Contoh Soal 3 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{l}\\ 11.& \text{Turunan pertama fungsi}\\ & h(x)=5\sin x\cos x\text{adalah}{h}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& 5\sin 2x\\ \text{b}.& 5\cos 2x\\ \text{c}.& 5{\sin }^{2}x\cos x\\ \text{d}.& 5{\sin }^{2}x{\cos }^{2}x\\ \text{e}.& 5\sin 2x\cos x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}h(x)=5\sin x\cos x\\ h(x)& ={\displaystyle \frac{5}{2}(2\sin x\cos x)={\displaystyle \frac{5}{2}\sin 2x}}\\ h^{\prime }(x)& ={\displaystyle \frac{5}{2}(\cos 2x).(2)}\\ & =5\cos 2x\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Turunan pertama fungsi}\\ & k(x)=\cos x\tan x\text{adalah}{k}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& -\sin x\cot x+\cos x{\sec }^{2}x\\ \text{b}.& -\sin x\tan x+\cos x{\sec }^{2}x\\ \text{c}.& \sin x\tan x-\cos x{\sec }^{2}x\\ \text{d}.& -{\displaystyle \frac{1+{\sin }^{2}x}{\cos x}}\\ \text{e}.& {\displaystyle \frac{1+{\sin }^{2}x}{\cos x}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}k(x)=\cos x\tan x\\ \text{guna}& \text{kan formula}y=u.v\Rightarrow y^{\prime }=u^{\prime }v+u.v^{\prime }\\ u& =\cos x\Rightarrow u^{\prime }=-\sin x\\ v& =\tan x\Rightarrow v^{\prime }={\sec }^{2}x\\ \text{maka}& \\ k^{\prime }(x)& =(-\sin x)\tan x+\cos x.({\sec }^{2}x)\\ & =-\sin x\tan x+\cos x{\sec }^{2}x\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Jika diketahui}f(x)=|\tan x|,\text{maka}{\displaystyle \frac{dy}{dx}}\\ & \text{saat}x=k,\text{di mana}{\displaystyle \frac{1}{2}\pi <k<\pi }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -\sin k\\ \text{b}.& \cos k\\ \text{c}.& -{\sec }^{2}k\\ \text{d}.& {\sec }^{2}k\\ \text{e}.& \cot k\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}f(x)=|\tan x|\\ \text{saat}& x=k\text{dengan}{\displaystyle \frac{1}{2}\pi <k<\pi }\\ \text{adal}& \text{ah}:\\ f(x)& =|\tan x|,\text{maka saat}x=k\\ f(k)& =|\tan k|=-\tan k,\text{karena di}{\displaystyle \frac{1}{2}\pi <k<\pi }\\ {\displaystyle \frac{dy}{dx}}& =f^{\prime }(k)=-{\sec }^{2}k\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Turunan pertama}g(x)=|\cos x|\\ & \text{adalah}{g}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& -|\sin x|\\ \text{b}.& -\sin x\\ \text{c}.& {\displaystyle \frac{\sin 2x}{2|\cos x|}}\\ \text{d}.& -{\displaystyle \frac{\sin 2x}{2|\cos x|}}\\ \text{e}.& |\sin x|\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}g(x)=|\cos x|=\sqrt{{\cos }^{2}x}={({\cos }^{2}x)}^{{.}^{\frac{1}{2}}}\\ g^{\prime }(x)& ={\displaystyle \frac{1}{2}{({\cos }^{2}x)}^{{.}^{-\frac{1}{2}}}.(2\cos x).(-\sin x)}\\ & ={\displaystyle \frac{-2\sin x\cos x}{2{({\cos }^{2}x)}^{{.}^{\frac{1}{2}}}}}\\ & =-{\displaystyle \frac{\sin 2x}{2\sqrt{{\cos }^{2}x}}}\\ & =-{\displaystyle \frac{\sin 2x}{2|\cos x|}}\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Turunan pertama dari}f(x)={\displaystyle \frac{\sin x}{x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{x\cos x+\sin x}{x^2}}\\ \text{b}.& {\displaystyle \frac{x\cos x-\sin x}{x^2}}\\ \text{c}.& {\displaystyle \frac{-x\cos x-\sin x}{x^2}}\\ \text{d}.& {\displaystyle \frac{\cos x-x\sin x}{x^2}}\\ \text{e}.& {\displaystyle \frac{\cos x+x\sin x}{x^2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui}\\ f(x)& ={\displaystyle \frac{\sin x}{x}}\\ \text{Guna}& \text{kan formula}y={\displaystyle \frac{u}{v}\Rightarrow y^{\prime }={\displaystyle \frac{u^{\prime }v-u.v^{\prime }}{v^2}}}\\ u& =\sin x\Rightarrow u^{\prime }=\cos x\\ v& =x\Rightarrow v^{\prime }=1\\ \text{maka}& \\ f^{\prime }(x)& ={\displaystyle \frac{\cos x.(x)-\sin x.1}{x^2}}\\ & ={\displaystyle \frac{x\cos x-\sin x}{x^2}}\end{array}\end{array}$

Contoh Soal 2 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{l}\\ 6.& \text{Turunan pertama}q(x)={\sin }^{2}x+{\cos }^{2}x\\ & \text{adalah}{q}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& {\cos }^{2}x-{\sin }^{2}x\\ \text{b}.& 2{\cos }^{2}x-2{\sin }^{2}x\\ \text{c}.& \cos x-\sin x\\ \text{d}.& 2\cos x-2\sin x\\ \text{e}.& 0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}q(x)& ={\sin }^{2}x+{\cos }^{2}x\\ \text{guna}& \text{kan formula identitas}:{\sin }^{2}x+{\cos }^{2}x=1\\ \text{Sehi}& \text{ngga soal di atas dapat dituliskan menjadi}\\ q(x)& =1,\text{maka}\\ q^{\prime }(x)& =0\\ \text{inga}& \text{t bahwa}y=a\Rightarrow {\displaystyle \frac{dy}{dx}=0}\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Nilai dari}\underset{h\to 0}{\text{lim}}{\displaystyle \frac{\sin \left({\displaystyle \frac{\pi }{3}+h}\right)-\sin {\displaystyle \frac{\pi }{3}}}{h}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{b}.& -{\displaystyle \frac{1}{2}}\\ \text{c}.& 0\\ \text{d}.& {\displaystyle \frac{1}{2}}\\ \text{e}.& {\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dari}& \text{soal diketahui}:\\ f(x)& =\sin {\displaystyle \frac{\pi }{3}}\\ \text{Nila}& \text{i dari}\underset{h\to 0}{\text{lim}}{\displaystyle \frac{\sin \left({\displaystyle \frac{\pi }{3}+h}\right)-\sin {\displaystyle \frac{\pi }{3}}}{h}}\\ \text{arti}& \text{nya bermakna, berapkah}{f}^{\prime }\left(x\right)?\\ \text{maka}& \\ f^{\prime }\left(x\right)& =0\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Jika}f(x)=8x-{\sin }^{3}x,\\ & \text{maka nilai}\underset{h\to 0}{\text{lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 4x^2-3{\cos }^{2}x\\ \text{b}.& 8x-3{\sin }^{2}x\cos x\\ \text{c}.& 8-3{\sin }^{2}x\cos x\\ \text{d}.& 8+{\sin }^{2}x\cos x\\ \text{e}.& 3{\sin }^{2}x\cos x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui dari soal}f(x)=8x-{\sin }^{3}x\\ \text{maka}& \text{nilai dari}\underset{h\to 0}{\text{lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}=f^{\prime }(x)}\\ f^{\prime }(x)& =8-3{\sin }^{2}x\cos x\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Turunan pertama fungsi}f(x)=\sqrt{\sin x},\\ & \text{adalah}{f}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2\sqrt{\sin x}}}\\ \text{b}.& {\displaystyle \frac{\cos x}{\sqrt{\sin x}}}\\ \text{c}.& {\displaystyle \frac{\cos x}{2\sqrt{\sin x}}}\\ \text{d}.& -{\displaystyle \frac{\sin x}{2\sqrt{\cos x}}}\\ \text{e}.& {\displaystyle \frac{2\cos x}{\sqrt{\sin x}}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}f(x)=\sqrt{\sin x}={\sin }^{{.}^{\frac{1}{2}}}x\\ f^{\prime }(x)& ={\displaystyle \frac{1}{2}({\sin }^{{.}^{-\frac{1}{2}}}x).(\cos x)}\\ & ={\displaystyle \frac{\cos x}{2{\sin }^{{.}^{\frac{1}{2}}}x}}\\ & ={\displaystyle \frac{\cos x}{2\sqrt{\sin x}}}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Jika}{g}^{\prime }(x)\text{adalah turunan pertama}\\ & \text{fungsi}g(x)\text{dengan}g(x)=5{\tan }^{2}x,\\ & \text{maka}{g}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& 10{\cos }^{2}x\sin x\\ \text{b}.& 10{\sin }^{2}x\cos x\\ \text{c}.& {\displaystyle \frac{10\sin x}{{\cos }^{3}x}}\\ \text{d}.& {\displaystyle \frac{10{\cos }^{3}x}{\sin x}}\\ \text{e}.& {\displaystyle \frac{10}{{\sin }^{2}x-{\cos }^{2}x}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}g(x)=5{\tan }^{2}x\\ g^{\prime }(x)& =5(2\tan x).({\sec }^{2}x)\\ & =10\tan x\times \left({\displaystyle \frac{1}{{\cos }^{2}x}}\right)\\ & =10\left({\displaystyle \frac{\sin x}{\cos x}}\right)\times \left({\displaystyle \frac{1}{{\cos }^{2}x}}\right)\\ & ={\displaystyle \frac{10\sin x}{{\cos }^{3}x}}\end{array}\end{array}$

Contoh Soal 1 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{l}\\ 1.& \text{Diketahui}f(x)=2\cos x-2020\\ & \text{Turunan pertama fungsi}f(x)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2\sin x\\ \text{b}.& -2\sin x\\ \text{c}.& -2\sin x-2020x\\ \text{d}.& 2{\sin }^{2}x\\ \text{e}.& 2\cos x-2020x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}f(x)& =2\cos x-2020\\ f^{\prime }(x)& =-2\sin x\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Jika}{f}^{\prime }(x)\text{adalah turunan pertama dari}\\ & \text{fungsi}f(x)={\sin }^{7}x,\text{maka}{f}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& 7{\cos }^{6}x\\ \text{b}.& 7{\cos }^{7}x\\ \text{c}.& 7{\sin }^{6}x\cos x\\ \text{d}.& 7{\cos }^{6}x\sin x\\ \text{e}.& 7{\cos }^{6}x{\sin }^{6}x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}f(x)& ={\sin }^{7}x\\ \text{guna}& \text{kan formula}:y=a.u^n\Rightarrow y^{\prime }=n.a.u^{n-1}.u^{\prime }\\ f^{\prime }(x)& =7{\sin }^{6}x(\cos x)=7{\sin }^{6}x\cos x\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Turunan pertama fungsi}g(x)=-5{\sin }^{3}x\\ & \text{adalah}{g}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& -5{\sin }^{2}x\cos x\\ \text{b}.& -5{\sin }^2{\cos }^{2}x\\ \text{c}.& -15{\sin }^{2}x\cos x\\ \text{d}.& -15{\cos }^{3}x\\ \text{e}.& -15{\sin }^{4}x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}g(x)& =-5{\sin }^{3}x\\ \text{guna}& \text{kan formula}:y=a.u^n\Rightarrow y^{\prime }=n.a.u^{n-1}.u^{\prime }\\ g^{\prime }(x)& =-5(3{\sin }^{2}x)(\cos x)=-15{\sin }^{2}x\cos x\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Jika}h(x)=4x^3+\sin x+\cos x\\ & \text{maka}{h}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& 12x^2+\cos x-\sin x\\ \text{b}.& 12x^2-\cos x+\sin x\\ \text{c}.& 4x^3-\cos x-\sin x\\ \text{d}.& 4x^3-\sin x-\cos x\\ \text{e}.& 12x^3+\cos x+\sin x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}h(x)& =4x^3+\sin x+\cos x\\ \text{guna}& \text{kan formula}:y=a.u^n\Rightarrow y^{\prime }=n.a.u^{n-1}.u^{\prime }\\ \text{pada}& \text{fungsi aljabarnya, yaitu}:y=4x^3\Rightarrow y^{\prime }=12x^2\\ \text{seda}& \text{ngkan fungsi transendennya mengikuti}\\ \text{turu}& \text{nan fungsi trigonometri biasa. Sehingga}\\ f^{\prime }(x)& =12x^2+\cos x-\sin x\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Jika}p(x)=-{\cos }^{4}x,\text{maka nilai}\\ & \text{maka}{p}^{\prime }\left({\displaystyle \frac{\pi }{3}}\right)=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& \sqrt{3}\\ \text{c}.& {\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{d}.& {\displaystyle \frac{1}{4}\sqrt{3}}\\ \text{e}.& 1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}p(x)& =-{\cos }^{4}x\\ p^{\prime }(x)& =-4{\cos }^{3}x.(-\sin x)=4{\cos }^{3}x\sin x\\ p^{\prime }\left({\displaystyle \frac{\pi }{3}}\right)& =4{\cos }^3\left({\displaystyle \frac{\pi }{3}}\right).\sin \left({\displaystyle \frac{\pi }{3}}\right)\\ & =4{\cos }^3{60}^{\circ }\times \sin {60}^{\circ }\\ & =4{\left({\displaystyle \frac{1}{2}}\right)}^3\times \left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & ={\displaystyle \frac{4}{16}\sqrt{3}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}}\end{array}\end{array}$

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

 $\text{D. Aturan Rantai Turunan Fungsi Trigonometri}$

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

$\begin{array}{c}{y}^{\prime }={(f\circ g)}^{\prime }(x)=f^{\prime }(g(x))\times g^{\prime }(x)\\ \\ \mathbf{\text{atau}}\\ \\ {\displaystyle \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}}\end{array}$

Perluasan dari teorema di atas, adalah berikut:

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

$\begin{array}{c}{y}^{\prime }={(f\circ g\circ h)}^{\prime }(x)=f^{\prime }\left(g(h(x))\right)\times g^{\prime }(h(x))\times h^{\prime }(x)\\ \\ \mathbf{\text{atau}}\\ \\ {\displaystyle \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

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

$\begin{array}{l}\\ 2.& \text{Tentukan turunan pertama dari}\\ & g(x)=\sqrt[5]{{\cos }^3(x^2-\pi )}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}g(x)& =\sqrt[5]{{\cos }^3(x^2-\pi )}={\cos }^{{.}^{\frac{3}{5}}}(x^2-\pi )\\ \text{Dim}& \text{isalkan}\\ y& =u^{{.}^{\frac{3}{5}}},\text{dengan}\\ u& =\cos (x^2-\pi )\text{serta}u=\cos v\\ & \text{dan}v=(x^2-\pi ),\\ \text{mak}& \text{a}\\ {\displaystyle \frac{dy}{du}}& ={\displaystyle \frac{3}{5}{u}^{{.}^{-\frac{2}{5}}}=\frac{3}{5}{\cos }^{{.}^{-\frac{2}{5}}}(x^2-\pi ),}\\ {\displaystyle \frac{du}{dv}}& =-\sin v=-\sin (x^2-\pi ),\\ {\displaystyle \frac{dv}{dx}}& =2x\\ \text{Seh}& \text{ingga}\\ g^{\prime }(x)& ={\displaystyle \frac{dy}{dx}}\\ & ={\displaystyle \frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}}\\ & ={\displaystyle \frac{3}{5}{\cos }^{{.}^{-\frac{2}{5}}}(x^2-\pi )\times (-\sin (x^2-\pi ))\times (2x)}\\ & =-{\displaystyle \frac{6x\sin (x^2-\pi )}{5{\cos }^{{.}^{\frac{2}{5}}}(x^2-\pi )}}\\ & =-{\displaystyle \frac{6x\sin (x^2-\pi )}{5\sqrt[5]{{\cos }^2(x^2-\pi )}}}\end{array}\\ \\ & \mathbf{\text{atau kalau ingin langsungan saja}}\\ \\ & \begin{array}{rl}g(x)& =\sqrt[5]{{\cos }^3(x^2-\pi )}={\cos }^{{.}^{\frac{3}{5}}}(x^2-\pi )\\ g^{\prime }(x)& ={\displaystyle \frac{3}{5}{\cos }^{{.}^{-\frac{2}{5}}}(x^2-\pi )\times (-\sin (x^2-\pi ))\times (2x)}\\ & =-{\displaystyle \frac{6x\sin (x^2-\pi )}{5{\cos }^{{.}^{\frac{2}{5}}}(x^2-\pi )}}\\ & =-{\displaystyle \frac{6x\sin (x^2-\pi )}{5\sqrt[5]{{\cos }^2(x^2-\pi )}}}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Tentukan turunan pertama dari}\\ & h(x)=\cos (\sin x^{2020})\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}h(x)& =\cos (\sin x^{2020})\\ \text{Dim}& \text{isalkan}\\ y& =\cos (\sin x^{2020})=\cos u,\text{dengan}\\ u& =\sin x^{2020}=\sin v,\text{serta}v=x^{2020}\\ \text{mak}& \text{a}\\ {\displaystyle \frac{dy}{du}}& =-\sin u=-\sin (\sin x^{2020}),\\ & \text{atau}dy=-\sin udu\\ {\displaystyle \frac{du}{dv}}& =\cos v\text{atau}du=\cos vdv\\ {\displaystyle \frac{dv}{dx}}& =2020x^{2019}\text{atau}dv=2020x^{2019}dx\\ \text{Seh}& \text{ingga}\\ h^{\prime }(x)& ={\displaystyle \frac{dy}{dx}}\\ & ={\displaystyle \frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}}\\ & =-\sin (\sin x^{2020})\times \cos x^{2020}\times \left(2020x^{2019}\right)\\ & =-2020x^{2019}\sin (\sin x^{2020})\cos x^{2020}\end{array}\\ \\ & \mathbf{\text{atau}}\\ & \begin{array}{rl}dy& =-\sin udu\\ & =-\sin u\times \cos vdv\\ & =-\sin u\times \cos v\times \left(2020x^{2019}\right)dx\\ {\displaystyle \frac{dy}{dx}}& =-\sin u\times \cos v\times \left(2020x^{2019}\right)\\ & =.......(\text{tinggal dimasukkan})\end{array}\end{array}$

DAFTAR PUSTAKA

1. 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)

 $\text{C. Sifat-Sifat Turunan Fungsi Trigonometri}$

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

$\begin{array}{|cll|}\hline \text{No}& \text{Fungsi}& \text{Turunan Pertama}\\ 1.& y=k.u& y^{\prime }=k.u^{\prime }\\ 2.& y=u\pm v& y^{\prime }=u^{\prime }\pm v^{\prime }\\ 3.& y=u.v& y^{\prime }=v.u^{\prime }+u.v^{\prime }\\ 4.& y=k.u^n& =n.k.u^{(n-1)}.u^{\prime }\\ 5.& y={\displaystyle \frac{u}{v}}& y^{\prime }={\displaystyle \frac{u^{\prime }.v-u.v^{\prime }}{v^2}}\\ \hline\end{array}$

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

$y^{\prime }={\displaystyle \frac{dy}{dx}=f^{\prime }(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}}$

dan untuk turunan keduanya dari fungsi di atas adalah:

$y^{″}={\displaystyle \frac{d^{2}y}{d{x}^2}=f^{″}(x)\text{atau kadang dituliskan}{\displaystyle \frac{d{f}^{\prime }(x)}{dx}=\frac{d^{2}f}{d{x}^2}}}$

Selanjutnya perhatikanlah tabel berikut

$\begin{array}{|ll|}\hline \text{Turunan}& \text{Notasi}\\ \text{Pertama}& y^{\prime }=f^{\prime }(x)={\displaystyle \frac{dy}{dx}=\frac{df}{dx}}\\ & \\ \text{Kedua}& y^{″}=f^{″}(x)={\displaystyle \frac{d^{2}y}{d{x}^2}=\frac{d^{2}f}{d{x}^2}}\\ & \\ \text{Ketiga}& y^{‴}=f^{‴}(x)={\displaystyle \frac{d^{3}y}{d{x}^3}=\frac{d^{3}f}{d{x}^3}}\\ & \\ \cdots & \cdots \cdots \cdots \cdots \\ \text{Ke-n}& y^n=f^n(x)={\displaystyle \frac{d^{n}y}{d{x}^n}=\frac{d^{n}f}{d{x}^n}}\\ \hline\end{array}$

$\boxed{\text{CONTOH SOAL}}$

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

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

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

$.\begin{array}{rl}2.(\text{a})\text{dike}& \text{tahui}f(x)={\displaystyle \frac{1+\sin x}{\cos x}}\\ \text{gun}& \text{akan formula}y={\displaystyle \frac{u}{v}\Rightarrow y^{\prime }={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}}\\ f^{\prime }(x)& ={\displaystyle \frac{(\cos x)(\cos x)-(1+\sin x)(-\sin x)}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{{\cos }^{2}x+\sin x+{\sin }^{2}x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{{\cos }^{2}x+{\sin }^{2}x+\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{1+\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}+\frac{\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}+\frac{1}{\cos x}.\frac{\sin x}{\cos x}}\\ & ={\sec }^{2}x+\sec x\tan x\end{array}$

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

$.\begin{array}{rl}2.(\text{c})\text{dike}& \text{tahui}h(x)=\sin x\tan x\\ \text{gun}& \text{akan formula}y=u.v\Rightarrow y^{\prime }=u^{\prime }.v+u.v^{\prime }\\ h^{\prime }(x)& =\cos x.(\tan x)+\sin x.({\sec }^{2}x)\\ h\left({45}^{\circ }\right)& =\cos \left({45}^{\circ }\right)\tan \left({45}^{\circ }\right)+\sin \left({45}^{\circ }\right){\sec }^2\left({45}^{\circ }\right)\\ & =\left({\displaystyle \frac{1}{2}\sqrt{2}}\right).1+\left({\displaystyle \frac{1}{2}\sqrt{2}}\right){\left(\sqrt{2}\right)}^2\\ & ={\displaystyle \frac{1}{2}\sqrt{2}+\sqrt{2}}\\ & ={\displaystyle \frac{3}{2}\sqrt{2}}\end{array}$

$.\begin{array}{rl}2.(\text{d})\text{dike}& \text{tahui}k(x)=\sin x+n\cos x\\ k^{\prime }(x)& =\cos x-n\sin x,\text{dengan}{k}^{\prime }\left({\displaystyle \frac{\pi }{3}}\right)=0\\ k^{\prime }\left({\displaystyle \frac{\pi }{6}}\right)& =\cos \left({\displaystyle \frac{\pi }{3}}\right)-n\sin \left({\displaystyle \frac{\pi }{3}}\right)\\ 0& =\cos {60}^{\circ }-n\sin {60}^{\circ }={\displaystyle \frac{1}{2}-n\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}\\ \iff & n\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)={\displaystyle \frac{1}{2}}\\ \iff & n={\displaystyle \frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}\sqrt{3}}}}\\ \iff & n={\displaystyle \frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}}\\ \iff & n={\displaystyle \frac{1}{3}\sqrt{3}}\end{array}$

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

DAFTAR PUSTAKA

1. Kurnia, N. 2018. Jelajah Matematika 3 SMA Kelas XII Peminatan MIPA. Bogor: Yudhistira
2. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
3. 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.
4. Tasari, Aksin, N., Miyanto & Muklis. 2016. Matematika untuk SMA/MA Kelas XII Peminatan Matematika dan Ilmu-Ilmu Alam. Klaten: INTAN PARIWARA.
5. 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 $\sin (x+h)\cos x$  dan $\cos (x+h)\sin x$, maka saat ketemu bentuk itu kita gunakan rumus:

$\{\begin{array}{ll}\sin A\cos B& ={\displaystyle \frac{1}{2}(\sin (A+B)+\sin (A-B))}\\ \\ \cos A\sin B& ={\displaystyle \frac{1}{2}(\sin (A+B)-\sin (A-B))}\end{array}$

Coba perhatikanlah uraian turunan fungsi tangen berikut:

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\tan (x+h)-\tan x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (x+h)}{\cos (x+h)}-{\displaystyle \frac{\sin x}{\cos x}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (x+h)\cos x}{\cos (x+h)\cos x}-{\displaystyle \frac{\cos (x+h)\sin x}{\cos (x+h)\cos x}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h).\cos x.h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{...+{\displaystyle \frac{1}{2}\sin h-...+{\displaystyle \frac{1}{2}\sin h}}}{\cos (x+h).\cos x.h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin h}{h.\cos (x+h).\cos x}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \left({\displaystyle \frac{\sin h}{h}}\right)\left({\displaystyle \frac{1}{\cos (x+h)\cos x}}\right)}\\ & =1\times {\displaystyle \frac{1}{\cos (x+0).\cos x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}}\\ & ={\sec }^{2}x\end{array}$

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

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\tan (x+h)-\tan x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (x+h)}{\cos (x+h)}-{\displaystyle \frac{\sin x}{\cos x}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (x+h)\cos x}{\cos (x+h)\cos x}-{\displaystyle \frac{\cos (x+h)\sin x}{\cos (x+h)\cos x}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h).\cos x.h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin ((x+h)-x)}{\cos (x+h).\cos x.h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin h}{h.\cos (x+h).\cos x}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \left({\displaystyle \frac{\sin h}{h}}\right)\left({\displaystyle \frac{1}{\cos (x+h)\cos x}}\right)}\\ & =1\times {\displaystyle \frac{1}{\cos (x+0).\cos x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}}\\ & ={\sec }^{2}x\end{array}$

Berikut hasil turunan pertama untuk fungsi trigonometri yang perlu diingat:

$\begin{array}{rl}1.& f(x)=\sin x\Rightarrow f^{\prime }(x)=\cos x\\ 2.& f(x)=\cos x\Rightarrow f^{\prime }(x)=-\sin x\\ 3.& f(x)=\tan x\Rightarrow f^{\prime }(x)={\sec }^{2}x\\ 4.& f(x)=\cot x\Rightarrow f^{\prime }(x)=-{\csc }^{2}x\\ 5.& f(x)=\sec x\Rightarrow f^{\prime }(x)=\sec x\tan x\\ 6.& f(x)=\csc x\Rightarrow f^{\prime }(x)=-\csc x\cot x\end{array}$

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

$\text{B. Turunan Fungsi Trigonometri}$

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  $f^{\prime }(x)$ adalah:

$f^{\prime }(x)=\underset{h\to 0}{\text{lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}$

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

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

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (x+h)-\sin x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2\cos {\displaystyle \frac{1}{2}(2x+h)\sin {\displaystyle \frac{1}{2}h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}2\cos {\displaystyle \frac{1}{2}(2x+h).{\displaystyle \frac{\sin {\displaystyle \frac{1}{2}h}}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle 2\cos {\displaystyle \frac{1}{2}(2x+h)\times {\displaystyle \frac{1}{2}}}}\\ & =2\cos {\displaystyle \frac{1}{2}(2x+0)\times {\displaystyle \frac{1}{2}}}\\ & =\cos {\displaystyle \frac{1}{2}(2x)}\\ & =\cos x\end{array}$

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

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

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

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

Coba perhatikan penggunaanya berikut, tapi malah agak panjang dikit jadinya

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (x+h)-\sin x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin x\cos h+\cos x\sin h-\sin x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin x(\cos h-1)+\cos x\sin h}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin x(\cos h-1)}{h}+\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\cos x\sin h}{h}}}\\ & =\sin x.\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\cos h-1}{h}+\cos x.\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin h}{h}}}\\ & =\sin x.\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{-2{\sin }^2{\displaystyle \frac{1}{2}h}}{h}+\cos x.1}\\ & =\sin x.0+\cos x\\ & =\cos x\end{array}$

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

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\cos (x+h)-\cos x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{-2\sin {\displaystyle \frac{1}{2}(2x+h)\sin {\displaystyle \frac{1}{2}h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle -2\sin {\displaystyle \frac{1}{2}(2x+h).\frac{\sin {\displaystyle \frac{1}{2}h}}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle -2\sin {\displaystyle \frac{1}{2}(2x+h)\times \frac{1}{2}}}\\ & =-2\sin {\displaystyle \frac{1}{2}(2x+0)\times \frac{1}{2}}\\ & =-\sin {\displaystyle \frac{1}{2}(2x)}\\ & =-\sin x\end{array}$

Contoh Soal 6 Fungsi Logaritma (Uraian)

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

$\begin{array}{l}\\ 27.& \mathbf{\text{(AIME 1984)}}\\ & \text{Diketahui bahwa}x\text{dan}y\\ & \text{adalah bilangan real yang memenuhi}\\ \\ & \{\begin{array}{c}{}^{{}^8}\log x+{}^{{}^4}\log y^2=5\\ \\ {}^{{}^8}\log y+{}^{{}^4}\log x^2=7\end{array}\\ \\ & \text{Tentukanlah nilai dari}xy\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {}^{{}^{{}^8}}\log x+{}^{{}^4}\log y^2=5\\ & \iff {}^{{}^{{}^{2^3}}}\log x+{}^{{}^{{}^{2^2}}}\log y^2=5....(1)\\ & {}^{{}^{{}^8}}\log y+{}^{{}^4}\log x^2=7\\ & \iff {}^{{}^{{}^{2^3}}}\log y+{}^{{}^{{}^{2^2}}}\log x^2=7....(2)\\ & \text{selanjutnya},& \\ & \frac{1}{3}{.}^{{}^{{}^2}}\log x+{}^{{}^{{}^2}}\log y=5|\times \frac{1}{3}|\\ & \Rightarrow \frac{1}{9}{.}^{{}^{{}^2}}\log x+\frac{1}{3}.{}^{{}^{{}^2}}\log y=\frac{5}{3}....(3)\\ & {}^{{}^{{}^2}}\log x+\frac{1}{3}{.}^{{}^{{}^2}}\log y=7|\times 1|\\ & {\Rightarrow }^{{}^{{}^2}}\log x+\frac{1}{3}{.}^{{}^{{}^2}}\log y=7....(4)\\ & \text{saat persamaan}(3)-(4)\\ & =-\frac{8}{9}.{}^{{}^{{}^2}}\log x=\frac{5}{3}-7=-\frac{16}{3}\\ & \text{maka}\\ & {}^{{}^{{}^2}}\log x=(-\frac{16}{3})(-\frac{9}{8})\\ & {}^{{}^{{}^2}}\log x=6\iff x=2^6\iff x=64\\ & \text{Pada persamaan 1 selanjutnya}\\ & \frac{1}{3}.{}^{{}^{{}^2}}\log x+{}^{{}^{{}^2}}\log y=5\\ & \iff \frac{1}{3}.{}^{{}^{{}^2}}\log 2^6+{}^{{}^{{}^2}}\log y=5\\ & \iff \frac{1}{3}.6+{}^{{}^{{}^2}}\log y=5\\ & \iff 2+{}^{{}^{{}^2}}\log y=5\\ & \iff {}^{{}^{{}^2}}\log y=5-2=3\iff y=2^3=8\\ & \text{Jadi},x.y=64.8=512\end{array}\end{array}$

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

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

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

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

$.\begin{array}{rl}& \text{dari persamaan}(4)\text{dan}(5)\\ & \log (2y)={\displaystyle \frac{\log x\times \log (2z)}{\log 2}}\\ & \log \left({\displaystyle \frac{x}{z}}\right)={\displaystyle \frac{\log x\times \log (2z)}{\log 2}}\\ & \log 2(\log x-\log z)=\log x\times \log (2z)\\ & \log 2\times \log x-\log 2\times \log z=\log x\times (\log 2+\log z)\\ & \log 2\times \log x-\log 2\times \log z=\log x\times \log 2+\log x\times \log z\\ & -\log 2\times \log z=\log x\times \log z\\ & \log 2^{-1}=\log x\\ & {\displaystyle \frac{1}{2}=x.....(6)}\end{array}$

$.\begin{array}{|cc|}\hline \text{persamaan}(2)& \text{Menentukan nilai}z\\ \begin{array}{rl}\log 2y\times \log (4x)& =\log x\times \log (8yz)\\ \log 2y\times \log (4(2yz))& =\log x\times \log (8yz)\\ \log 2y\times \log (8yz)& =\log x\times \log (8yz)\\ \log (2y)& =\log x\\ 2y& =x\\ y& ={\displaystyle \frac{1}{2}x}\\ & ={\displaystyle \frac{1}{2}\times \frac{1}{2}}\\ & ={\displaystyle \frac{1}{4}.....(7)}\\ & \end{array}& \begin{array}{rl}& \\ x& =2yz\\ {\displaystyle \frac{1}{2}}& =2\left({\displaystyle \frac{1}{4}}\right)z\\ 1& =z\\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}$

$.\begin{array}{rl}& \text{maka nilai untuk}x{y}^{5}z\text{adalah}\\ & x{y}^{5}z=\left({\displaystyle \frac{1}{2}}\right).{\left({\displaystyle \frac{1}{4}}\right)}^5.1\\ & ={\displaystyle \frac{1}{2\times 4^5}}\\ & ={\displaystyle \frac{1}{2\times {\left(2^2\right)}^5}={\displaystyle \frac{1}{2^{1+10}}}}\\ & ={\displaystyle \frac{1}{2^{11}}={\displaystyle \frac{1}{2^{{}^{\frac{11}{1}}}}={\displaystyle \frac{1}{2^{{}^{\frac{p}{q}}}}}}}\\ & \{\begin{array}{ll}p& =11\\ q& =1\end{array}\text{dan jelas bahwa}p\text{dan}q\text{saling prima}\\ & \text{Jadi},\\ & p+q=11+1=12\end{array}$

DAFTAR PUSTAKA

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

Contoh Soal 5 Fungsi Logaritma

$\begin{array}{l}\\ 21.& \mathbf{\text{(SPMB '04)}}\\ & \text{Jika}a>1,\text{maka penyelesaian untuk}\\ & ({}^a\log (2x+1))({}^3\log \sqrt{a})=1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 1}\\ \text{b}.& {\displaystyle 2}\\ \text{c}.& {\displaystyle 3}\\ \text{d}.& {\displaystyle 4}\\ \text{e}.& {\displaystyle 5}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}({}^a\log (2x+1))({}^3\log \sqrt{a})& =1\\ ({}^3\log \sqrt{a})({}^a\log (2x+1))& =1\\ ({}^3\log a^{{.}^{{}^{\frac{1}{2}}}})({}^a\log (2x+1))& =1\\ {\displaystyle \frac{1}{2}({}^3\log a)({}^a\log (2x+1))}& =1\\ {}^3\log (2x+1)& =2\\ 2x+1& =3^2\\ 2x& =9-1\\ 2x& =8\\ x& =4\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \mathbf{\text{(SPMB '04)}}\\ & \text{Nilai}{\displaystyle \frac{{({}^5\log 10)}^2-{({}^5\log 2)}^2}{{}^5\log \sqrt{20}}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}}\\ \text{b}.& {\displaystyle 1}\\ \text{c}.& {\displaystyle 2}\\ \text{d}.& {\displaystyle 4}\\ \text{e}.& {\displaystyle 5}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {\displaystyle \frac{{({}^5\log 10)}^2-{({}^5\log 2)}^2}{{}^5\log \sqrt{20}}}\\ & ={\displaystyle \frac{({}^5\log 10+{}^5\log 2)({}^5\log 10-{}^5\log 2)}{{}^5\log (20{)}^{{.}^{\frac{1}{2}}}}}\\ & ={\displaystyle \frac{{}^5\log (10.2){\times }^5\log \left(\frac{10}{2}\right)}{{\displaystyle \frac{1}{2}\times {}^5\log 20}}}\\ & =2\times \left({\displaystyle \frac{{}^5\log 20}{{}^5\log 20}}\right)\times {}^5\log 5\\ & =2.1.1\\ & =2\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \mathbf{\text{(SPMB '03)}}\\ & \text{Jika diketahui bahwa}\\ & {}^4{\log }^4\log x-{}^4{\log }^4{\log }^4\log 16=2\\ & \text{maka}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {}^2\log x=8}\\ \text{b}.& {\displaystyle {}^2\log x=4}\\ \text{c}.& {\displaystyle {}^4\log x=8}\\ \text{d}.& {\displaystyle {}^4\log x=16}\\ \text{e}.& {\displaystyle {}^{16}\log x=8}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^4\log {}^4\log x-{}^4\log {}^4\log {}^4\log 16=2\\ & \iff {}^4\log {}^4\log x-{}^4\log {}^4\log {}^4\log 4^2=2\\ & \iff {}^4\log {}^4\log x-{}^4\log {}^4\log 2=2\\ & \iff {}^4\log {}^4\log x-{}^4\log {}^{2^2}\log 2^1=2\\ & \iff {}^4\log {}^4\log x-{}^4\log \left({\displaystyle \frac{1}{2}}\right)=2\\ & \iff {}^4\log {}^4\log x-{}^{2^2}\log 2^{-1}=2\\ & \iff {}^4\log {}^4\log x-(-{\displaystyle \frac{1}{2}})=2\\ & \iff {}^4\log {}^4\log x+\frac{1}{2}=2\\ & \iff {}^4\log {}^4\log x=2-{\displaystyle \frac{1}{2}=\frac{3}{2}}\\ & \iff {}^4\log x=4^{.\frac{3}{2}}\\ & \iff {}^4\log x={\left(2^2\right)}^{{.}^{\frac{3}{2}}}\\ & \iff {}^4\log x=2^3\\ & \iff {}^4\log x=8\end{array}\end{array}$

$\begin{array}{l}\\ 24.& \mathbf{\text{(UMPTN '92)}}\\ & \text{Jika}x\text{memenuhi persamaan}\\ & {}^4{\log }^4\log x-{}^4{\log }^4{\log }^4\log 16=2\\ & \text{maka nilai}{}^{16}\log x=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 4}\\ \text{b}.& {\displaystyle 2}\\ \text{c}.& {\displaystyle 1}\\ \text{d}.& {\displaystyle -2}\\ \text{e}.& {\displaystyle -4}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^4\log {}^4\log x-{}^4\log {}^4\log {}^4\log 16=2\\ & \text{menyebabkan}\\ & {}^4\log x=8\Rightarrow x=4^8\\ & (\text{lihat pembahasan no.23})\\ & \text{maka},\\ & {}^{16}\log x={}^{16}\log 4^8={}^{4^2}\log 4^8\\ & ={\displaystyle \frac{8}{2}}\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \mathbf{\text{(UMPTN '94)}}\\ & \text{Hasil kali akar-akar persamaan}\\ & {}^3\log x^{.\left({}^{2+{}^3\log x}\right)}=15\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{9}}\\ \text{b}.& {\displaystyle \frac{1}{3}}\\ \text{c}.& {\displaystyle 1}\\ \text{d}.& {\displaystyle 3}\\ \text{e}.& {\displaystyle 9}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^3\log x^{.\left({}^{2+{}^3\log x}\right)}=15\\ & \iff {\left(2+{}^3\log x\right)}^3\log x-15=0\\ & \iff 2{}^3\log x+{({}^3\log x)}^2-15=0\\ & \iff {({}^3\log x)}^2+2{}^3\log x-15=0\\ & \iff ({}^3\log x_1+5)({}^3\log x_2-3)=0\\ & \iff {}^3\log x_1+5=0\text{atau}{}^3\log x_1-3=0\\ & \iff {}^3\log x_1=-5\text{atau}{}^3\log x_2=3\\ & \iff x_1=3^{-5}\text{atau}{x}_2=3^3\\ & \text{maka}\\ & \iff x_1\times x_2=3^{-5}\times 3^3=3^{-5+3}=3^{-2}\\ & \iff ={\displaystyle \frac{1}{3^2}}\\ & \iff =\frac{1}{9}\end{array}\end{array}$

Contoh Soal 4 Fungsi Logaritma

$\begin{array}{l}\\ 16.& \mathbf{\text{(UMPTN '01)}}\\ & \text{Jika}{}^{10}\log x=b,\text{maka}{}^{10x}\log 100=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{b+1}}\\ \text{b}.& {\displaystyle \frac{2}{b+1}}\\ \text{c}.& {\displaystyle \frac{1}{b}}\\ \text{d}.& {\displaystyle \frac{2}{b}}\\ \text{e}.& {\displaystyle \frac{2}{10b}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {}^{10x}\log 100\\ & ={\displaystyle \frac{\log 100}{\log 10x}}\\ & ={\displaystyle \frac{{}^{10}\log 100}{{}^{10}\log 10x},\text{pilih basis 10}}\\ & \text{alasannya: supaya sama dengan soal}\\ & ={\displaystyle \frac{{}^{10}\log {10}^2}{{}^{10}\log 10+{}^{10}\log x}}\\ & ={\displaystyle \frac{2}{1+b}\text{atau}}\\ & =\frac{2}{b+1}\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \mathbf{\text{(UM UGM '03)}}\\ & \text{Jika}{}^4\log 6=m+1,\text{maka}{}^9\log 8=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{3}{4m-2}}\\ \text{b}.& {\displaystyle \frac{3}{4m+2}}\\ \text{c}.& {\displaystyle \frac{3}{2m+4}}\\ \text{d}.& {\displaystyle \frac{3}{2m-4}}\\ \text{e}.& {\displaystyle \frac{3}{2m+2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \text{Sebelumnya perhatikanlah}\\ & {}^4\log 6=m+1\\ & \iff {}^{2^2}\log (2.3{)}^1=m+1\\ & \iff {\displaystyle \frac{1}{2}\times {}^2\log (2.3)=m+1}\\ & \iff {\displaystyle \frac{1}{2}\times ({}^2\log 2+{}^2\log 3)=m+1}\\ & \iff {\displaystyle \frac{1}{2}\times (1+{}^2\log 3)=m+1}\\ & \iff 1+{}^2\log 3=2m+2\\ & \iff {}^2\log 3=2m+1\\ & \text{Selanjutnya adalah}:\\ & {}^9\log 8={\displaystyle \frac{1}{{}^8\log 9}}\\ & ={\displaystyle \frac{1}{{}^{2^3}\log 3^2}}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{2}{3}{}^2\log 3}}}\\ & ={\displaystyle \frac{3}{2{}^2\log 3}}\\ & ={\displaystyle \frac{3}{2(2m+1)}}\\ & ={\displaystyle \frac{3}{4m+2}}\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \mathbf{\text{(UMPTN '00)}}\\ & \text{Jika}{}^3\log 5=p\text{dan}{}^3\log 4=q,\\ & \text{maka}{}^4\log 15=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{pq}{1+p}}\\ \text{b}.& {\displaystyle \frac{p+q}{pq}}\\ \text{c}.& {\displaystyle \frac{p+1}{pq}}\\ \text{d}.& {\displaystyle \frac{p+1}{q+1}}\\ \text{e}.& {\displaystyle \frac{pq}{1-p}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^4\log 15\\ & ={\displaystyle \frac{{}^{...}\log 15}{{}^{...}\log 4},\text{pilih basis 5}}\\ & \text{mengapa tidak pilih basis selain 5}\\ & \text{lihat penyebut, di sana terdapat numerus 4}\\ & \text{pada soal, pasangan numerus 4 adalah 5},\\ & \text{makanya basis 5 dipilih, bukan yang lain}\\ & ={\displaystyle \frac{{}^5\log 15}{{}^5\log 4}={\displaystyle \frac{{}^5\log (3.5)}{{}^5\log 4}}}\\ & ={\displaystyle \frac{{}^5\log 3+{}^5\log 5}{{}^5\log 4}={\displaystyle \frac{{\displaystyle \frac{1}{{}^3\log 5}+{}^5\log 5}}{{}^5\log 4}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{p}+1}}{q}={\displaystyle \frac{1+p}{pq},\text{atau}}}\\ & ={\displaystyle \frac{p+1}{pq}}\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \mathbf{\text{(UMPTN '94)}}\\ & \text{Jika}{}^6\log 5=a\text{dan}{}^5\log 4=b,\\ & \text{maka}{}^4\log 0,24=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{a+2}{ab}}\\ \text{b}.& {\displaystyle \frac{2a+1}{ab}}\\ \text{c}.& {\displaystyle \frac{a-2}{ab}}\\ \text{d}.& {\displaystyle \frac{2a+1}{2ab}}\\ \text{e}.& {\displaystyle \frac{1-2a}{ab}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {}^4\log 0,24\\ & ={\displaystyle \frac{{}^{...}\log 0,24}{{}^{...}\log 4}=\frac{{}^{...}\log {\displaystyle \frac{6}{25}}}{{}^{...}\log 4},\text{pilih basis 5}}\\ & \text{mengapa tidak pilih basis selain 5}\\ & \text{lihat penyebut, di sana terdapat numerus 4}\\ & \text{pada soal, pasangan numerus 4 adalah 5},\\ & \text{makanya basis 5 dipilih, bukan yang lain}\\ & =\frac{{}^5\log {\displaystyle \frac{6}{25}}}{{}^5\log 4}={\displaystyle \frac{{}^5\log 6-{}^5\log 25}{{}^5\log 4}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{{}^6\log 5}-{}^5\log 5^2}}{{}^5\log 4}={\displaystyle \frac{{\displaystyle \frac{1}{a}-2}}{b}=\frac{1-2a}{ab}}}\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \mathbf{\text{(SPMB '05)}}\\ & \text{Jika}{}^3\log 2=p\text{dan}{}^2\log 7=q,\\ & \text{maka}{}^{14}\log 54=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{p+3}{p+q}}\\ \text{b}.& {\displaystyle \frac{2p}{p+q}}\\ \text{c}.& {\displaystyle \frac{p+3}{p(q+1)}}\\ \text{d}.& {\displaystyle \frac{p+q}{p(q+1)}}\\ \text{e}.& {\displaystyle \frac{p(q+1)}{p+q}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& {}^{14}\log 54\\ & ={\displaystyle \frac{{}^{...}\log 54}{{}^{...}\log 14}=\frac{{}^{...}\log (2.27)}{{}^{...}\log (2.7)},\text{pilih basis 2}}\\ & \text{mengapa tidak pilih basis selain 2}\\ & \text{lihat penyebut, di sana terdapat numerus 7}\\ & \text{pada soal, pasangan numerus 7 adalah 2},\\ & \text{makanya basis 2 dipilih, bukan yang lain}\\ & =\frac{{}^2\log (2.27)}{{}^2\log (2.7)}={\displaystyle \frac{{}^2\log 2+{}^2\log 27}{{}^2\log 2+{}^2\log 7}}\\ & ={\displaystyle \frac{{}^2\log 2+{}^2\log 3^3}{{}^2\log 2+{}^2\log 7}={\displaystyle \frac{{}^2\log 2+(3\times {\displaystyle \frac{1}{{}^3\log 2}})}{{}^2\log 2+{}^2\log 7}}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{3}{p}}}{1+q}}\\ & ={\displaystyle \frac{p+3}{p(q+1)}}\end{array}\end{array}$

Contoh Soal 3 Fungsi Logaritma

$\begin{array}{l}\\ 11.& \text{Nilai dari}\\ & {\displaystyle \frac{1}{6}.{}^2\log 25-\frac{1}{3}.{}^2\log 10\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -3}\\ \text{b}.& {\displaystyle -2}\\ \text{c}.& -2{\displaystyle \frac{1}{2}}\\ \text{d}.& -{\displaystyle \frac{1}{2}}\\ \text{e}.& -{\displaystyle \frac{1}{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& ={\displaystyle \frac{1}{6}.{}^2\log 25-\frac{1}{3}.{}^2\log 10}\\ & ={\displaystyle \frac{1}{3}.\frac{1}{2}.{}^2\log 25-\frac{1}{3}.{}^2\log 10}\\ & =\frac{1}{3}({}^2\log {25}^{\frac{1}{2}}-{}^2\log 10)\\ & =\frac{1}{3}({}^2\log 5-{}^2\log 10)\\ & =\frac{1}{3}({}^2\log \frac{5}{10})\\ & =\frac{1}{3}({}^2\log \frac{1}{2})\\ & =\frac{1}{3}({}^2\log 2^{-1})\\ & =-\frac{1}{3}\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Nilai dari}\\ & {}^2\log ({}^2\log \sqrt{\sqrt{2}})\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -4}\\ \text{b}.& {\displaystyle -2}\\ \text{c}.& -1{\displaystyle \frac{1}{2}}\\ \text{d}.& -{\displaystyle \frac{1}{2}}\\ \text{e}.& -{\displaystyle \frac{1}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& ={}^2\log ({}^2\log \sqrt{\sqrt{2}})\\ & ={}^2\log ({}^2\log 2^{\frac{1}{4}})\\ & ={}^2\log \frac{1}{4}\\ & ={}^2\log {\left(2\right)}^{-2}\\ & =-2\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \mathbf{\text{(UMPTN '99)}}\\ & \text{Diketahui}\log 2=0,3010\text{dan}\log 3=0,4771\\ & \text{maka}\log (\sqrt[3]{2}\times \sqrt{3})\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 0,1505}\\ \text{b}.& {\displaystyle 0,1590}\\ \text{c}.& {\displaystyle 0,2007}\\ \text{d}.& {\displaystyle 0,3389}\\ \text{e}.& {\displaystyle 0,3891}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \log (\sqrt[3]{2}\times \sqrt{3})\\ & =\log \sqrt[3]{2}+\log \sqrt{3}\\ & =\log 2^{\frac{1}{3}}+\log 3^{\frac{1}{2}}\\ & ={\displaystyle \frac{1}{3}\log 2+{\displaystyle \frac{1}{2}\log 3}}\\ & ={\displaystyle \frac{1}{3}(0,3010)+{\displaystyle \frac{1}{2}(0,4771)}}\\ & =0,1003+0,2386\\ & =0,3389\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \mathbf{\text{(UMPTN '98)}}\\ & \text{Nilai}{}^a\log {\displaystyle \frac{1}{b}\times {}^b\log {\displaystyle \frac{1}{c^2}\times {}^c\log {\displaystyle \frac{1}{a^3}=....}}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -6}\\ \text{b}.& {\displaystyle 6}\\ \text{c}.& {\displaystyle \frac{b}{a^{2}c}}\\ \text{d}.& {\displaystyle \frac{a^{2}c}{b}}\\ \text{e}.& {\displaystyle -\frac{1}{6}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {}^a\log {\displaystyle \frac{1}{b}\times {}^b\log {\displaystyle \frac{1}{c^2}\times {}^c\log {\displaystyle \frac{1}{a^3}}}}\\ & ={}^a\log {\displaystyle b^{-1}\times {}^b\log {\displaystyle c^{-2}\times {}^c\log {\displaystyle a^{-3}}}}\\ & =(-1).(-2).(-3)\times {}^a\log {\displaystyle a\times {}^b\log {\displaystyle c\times {}^c\log {\displaystyle a}}}\\ & =-6\times {}^a\log a\\ & =-6\times 1\\ & =-6\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \mathbf{\text{(UMPTN '01)}}\\ & \text{Jika}{\displaystyle \frac{{}^2\log a}{{}^3\log b}=m\text{dan}{\displaystyle \frac{{}^3\log a}{{}^2\log b}=n}}\\ & \text{dengan}a>1,b>1,\text{maka}{\displaystyle \frac{m}{n}=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle {}^2\log 3}\\ \text{b}.& {\displaystyle {}^3\log 2}\\ \text{c}.& {\displaystyle {}^4\log 9}\\ \text{d}.& {\displaystyle {({}^3\log 2)}^2}\\ \text{e}.& {\displaystyle {({}^2\log 3)}^2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{m}{n}}& ={\displaystyle \frac{{\displaystyle \frac{{}^2\log a}{{}^3\log b}}}{{\displaystyle \frac{{}^3\log a}{{}^2\log b}}}}\\ & ={\displaystyle \frac{{}^2\log a\times {}^2\log b}{{}^3\log b\times {}^3\log a}}\\ & ={\displaystyle \frac{{}^2\log a\times {\displaystyle \frac{1}{{}^b\log 2}}}{{}^3\log b\times {\displaystyle \frac{1}{{}^a\log 3}}}}\\ & ={\displaystyle \frac{{}^2\log a\times {}^a\log 3}{{}^3\log b\times {}^b\log 2}}\\ & ={\displaystyle \frac{{}^2\log 3}{{}^3\log 2}={\displaystyle \frac{{}^2\log 3}{{\displaystyle \frac{1}{{}^2\log 3}}}}}\\ & ={({}^2\log 3)}^2\end{array}\end{array}$

Contoh Soal 2 Fungsi Logaritma

$\begin{array}{l}\\ 6.& \text{Nilai dari}{}^{\sqrt{2}}\log 16\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 10}\\ \text{b}.& {\displaystyle 9}\\ \text{c}.& {\displaystyle 8}\\ \text{d}.& {\displaystyle 6}\\ \text{e}.& {\displaystyle 4}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& ={}^{\sqrt{2}}\log 16\\ & ={}^{{\displaystyle 2^{\frac{1}{2}}}}\log 2^4\\ & ={\displaystyle \frac{4}{\frac{1}{2}}\times {}^2\log 2}\\ & =8\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Nilai dari}{}^{\sqrt{5}}\log \sqrt{125}\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -3}\\ \text{b}.& {\displaystyle -2}\\ \text{c}.& {\displaystyle 2}\\ \text{d}.& {\displaystyle 3}\\ \text{e}.& {\displaystyle 5}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& ={}^{\sqrt{5}}\log \sqrt{125}\\ & ={}^{{\sqrt{5}}^1}\log {\left(\sqrt{5}\right)}^3\\ & ={\displaystyle \frac{3}{1}\times {}^{\sqrt{5}}\log \sqrt{5}}\\ & =3\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Nilai dari}{}^{\sqrt{\sqrt{2}}}\log \sqrt{8\sqrt{8}}\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 4}\\ \text{b}.& {\displaystyle 6}\\ \text{c}.& {\displaystyle 8}\\ \text{d}.& {\displaystyle 9}\\ \text{e}.& {\displaystyle 12}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& ={}^{\sqrt{\sqrt{2}}}\log \sqrt{8\sqrt{8}}\\ & ={}^{\sqrt[4]{2}}\log {\left(8{\left(8\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}}\\ & ={}^{2^{\frac{1}{4}}}\log 8^{(\frac{1}{2}+\frac{1}{4})}\\ & ={}^{2^{\frac{1}{4}}}\log 2^{3\left(\frac{3}{4}\right)}\\ & ={\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}\times {}^2\log 2}\\ & =9\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Nilai dari}\\ & {}^6\log 8+{}^6\log {\displaystyle \frac{9}{2}\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 4}\\ \text{b}.& {\displaystyle 3}\\ \text{c}.& 3{\displaystyle \frac{1}{2}}\\ \text{d}.& 2{\displaystyle \frac{1}{2}}\\ \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& ={}^6\log 8+{}^6\log {\displaystyle \frac{9}{2}}\\ & ={}^6\log 8\times \frac{9}{2}\\ & ={}^6\log 36\\ & ={}^6\log 6^2\\ & =2\\ & \\ & \\ & \end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Nilai dari}\\ & {}^2\log {\displaystyle \frac{4}{3}+2.{}^2\log \sqrt{12}\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 6}\\ \text{b}.& {\displaystyle 4}\\ \text{c}.& 3{\displaystyle \frac{1}{2}}\\ \text{d}.& 2{\displaystyle \frac{1}{2}}\\ \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& ={}^2\log {\displaystyle \frac{4}{3}+2.{}^2\log \sqrt{12}}\\ & ={}^2\log \frac{4}{3}+{}^2\log {\left(\sqrt{12}\right)}^2\\ & ={}^2\log \frac{4}{3}+{}^2\log 12\\ & ={}^2\log \frac{4}{3}\times 12\\ & ={}^2\log 16\\ & ={}^2\log 2^4\\ & =4\end{array}\end{array}$

Contoh Soal 1 Fungsi Logaritma

$\begin{array}{l}\\ 1.& \text{Nilai dari}{}^2\log {\displaystyle \frac{1}{32}\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -7}\\ \text{b}.& {\displaystyle -5}\\ \text{c}.& {\displaystyle -3}\\ \text{d}.& {\displaystyle -2}\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& ={}^2\log {\displaystyle \frac{1}{32}}\\ & ={}^{2^1}\log 2^{-5}\\ & ={\displaystyle \frac{-5}{1}\times {}^2\log 2}\\ & =-5\\ & \\ & \end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Nilai dari}{}^{0,333...}\log 0,111....\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{3}}\\ \text{b}.& {\displaystyle \frac{1}{2}}\\ \text{c}.& 2\\ \text{d}.& 3\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& ={}^{0,333...}\log 0,111...\\ & ={}^{\frac{1}{3}}\log \frac{1}{9}\\ & ={}^{\frac{1}{3}}\log {\left(\frac{1}{3}\right)}^2\\ & =2\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Nilai dari}{}^5\log 25\sqrt{5}\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5}{2}}\\ \text{b}.& {\displaystyle \frac{3}{2}}\\ \text{c}.& {\displaystyle \frac{1}{2}}\\ \text{d}.& {\displaystyle 2}\\ \text{e}.& 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& ={}^5\log 25\sqrt{5}\\ & ={}^{5^1}\log 5^2{.5}^{\frac{1}{2}}\\ & ={}^{5^1}\log 5^{\frac{5}{2}}\\ & ={\displaystyle \frac{\frac{5}{2}}{1}\times {}^5\log 5}\\ & ={\displaystyle \frac{5}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Nilai dari}{}^{\sqrt{3}}\log 81\text{adalah}...\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 12}\\ \text{b}.& {\displaystyle 10}\\ \text{c}.& {\displaystyle 9}\\ \text{d}.& {\displaystyle 8}\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& ={}^{\sqrt{3}}\log 81\\ & ={}^{{\displaystyle 3^{\frac{1}{2}}}}\log 3^4\\ & ={\displaystyle \frac{4}{\frac{1}{2}}\times {}^3\log 3}\\ & =8\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Nilai dari}{}^{\frac{1}{3}}\log {\displaystyle \frac{1}{243}\text{adalah}...}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 6}\\ \text{b}.& {\displaystyle 5}\\ \text{c}.& {\displaystyle 4}\\ \text{d}.& {\displaystyle 3}\\ \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& ={}^{\frac{1}{3}}\log {\displaystyle \frac{1}{243}}\\ & ={}^{{\left(\frac{1}{3}\right)}^1}\log {\displaystyle {\left(\frac{1}{3}\right)}^5}\\ & ={\displaystyle \frac{5}{1}\times {}^{\frac{1}{3}}\log \frac{1}{3}}\\ & =5\end{array}\end{array}$