle \frac{^{2}\log 3}{\displaystyle \frac{1}{^{2}\log 3}}\\ &=\left ( ^{2}\log 3 \right )^{2} \end{aligned} \end{array}$

Contoh Soal 2 Fungsi Logaritma

$\begin{array}{ll}\\ 6.&\textrm{Nilai dari}\: \: ^{\sqrt{2}}\log 16\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 10\\ \textrm{b}.&\displaystyle 9\\ \color{red}\textrm{c}.& \displaystyle 8\\ \textrm{d}.& \displaystyle 6\\ \textrm{e}.& \displaystyle 4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&=\: ^{\sqrt{2}}\log 16\\ &=\: ^{\displaystyle 2^{\frac{1}{2}}}\log 2^{4}\\ &=\displaystyle \frac{4}{\frac{1}{2}}\times \: ^{2}\log 2\\ &=8\end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Nilai dari}\: \: ^{\sqrt{5}}\log \sqrt{125}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -3\\ \textrm{b}.&\displaystyle -2\\ \textrm{c}.& \displaystyle 2\\ \color{red}\textrm{d}.& \displaystyle 3\\ \textrm{e}.& \displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&=\: ^{\sqrt{5}}\log \sqrt{125}\\ &=\: ^{\sqrt{5}^{1}}\log \left ( \sqrt{5} \right )^{3}\\ &=\displaystyle \frac{3}{1}\times \: ^{\sqrt{5}}\log \sqrt{5}\\ &=3\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Nilai dari}\: \: ^{\sqrt{\sqrt{2}}}\log \sqrt{8\sqrt{8}}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 4\\ \textrm{b}.&\displaystyle 6\\ \textrm{c}.& \displaystyle 8\\ \color{red}\textrm{d}.& \displaystyle 9\\ \textrm{e}.& \displaystyle 12 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&=\: ^{\sqrt{\sqrt{2}}}\log \sqrt{8\sqrt{8}}\\ &=\: ^{\sqrt[4]{2}}\log \left ( 8\left ( 8 \right )^{\frac{1}{2}} \right )^{\frac{1}{2}}\\ &=\: ^{2^{\frac{1}{4}}}\log 8^{\left (\frac{1}{2}+\frac{1}{4} \right )}\\ &=\: ^{2^{\frac{1}{4}}}\log 2^{3\left ( \frac{3}{4} \right )}\\ &=\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}\times \: ^{2}\log 2\\ &=9 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Nilai dari}\\ & ^{6}\log 8 +\: ^{6}\log \displaystyle \frac{9}{2}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 4\\ \textrm{b}.&\displaystyle 3\\ \textrm{c}.&3 \displaystyle \frac{1}{2}\\ \textrm{d}.& 2\displaystyle \frac{1}{2}\\ \color{red}\textrm{e}.& \displaystyle 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}&=\: ^{6}\log 8 +\: ^{6}\log \displaystyle \frac{9}{2}\\ &=\: ^{6}\log 8\times \frac{9}{2}\\ &=\: ^{6}\log 36\\ &=\: ^{6}\log 6^{2}\\ &=2\\ &\\ &\\ & \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Nilai dari}\\ & ^{2}\log \displaystyle \frac{4}{3}+\: 2.\, ^{2}\log \sqrt{12}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 6\\ \color{red}\textrm{b}.&\displaystyle 4\\ \textrm{c}.&3 \displaystyle \frac{1}{2}\\ \textrm{d}.& 2\displaystyle \frac{1}{2}\\ \textrm{e}.& \displaystyle 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&=\: ^{2}\log \displaystyle \frac{4}{3}+\: 2.\, ^{2}\log \sqrt{12}\\ &=\: ^{2}\log \frac{4}{3}+\: ^{2}\log \left ( \sqrt{12} \right )^{2}\\ &=\: ^{2}\log \frac{4}{3}+\: ^{2}\log 12\\ &=\: ^{2}\log \frac{4}{3}\times 12\\ &=\: ^{2}\log 16\\ &=\: ^{2}\log 2^{4}\\ &=4 \end{aligned} \end{array}$

Contoh Soal 1 Fungsi Logaritma

$\begin{array}{ll}\\ 1.&\textrm{Nilai dari}\: \: ^{2}\log \displaystyle \frac{1}{32}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -7\\ \color{red}\textrm{b}.&\displaystyle -5\\ \textrm{c}.& \displaystyle -3\\ \textrm{d}.& \displaystyle -2\\ \textrm{e}.& 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&=\: ^{2}\log \displaystyle \frac{1}{32}\\ &=\: ^{2^{1}}\log 2^{-5}\\ &=\displaystyle \frac{-5}{1}\times \: ^{2}\log 2\\ &=-5\\ &\\ & \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Nilai dari}\: \: ^{0,333...}\log 0,111....\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\\ \textrm{b}.&\displaystyle \frac{1}{2}\\ \color{red}\textrm{c}.& 2\\ \textrm{d}.& 3\\ \textrm{e}.& 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&=\: ^{0,333...}\log 0,111...\\ &=\: ^{\frac{1}{3}}\log \frac{1}{9}\\ &=\: ^{\frac{1}{3}}\log \left (\frac{1}{3} \right )^{2}\\ &=2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Nilai dari}\: \: ^{5}\log 25\sqrt{5}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{5}{2}\\ \textrm{b}.&\displaystyle \frac{3}{2}\\ \textrm{c}.& \displaystyle \frac{1}{2}\\ \textrm{d}.& \displaystyle 2\\ \textrm{e}.& 3 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&=\: ^{5}\log 25\sqrt{5}\\ &=\: ^{5^{1}}\log 5^{2}.5^{\frac{1}{2}}\\ &=\: ^{5^{1}}\log 5^{\frac{5}{2}}\\ &=\displaystyle \frac{\frac{5}{2}}{1}\times \: ^{5}\log 5\\ &=\displaystyle \frac{5}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Nilai dari}\: \: ^{\sqrt{3}}\log 81\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 12\\ \textrm{b}.&\displaystyle 10\\ \textrm{c}.& \displaystyle 9\\ \color{red}\textrm{d}.& \displaystyle 8\\ \textrm{e}.& 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&=\: ^{\sqrt{3}}\log 81\\ &=\: ^{\displaystyle 3^{\frac{1}{2}}}\log 3^{4}\\ &=\displaystyle \frac{4}{\frac{1}{2}}\times \: ^{3}\log 3\\ &=8 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Nilai dari}\: \: ^{\frac{1}{3}}\log \displaystyle \frac{1}{243}\: \: \textrm{adalah}\: ...\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 6\\ \color{red}\textrm{b}.&\displaystyle 5\\ \textrm{c}.& \displaystyle 4\\ \textrm{d}.& \displaystyle 3\\ \textrm{e}.& \displaystyle 2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&=\: ^{\frac{1}{3}}\log \displaystyle \frac{1}{243}\\ &=\: ^{\left (\frac{1}{3} \right )^{1}}\log \displaystyle \left (\frac{1}{3} \right )^{5}\\ &=\displaystyle \frac{5}{1}\times \: ^{\frac{1}{3}}\log \frac{1}{3}\\ &=5 \end{aligned} \end{array}$

Lanjutan Materi Fungsi Logaritma

$\color{blue}\textrm{B. Sifat-Sifat Logaritma}$

Jika syarat logaritma memenuhi untuk bilangan yang diposisikan sebagai basis dan numerus, maka akan berlaku sifat-sifat loaritma berikut:

$\color{purple}\begin{aligned}(1)\quad&a^{{^{a}}\textrm{log b}}=b\\ (2)\quad&^{a}\log (b.c)=\: ^{a}\log b+\: ^{a}\log c\\ (3)\quad&^{a}\log \left ( \displaystyle \frac{b}{c} \right )=\: ^{a}\log b-\: ^{a}\log c\\ (4)\quad&^{a}\log b=\: \displaystyle \frac{^{x}\log b}{^{x}\log c}\\ (5)\quad&^{a}\log b=\: \displaystyle \frac{1}{^{b}\log a}\\ (6)\quad&^{a}\log b=n\Rightarrow \: ^{b}\log a=\displaystyle \frac{1}{n}\\ (7)\quad&^{a^{m}}\log b^{n}=\displaystyle \frac{n}{m}\times \: ^{a}\log b\\ (8)\quad&^{a}\log b\times \: ^{b}\log c\times \: ^{c}\log p=\: ^{a}\log p\\ (9)\quad&^{a}\log a=1\\ (10)\quad&^{a}\log a^{n}=\: n\\ (11)\quad&^{a}\log 1=\: 0\\ (12)\quad&^{.}\log b=\: ^{10}\log b\\ \end{aligned}$

ada yang tak kalah penting untuk diketahui walaupun kadang sebagian orang menganggap tidak perlu dituliskan, di sini saya tuliskan, yaitu:

$\color{blue}\begin{aligned}(\textrm{a})\quad&\log 2=0,3010\\ (\textrm{b})\quad&\log 3=0,4771\\ (\textrm{c})\quad&\log 5=0,6990\\ (\textrm{d})\quad&\log 7=0,8451\\ \end{aligned}$

$\LARGE\color{blue}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&^{2}\log 3+\: ^{2}\log 8-\: ^{2}\log 24=...\: .\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}^{2}\log 3+\: ^{2}\log 8-\: ^{2}\log 24&=\: ^{2}\log \left ( \displaystyle \frac{3\times 8}{24} \right )\\ &=\: ^{2}\log 1\\ &=\: ^{2}\log 2^{0}\\ &=0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&^{2}\log 12+\: ^{2}\log 8-\: ^{2}\log 24=...\: .\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}^{2}\log 12+\: ^{2}\log 8-\: ^{2}\log 24&=\: ^{2}\log \left ( \displaystyle \frac{12\times 8}{24} \right )\\ &=\: ^{2}\log 4\\ &=\: ^{2}\log 2^{2}\\ &=2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: ^{3}\log 7=a,\: \: ^{5}\log 2=b,\: \: \textrm{dan}\: \: ^{2}\log 3=c\\ &\textrm{Nyatakanlah logaritma berikut dalam bentuk}\: \: a,\: b,\: \textrm{dan}\: \: c,\: \: \textrm{yaitu}:\\ &\textrm{a}.\quad ^{7}\log 3\\ &\textrm{b}.\quad ^{4}\log 5\\ &\textrm{c}.\quad ^{21}\log 5\\ &\textrm{d}.\quad ^{6}\log 7\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{array}{|l|l|}\hline \begin{aligned}&\\ ^{7}\log 3&=\displaystyle \frac{1}{^{3}\log 7}\\ &=\displaystyle \frac{1}{a}\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}^{4}\log 5&=\displaystyle \frac{1}{^{5}\log 4}\\ &=\displaystyle \frac{1}{^{5}\log 2^{2}}\\ &=\displaystyle \frac{1}{2\: ^{5}\log 2}\\ &=\displaystyle \frac{1}{2b} \end{aligned}\\\hline \begin{aligned}&\\ ^{21}\log 5&=\displaystyle \frac{^{...}\log 5}{^{...}\log 21}\\ &=\displaystyle \frac{^{2}\log 5}{^{2}\log 21}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{5}\log 2}}{\displaystyle \frac{^{3}\log 21}{^{3}\log 2}}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{5}\log 2}}{\displaystyle \frac{^{3}\log 3\times 7}{^{3}\log 2}}\\ &=\displaystyle \frac{1}{bc(1+a)} \end{aligned}&\begin{aligned}&\\ ^{6}\log 7&=\displaystyle \frac{^{3}\log 7}{^{3}\log 6}\\ &=\displaystyle \frac{^{3}\log 7}{^{3}\log 2\times 3}\\ &=\displaystyle \frac{^{3}\log 7}{^{3}\log 2+\: ^{3}\log 3}\\ &=\displaystyle \frac{^{3}\log 7}{\displaystyle \frac{1}{^{2}\log 3}+\: ^{3}\log 3}\\ &=\displaystyle \frac{a}{\displaystyle \frac{1}{c}+1}\\ &=\displaystyle \frac{ac}{1+c}\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Diketahui bahwa}\: \: \: ^{4}\log 5=a\\ &\textrm{a}.\quad \textrm{Carilah nilai}\: \: \: ^{4}\log 10\\ &\textrm{b}.\quad \textrm{Tunjukkan bahwa}\: \: \: ^{0,1}\log 1,25=\displaystyle \frac{2-2a}{2a+1}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{array}{|l|l|}\hline \begin{aligned}&\\ ^{4}\log 10&=\: ^{4}\log (2\times 5)\\ &=\: ^{4}\log 2+\: ^{4}\log 5\\ &=\: ^{2^{2}}\log 2+\: a\\ &=\: \displaystyle \frac{1}{2}.\: ^{2}\log 2+\: a\\ &=\: \displaystyle \frac{1}{2}+a\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\\ &^{0,1}\log 1,25\\ &=\displaystyle \frac{^{4}\log 1,25}{^{4}\log 0,1}\\ &=\displaystyle \frac{^{4}\log \displaystyle \frac{125}{100}}{^{4}\log \displaystyle \frac{1}{10}}\\ &=\displaystyle \frac{^{4}\log 125-\: ^{4}\log 100}{^{4}\log 10^{-1}}\\ &=\displaystyle \frac{^{4}\log 5^{3}-\: ^{4}\log 10^{2}}{-\: ^{4}\log 10}\\ &=\displaystyle \frac{3.\: ^{4}\log 5-\: 2.\: ^{4}\log 10}{-\: ^{4}\log 10}\\ &=\displaystyle \frac{3a-2\left ( \displaystyle \frac{1}{2}+a \right )}{-\left ( \displaystyle \frac{1}{2}+a \right )}\\ &=\displaystyle \frac{a-1}{-a-\displaystyle \frac{1}{2}}\times \displaystyle \frac{-2}{-2}\\ &=\displaystyle \frac{2-2a}{2a+1}\qquad \blacksquare \end{aligned}\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: ^{2017}\log \displaystyle \frac{1}{x}=\: ^{x}\log \displaystyle \frac{1}{y}=\: ^{y}\log \displaystyle \frac{1}{2017}\\ &\textrm{maka hasil dari}\: \: \left ( 2x-3y \right )\\\\ &\textrm{Jawab}:\\\\ &\color{blue}\begin{aligned}^{2017}\log \displaystyle \frac{1}{x}&=\: ^{x}\log \displaystyle \frac{1}{y}=\: ^{y}\log \displaystyle \frac{1}{2017}\\ ^{2017}\log \displaystyle \frac{1}{x}&=\: ^{y}\log \displaystyle \frac{1}{2017}\\ ^{2017}\log \displaystyle x^{-1}&=\: ^{y}\log \displaystyle (2017)^{-1}\\ -\:\: ^{2017}\log x&=-\: \: ^{y}\log 2017\\ ^{2017}\log x&=\: ^{y}\log 2017\\ &\textrm{dipenuhi saat}\\ x&=y=2017 \end{aligned}\\\\ &(2x-3y)=2x-3x=-x=-2017 \end{array}$

Fungsi Logaritma

$\color{blue}\textrm{A. Pendahuluan}$

Logaritma merupakan invers(balikan) dari perpangkatan

Secara definisi:

$\LARGE\color{purple}\boxed{a^{c}=b\Rightarrow ^{a}\log b=c}$, tetapi di sini diberikan syarat bahwa bilangan basis/dasar perpangkatannya harus berupa bilangan real positif dan tidak sama dengan satu serta bilangan pangkatnya(ekponen) harus berupa bilangan real positif juga.

Perhatikanlah ringkasannya

$^{a}\log b=c\: \: \: \begin{cases} a & \textrm{syaratnya}:a> 0,\: a\neq 1 \\ &\color{magenta}\textrm{selanjutnya disebut basis}\\ b & \textrm{syaratnya}:b>0 \\ & \color{magenta}\textrm{selanjutnya disebut}\: \: \color{blue}\textbf{numerus}\\ c&\textrm{tidak ada syarat apapun}\\ &\color{magenta}\textrm{selanjutnya disebut hasil logaritma} \end{cases}$

Contoh berikut adalah mengubah bentuk perpangkatan ke dalam logaritma yang memenuhi persyaratan

$\color{purple}\begin{aligned} (1)\quad&2^{4}=16\Rightarrow \: ^{2}\log 16=4\\ (2)\quad&2^{3}=8\Rightarrow \: ^{2}\log 8=3\\ (3)\quad&2^{2}=4\Rightarrow \: ^{2}\log 4=2\\ (4)\quad&2^{1}=2\Rightarrow \: ^{2}\log 2=1\\ (5)\quad&2^{0}=1\Rightarrow \: ^{2}\log 1=0\\ (6)\quad&2^{-1}=\displaystyle \frac{1}{2}=0,5\Rightarrow \: ^{2}\log \displaystyle \frac{1}{2}=-1\\ (7)\quad&2^{-2}=\displaystyle \frac{1}{4}=0,25\Rightarrow \: ^{2}\log \displaystyle \frac{1}{4}=-2\\ (8)\quad&2^{-3}=\displaystyle \frac{1}{8}=0,125\Rightarrow \: ^{2}\log \displaystyle \frac{1}{8}=-3\\ (9)\quad&2^{-4}=\displaystyle \frac{1}{16}=0,0625\Rightarrow \: ^{2}\log \displaystyle \frac{1}{16}=-4\\ \end{aligned}$

Berikut contoh kebalikan di atas yang tidak memenuhi definisi logaritma yang ada, yaitu:

$\color{blue}\begin{aligned} (1)\quad&(-2)^{4}=16\Rightarrow \: ^{(-2)}\log 16=\cdots \\ (2)\quad&(-2)^{3}=-8\Rightarrow \: ^{(-2)}\log (-8)=\cdots \\ (3)\quad&(-2)^{2}=4\Rightarrow \: ^{(-2)}\log 4=\cdots \\ (4)\quad&(-2)^{1}=-2\Rightarrow \: ^{(-2)}\log (-2)=\cdots \\ (5)\quad&(-2)^{0}=1\Rightarrow \: ^{(-2)}\log 1=\cdots \\ (6)\quad&(-2)^{-1}=-\displaystyle \frac{1}{2}\Rightarrow \: ^{(-2)}\log \left (-\displaystyle \frac{1}{2} \right )=\cdots \\ (7)\quad&(-2)^{-2}=\displaystyle \frac{1}{4}\Rightarrow \: ^{(-2)}\log \displaystyle \frac{1}{4}=\cdots \\ (8)\quad&(-2)^{-3}=-\displaystyle \frac{1}{8}\Rightarrow \: ^{(-2)}\log \left (-\displaystyle \frac{1}{8} \right )=\cdots \\ (9)\quad&(-2)^{-4}=\displaystyle \frac{1}{16}\Rightarrow \: ^{(-2)}\log \displaystyle \frac{1}{16}=\cdots \\ \end{aligned}$

Statistika (Matematika Wajib kelas XII)

$\color{blue}\textrm{A. Pendahuluan}$

$\color{purple}\begin{array}{|c|l|l|}\hline \textrm{No}&\: \: \: \textrm{Istilah}&\textrm{Pengertian}\\\hline 1.&\textrm{Statistika}&\textrm{Cabang ilmu tentang cara mengumpulkan,} \\ &&\textrm{menyusun, penyajian, dan}\\ &&\textrm{penganalisaan dari suatu data}\\\hline 2.&\textrm{Statistik}&\textrm{Data yang telah tersusun ke dalam}\\ &&\textrm{daftar atau diagram}\\\hline 3.&\textrm{Populasi}&\textrm{Keseluruhan objek dari hasil penelitian}\\ &&\textrm{yang memenuhi syarat tertentu}\\\hline 4.&\textrm{Sampel}&\textrm{Bagian dari populasi yang dapat mewakili}\\ &&\textrm{seluruh populasi}\\\hline \end{array}$

Sebagai tambahan penjelasan

$\color{purple}\begin{array}{|l|l|}\hline .\: \: \: \: \: \qquad \textrm{Istilah}&\textrm{Pengertian dan atau Penjelasan}\\\hline \textrm{Statistika}&\textrm{Lihat pengertian di atas}\\ \textrm{Statistik}&\textrm{Hasil pengolahan data}\\ \textrm{Statistika deskriptif}&\textrm{Statistika baik yang berkenaan dengan}\\ &\textrm{kegiatan pengumpulan, penyajian},\\ & \textrm{penyederhanaan atau penganalisaan},\\ & \textrm{serta penentuan khusus dari suatu data}\\ & \textrm{tanpa penarikan suatu kesimpulan}\\ \textrm{populasi}&\textrm{Keseluruhan objek yang akan diteliti}\\ \color{magenta}\textrm{Sampel (Contoh)}&\color{magenta}\textrm{Bagian dari populasi yang diamati}\\ \textrm{Data}&\textrm{Kumpulan dari datum}\\ \textrm{Datum}&\textrm{Informasi atau catatan keterangan dari}\\ & \textrm{penelitian}\\ \textrm{Data kualitatif}&\textrm{Data yang menunjukkan sifat atau}\\ & \textrm{kondisi objek}\\ \textrm{Data kuantitatif}&\textrm{Data yang menunjukkan jumlah objek}\\ \textrm{Data ukuran}&\textrm{Data yang diperoleh dengan cara}\\ \textrm{(Data kontinu)}& \textrm{mengukur besaran objek}\\ \textrm{Data cacahan}&\textrm{Data yang diperoleh dengan cara}\\ \textrm{(Data diskrit)}& \textrm{mencacah, membilang atau menghitung}\\ &\textrm{banyak objek}\\\hline \end{array}$

$\color{blue}\textrm{B. Penyajian Data}$

$\color{purple}\begin{cases} 1.&\textrm{Daftar bilangan} \\ 2.&\textrm{Tabel distribusi frekuensi} \\ 3.&\textrm{Diagram batang} \\ 4.&\textrm{Diagram garis} \\ 5.&\textrm{Diagram lingkaran} \\ 6.&\textrm{Piktogram} \\ 7.&\textrm{Histogram} \\ 8.&\textrm{Poligon distribusi frekuensi} \\ 9.&\textrm{Ogive} \end{cases}$

$\color{blue}\textrm{C. Data Tunggal}$

$\color{blue}\textrm{C. 1 Ukuran Pemusatan Data (Tendesi Sentral)}$

$\color{purple}\begin{array}{|c|l|l|}\hline \textrm{No}&\textrm{Nilai}&\textrm{Ukuran Pemusatan Data}\\\hline 1.&\textrm{Mean}\: \: \left ( \bar{x} \right )&\bar{x}=\displaystyle \frac{x_{1}+x_{2}+x_{3}+...+x_{n}}{n}\\\hline 2.&\textrm{Median}\: \: \left (M_{e} \right )&\begin{cases} \textrm{Ganjil} & M_{e}=x_{\frac{n+1}{2}} \\ \textrm{Genap} & M_{e}=\displaystyle \frac{1}{2}\left ( x_{\frac{n}{2}}+x_{\frac{n}{2}+1} \right ) \end{cases}\\\hline 3.&\textrm{Modus}\: \: \left ( M_{o} \right )&\textrm{Nilai yang sering muncul}\\\hline 4.&\textrm{Kuartil}\: \: \left ( Q \right )&\begin{cases} \textrm{Ganjil} &\begin{cases} Q_{1}= & x_{\frac{1}{4}(n+1)}\\ Q_{2}= & x_{\frac{2}{4}(n+1)}\\ Q_{3}= & x_{\frac{3}{4}(n+1)} \end{cases} \\\\ \textrm{Genap} & \begin{cases} Q_{1}= & x_{\frac{1}{4}n+\frac{1}{2}}\\ Q_{2}= & x_{\frac{2}{4}n+\frac{1}{2}}\\ Q_{3}= & x_{\frac{3}{4}n+\frac{1}{2}} \end{cases} \end{cases}\\\hline \end{array}$

$\color{blue}\textrm{C. 2 Ukuran Penyebaran Data (Dispersi)}$

$\color{purple}\begin{array}{|c|l|l|}\hline \textrm{No}&\textrm{Nilai}&\textrm{Ukuran Penyebaran Data}\\\hline 1.&\textrm{Jangkauan}\: \: \left ( J \right )&J=R\\ &\textrm{atau Rentang}\: \: (R)&=x_{datum\: max}-x_{datum\: min}\\\hline 2.&\textrm{Hamparan}\: \: (H)&\\ &\textrm{Atau Jangkauan}&H=Q_{3}-Q_{1}\\ &\textrm{antar kuartil}&\\\hline 3.&\textrm{Simpangan}&Q_{d}=\displaystyle \frac{1}{2}H\\ &\textrm{Kuartil}\: \: \left (Q_{d} \right )&\\\hline 4.&\textrm{Langkah}\: \: \left ( L \right )&L=\displaystyle \frac{3}{2}H\\\hline 5.&\textrm{Pagar}&\begin{cases} \textrm{Dalam} &=Q_{1}-L \\ \textrm{Luar} &=Q_{3}+L \end{cases}\\\hline 6&\textrm{Data}&\begin{cases} &\textrm{Normal} \\ &:Q_{1}-L\leq x_{i}\leq Q_{3}+L \\ &\textrm{Tidak Normal} \\ &:\begin{cases} x_{i}<Q_{1}-L \\ x_{i}>Q_{3}+L \end{cases} \end{cases}\\\hline 7.&\textrm{Simpangan}&SR=\displaystyle \frac{\sum \left | x_{i}-\bar{x} \right |}{n}\\ &\textrm{Rata-rata}\: \: (SR)&\\\hline 8.&\textrm{Ragam}\: \: \left ( s^{2} \right )&s^{2}=\displaystyle \frac{\sum \left ( x_{i}-\bar{x} \right )^{2}}{n}\\ &\textrm{atau Varian}&\\\hline 9.&\textrm{Simpangan}&s=\sqrt{s^{2}}\\ &\textrm{Baku}\: \: (s)&\\\hline \end{array}$

$\color{blue}\textrm{D. Data Berkelompok}$

Untuk tipe ini antara lain

$\begin{cases} \textrm{(1) Mean},& \bar{\textrm{x}}=\bar{x}_{s}+\displaystyle \frac{\sum f_{i}.d_{i}}{\sum f_{i}} \\\\ \textrm{(2) Modus},& \textrm{M}_{o}=t_{p}+\displaystyle p\left ( \frac{d_{1}}{d_{1}+d_{2}} \right ) \\\\ \textrm{(3) Kuartil},& \textrm{Q}_{i}=t_{p}+\displaystyle p\left ( \frac{\displaystyle \frac{i.n}{4}-\sum f_{i}}{f_{b}} \right ) \end{cases}$

Berikut keterangannya untuk beberapa istilah pada formula di atas baik poin 1, poin 2, maupun poin 3

$\begin{cases} \begin{aligned}&(1)\\ &\\ &\\ &\\ & \end{aligned}&\begin{array}{|l|}\hline \begin{aligned}\bar{x}_{s}=&\textrm{rataan sementara}\\ x_{i}=&\textrm{titik tengahinterval}\\ &\textrm{kelas ke}-i\\ d_{i}=&x_{i}-\bar{x}_{s}\\ f_{i}=&\textrm{frekuensi kelas ke}-i\\ \end{aligned}\\\hline \end{array} \\\\ \begin{aligned}&(2)\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{array}{|l|}\hline \begin{aligned}t_{p}=&\textrm{tepi bawah kelas modus}\\ p=&\textrm{panjang interval kelas}\\ d_{1}=&f_{0}-f_{-1}\\ d_{2}=&f_{0}-f_{+1}\\ f_{0}=&\textrm{frekuensi kelas modus}\\ f_{-1}=&\textrm{frekuensi sebelum kelas modus}\\ f_{+1}=&\textrm{frekuensi setelah kelas modus} \end{aligned}\\\hline \end{array} \\\\ \begin{aligned}&(3)\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{array}{|l|}\hline \begin{aligned}tp=&\textrm{tepi bawah kuartil ke}-i\\ p=&\textrm{panjang interval kelas}\\ n=&\textrm{banyaknya data}\\ \sum f_{i}=&\textrm{jumlah semua frekuensi}\\ &\textrm{sebelum kelas kurtil ke}-i\\ f_{q}=&\textrm{frekuensi kelas kuartil ke}-i\\ Q_{i}=&\textrm{kuartil ke}-i\\ Q_{1}=&\textrm{kuartil bawah}\\ Q_{2}=&\textrm{kuatil tengah/median}\\ Q_{3}=&\textrm{kuatil atas} \end{aligned}\\\hline \end{array} \end{cases}$

DAFTAR PUSTAKA

Johanes, Kastolan, dan Sulasim. 2004. Kompetensi Matematika SMA Kelas 2 Semester 1 Program Ilmu Sosial Kurikulum Berbasis Kompetensi 2004. Jakarta: Yudistira
Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI. Bandung SEWU.
Sobirin. 2006. Kompas Matematika: Strategi Praktis Menguasai Tes Matematika SMA Kelas 2 IPA. Jakarta: Kawan Pustaka.
Tampomas, H. 1999. Seribu Pena Matematika SMU Jilid 2 Kelas 2. Jakarta: Erlangga.
Wirodikromo, S. 2007. Matematika untuk SMA Kelas XI. Jakarta: Erlangga.