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

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

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

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

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

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

Lanjutan Materi Fungsi Logaritma

 $\text{B. Sifat-Sifat Logaritma}$

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

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

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

$\begin{array}{rl}(\text{a})& \log 2=0,3010\\ (\text{b})& \log 3=0,4771\\ (\text{c})& \log 5=0,6990\\ (\text{d})& \log 7=0,8451\end{array}$

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

$\begin{array}{l}\\ 1.& {}^2\log 3+{}^2\log 8-{}^2\log 24=....\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{}^2\log 3+{}^2\log 8-{}^2\log 24& ={}^2\log \left({\displaystyle \frac{3\times 8}{24}}\right)\\ & ={}^2\log 1\\ & ={}^2\log 2^0\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 2.& {}^2\log 12+{}^2\log 8-{}^2\log 24=....\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{}^2\log 12+{}^2\log 8-{}^2\log 24& ={}^2\log \left({\displaystyle \frac{12\times 8}{24}}\right)\\ & ={}^2\log 4\\ & ={}^2\log 2^2\\ & =2\end{array}\end{array}$

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

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

$\begin{array}{l}\\ 5.& \text{Jika}{}^{2017}\log {\displaystyle \frac{1}{x}={}^x\log {\displaystyle \frac{1}{y}={}^y\log {\displaystyle \frac{1}{2017}}}}\\ & \text{maka hasil dari}(2x-3y)\\ \\ & \text{Jawab}:\\ \\ & \begin{array}{rl}{}^{2017}\log {\displaystyle \frac{1}{x}}& ={}^x\log {\displaystyle \frac{1}{y}={}^y\log {\displaystyle \frac{1}{2017}}}\\ {}^{2017}\log {\displaystyle \frac{1}{x}}& ={}^y\log {\displaystyle \frac{1}{2017}}\\ {}^{2017}\log {\displaystyle x^{-1}}& ={}^y\log {\displaystyle (2017{)}^{-1}}\\ -{}^{2017}\log x& =-{}^y\log 2017\\ {}^{2017}\log x& ={}^y\log 2017\\ & \text{dipenuhi saat}\\ x& =y=2017\end{array}\\ \\ & (2x-3y)=2x-3x=-x=-2017\end{array}$

Fungsi Logaritma

$\text{A. Pendahuluan}$

Logaritma merupakan invers(balikan) dari perpangkatan

Secara definisi:

$\boxed{{\displaystyle a^c=b{\Rightarrow }^a\log b=c}}$, 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

${}^a\log b=c\{\begin{array}{ll}a& \text{syaratnya}:a>0,a\ne 1\\ & \text{selanjutnya disebut basis}\\ b& \text{syaratnya}:b>0\\ & \text{selanjutnya disebut}\mathbf{\text{numerus}}\\ c& \text{tidak ada syarat apapun}\\ & \text{selanjutnya disebut hasil logaritma}\end{array}$

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

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

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

$\begin{array}{rl}(1)& (-2{)}^4=16\Rightarrow {}^{(-2)}\log 16=\cdots \\ (2)& (-2{)}^3=-8\Rightarrow {}^{(-2)}\log (-8)=\cdots \\ (3)& (-2{)}^2=4\Rightarrow {}^{(-2)}\log 4=\cdots \\ (4)& (-2{)}^1=-2\Rightarrow {}^{(-2)}\log (-2)=\cdots \\ (5)& (-2{)}^0=1\Rightarrow {}^{(-2)}\log 1=\cdots \\ (6)& (-2{)}^{-1}=-{\displaystyle \frac{1}{2}\Rightarrow {}^{(-2)}\log (-{\displaystyle \frac{1}{2}})=\cdots }\\ (7)& (-2{)}^{-2}={\displaystyle \frac{1}{4}\Rightarrow {}^{(-2)}\log {\displaystyle \frac{1}{4}=\cdots }}\\ (8)& (-2{)}^{-3}=-{\displaystyle \frac{1}{8}\Rightarrow {}^{(-2)}\log (-{\displaystyle \frac{1}{8}})=\cdots }\\ (9)& (-2{)}^{-4}={\displaystyle \frac{1}{16}\Rightarrow {}^{(-2)}\log {\displaystyle \frac{1}{16}=\cdots }}\end{array}$

Statistika (Matematika Wajib kelas XII)

 $\text{A. Pendahuluan}$

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

Sebagai tambahan penjelasan

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

$\text{B. Penyajian Data}$

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

$\text{C. Data Tunggal}$

$\text{C. 1 Ukuran Pemusatan Data (Tendesi Sentral)}$

$\begin{array}{|cll|}\hline \text{No}& \text{Nilai}& \text{Ukuran Pemusatan Data}\\ 1.& \text{Mean}\left(\overline{x}\right)& \overline{x}={\displaystyle \frac{x_1+x_2+x_3+...+x_n}{n}}\\ 2.& \text{Median}\left(M_e\right)& \{\begin{array}{ll}\text{Ganjil}& M_e=x_{\frac{n+1}{2}}\\ \text{Genap}& M_e={\displaystyle \frac{1}{2}(x_{\frac{n}{2}}+x_{\frac{n}{2}+1})}\end{array}\\ 3.& \text{Modus}\left(M_o\right)& \text{Nilai yang sering muncul}\\ 4.& \text{Kuartil}\left(Q\right)& \{\begin{array}{ll}\text{Ganjil}& \{\begin{array}{ll}{Q}_1=& x_{\frac{1}{4}(n+1)}\\ Q_2=& x_{\frac{2}{4}(n+1)}\\ Q_3=& x_{\frac{3}{4}(n+1)}\end{array}\\ \\ \text{Genap}& \{\begin{array}{ll}{Q}_1=& x_{\frac{1}{4}n+\frac{1}{2}}\\ Q_2=& x_{\frac{2}{4}n+\frac{1}{2}}\\ Q_3=& x_{\frac{3}{4}n+\frac{1}{2}}\end{array}\end{array}\\ \hline\end{array}$

$\text{C. 2 Ukuran Penyebaran Data (Dispersi)}$

$\begin{array}{|cll|}\hline \text{No}& \text{Nilai}& \text{Ukuran Penyebaran Data}\\ 1.& \text{Jangkauan}\left(J\right)& J=R\\ & \text{atau Rentang}(R)& =x_{datummax}-x_{datummin}\\ 2.& \text{Hamparan}(H)& \\ & \text{Atau Jangkauan}& H=Q_3-Q_1\\ & \text{antar kuartil}& \\ 3.& \text{Simpangan}& Q_d={\displaystyle \frac{1}{2}H}\\ & \text{Kuartil}\left(Q_d\right)& \\ 4.& \text{Langkah}\left(L\right)& L={\displaystyle \frac{3}{2}H}\\ 5.& \text{Pagar}& \{\begin{array}{ll}\text{Dalam}& =Q_1-L\\ \text{Luar}& =Q_3+L\end{array}\\ 6& \text{Data}& \{\begin{array}{ll}& \text{Normal}\\ & :Q_1-L\le x_i\le Q_3+L\\ & \text{Tidak Normal}\\ & :\{\begin{array}{l}{x}_i<Q_1-L\\ x_i>Q_3+L\end{array}\end{array}\\ 7.& \text{Simpangan}& SR={\displaystyle \frac{\sum |x_i-\overline{x}|}{n}}\\ & \text{Rata-rata}(SR)& \\ 8.& \text{Ragam}\left(s^2\right)& s^2={\displaystyle \frac{\sum {(x_i-\overline{x})}^2}{n}}\\ & \text{atau Varian}& \\ 9.& \text{Simpangan}& s=\sqrt{s^2}\\ & \text{Baku}(s)& \\ \hline\end{array}$

$\text{D. Data Berkelompok}$

Untuk tipe ini antara lain

$\{\begin{array}{ll}\text{(1) Mean},& \overline{\text{x}}={\overline{x}}_s+{\displaystyle \frac{\sum f_i.d_i}{\sum f_i}}\\ \\ \text{(2) Modus},& {\text{M}}_o=t_p+{\displaystyle p\left(\frac{d_1}{d_1+d_2}\right)}\\ \\ \text{(3) Kuartil},& {\text{Q}}_i=t_p+{\displaystyle p\left(\frac{{\displaystyle \frac{i.n}{4}-\sum f_i}}{f_b}\right)}\end{array}$

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

$\{\begin{array}{ll}\begin{array}{rl}& (1)\\ & \\ & \\ & \\ & \end{array}& \begin{array}{|c|}\hline \begin{array}{rl}{\overline{x}}_s=& \text{rataan sementara}\\ x_i=& \text{titik tengahinterval}\\ & \text{kelas ke}-i\\ d_i=& x_i-{\overline{x}}_s\\ f_i=& \text{frekuensi kelas ke}-i\end{array}\\ \hline\end{array}\\ \\ \begin{array}{rl}& (2)\\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{|c|}\hline \begin{array}{rl}{t}_p=& \text{tepi bawah kelas modus}\\ p=& \text{panjang interval kelas}\\ d_1=& f_0-f_{-1}\\ d_2=& f_0-f_{+1}\\ f_0=& \text{frekuensi kelas modus}\\ f_{-1}=& \text{frekuensi sebelum kelas modus}\\ f_{+1}=& \text{frekuensi setelah kelas modus}\end{array}\\ \hline\end{array}\\ \\ \begin{array}{rl}& (3)\\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{|c|}\hline \begin{array}{rl}tp=& \text{tepi bawah kuartil ke}-i\\ p=& \text{panjang interval kelas}\\ n=& \text{banyaknya data}\\ \sum f_i=& \text{jumlah semua frekuensi}\\ & \text{sebelum kelas kurtil ke}-i\\ f_q=& \text{frekuensi kelas kuartil ke}-i\\ Q_i=& \text{kuartil ke}-i\\ Q_1=& \text{kuartil bawah}\\ Q_2=& \text{kuatil tengah/median}\\ Q_3=& \text{kuatil atas}\end{array}\\ \hline\end{array}\end{array}$

DAFTAR PUSTAKA

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

Lanjutan Sistem Persamaan Linear Tiga Variabel (Matematika Wajib Kelas X)

 $\text{2. Metode determinan Matriks}$

Perhatikan kemabil bentuk SPLDV dan SPLTV berikut:

$\{\begin{array}{l}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\end{array}$ 

dan

$\{\begin{array}{l}{a}_{1}x+b_{1}y+c_{1}z=d_1\\ a_{1}x+b_{1}y+c_{1}z=d_1\\ a_{1}x+b_{1}y+c_{1}z=d_1\end{array}$

Metode determinat matriks adalah penyelesaian nilai tidap variabel dengan menggunakan determinan berikut:

Misalkan saja diberikan:

$\begin{array}{rl}& \begin{array}{rl}ax+by& =p\\ cx+dy& =q\end{array}\\ \\ & \text{dan}\\ \\ & \begin{array}{rl}ax+by+cz& =r\\ dx+ey+fz& =s\\ gx+hy+iz& =t\end{array}\end{array}$

maka penyelesaian dengan model matriks adalah:

$\begin{array}{|ccc|}\hline \text{Metode}& \mathbf{\text{SPLDV}}& \mathbf{\text{SPLTV}}\\ \text{Determinan}& \begin{array}{rl}x& ={\displaystyle \frac{\left|\begin{array}{cc}p& b\\ q& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}}\\ & \text{dan}\\ y& ={\displaystyle \frac{\left|\begin{array}{cc}a& p\\ c& q\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}}\\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}x& ={\displaystyle \frac{\left|\begin{array}{ccc}r& b& c\\ s& e& f\\ t& h& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}}\\ & \text{dan}\\ y& ={\displaystyle \frac{\left|\begin{array}{ccc}a& r& c\\ d& s& f\\ g& t& i\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}}\\ & \text{serta}\\ z& ={\displaystyle \frac{\left|\begin{array}{ccc}a& b& r\\ d& e& s\\ g& h& t\end{array}\right|}{\left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|}}\end{array}\\ \hline\end{array}$

Sebagai catatan:

$\begin{array}{rl}& \left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc\\ & \text{dan}\\ & \left|\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right|=a\left|\begin{array}{cc}e& f\\ h& i\end{array}\right|-b\left|\begin{array}{cc}d& f\\ g& i\end{array}\right|+c\left|\begin{array}{cc}d& e\\ g& h\end{array}\right|\end{array}$

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

Mari kita buka lagi contoh sebelumnya dengan soal yang sama di SINI

dan kearang penyelesaian dari soal tersebut akan diselesaikan dengan cara determinan matriks (cara Cramer sesuai nama penemunya) berikut:

$\begin{array}{l}\\ 1.& \text{Tentukanlah dengan metode matriks}\\ & \text{(cara Cramer) SPLDV berikut}:\\ & \{\begin{array}{ll}2x-y& =7\\ x-y& =-1\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}x& ={\displaystyle \frac{\left|\begin{array}{cc}7& -1\\ -1& -1\end{array}\right|}{\left|\begin{array}{cc}2& -1\\ 1& -1\end{array}\right|}=\frac{7(-1)-(-1).(-1)}{2.(-1)-(-1).1}}\\ & ={\displaystyle \frac{-7-1}{-2+1}=\frac{-8}{-1}=8}\\ y& ={\displaystyle \frac{\left|\begin{array}{cc}2& 7\\ 1& -1\end{array}\right|}{\left|\begin{array}{cc}2& -1\\ 1& -1\end{array}\right|}=\frac{2(-1)-(7).1}{2.(-1)-(-1).1}}\\ & ={\displaystyle \frac{-2-7}{-2+1}=\frac{-9}{-1}=9}\\ \text{J}& \text{adi}(x,y)=(8,9)\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah dengan metode matriks}\\ & \text{(cara Cramer) SPLTV berikut}:\\ & \{\begin{array}{ll}2x-y+z& =-4\\ 2x-y-2z& =-3\\ x+3y-z& =0\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}x& ={\displaystyle \frac{\left|\begin{array}{ccc}-4& -1& 1\\ -3& -1& -2\\ 0& 3& -1\end{array}\right|}{\left|\begin{array}{ccc}2& -1& 1\\ 2& -1& -2\\ 1& 3& -1\end{array}\right|}}\\ x& ={\displaystyle \frac{-4\left|\begin{array}{cc}-1& -2\\ 3& -1\end{array}\right|+1\left|\begin{array}{cc}-3& -2\\ 0& -1\end{array}\right|+1\left|\begin{array}{cc}-3& -1\\ 0& 3\end{array}\right|}{2\left|\begin{array}{cc}-1& -2\\ 3& -1\end{array}\right|+1\left|\begin{array}{cc}2& -2\\ 1& -1\end{array}\right|+1\left|\begin{array}{cc}2& -1\\ 1& 3\end{array}\right|}}\\ & ={\displaystyle \frac{-4(1+6)+1(3-0)+1(-9-0)}{2(1+6)+1(-2+2)+1(6+1)}}\\ & ={\displaystyle \frac{-28+3-9}{14+0+7}}\\ & =\frac{-34}{21}\\ y& =....\\ z& =....\\ \text{J}& \text{adi}(x,y,y)=(-{\displaystyle \frac{34}{21},\frac{3}{7},-\frac{1}{3}})\end{array}\end{array}$

DAFTAR PUSTAKA

1. Johanes, Kastola & Sulasim. 2006. Kompetensi Matematika 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA

Sistem Persamaan Linear Tiga Variabel (Matematika Wajib Kelas X)

$\text{A. Sistem Persamaan Linear}$

Sistem persamaan linear adalah adalah kumpulan dari beberapa persamaan linear di mana koefisien-koefisien persamaannya berupa bilangan real dan anatar variabel saling ada keterkaitan

$\text{1. Sistem Persamaan Linear Dua Variabel}$

Sistem Persamaan Linear Dua Variabel yang selanjutnya disingkat dengan SPLDV memiliki bentuk umum sebagai berikut:

$\begin{array}{rl}& \{\begin{array}{c}{a}_{1}x+b_{1}y=c_1\\ a_{2}x+b_{2}y=c_2\end{array}\end{array}$

Keterangan:

• $x,y\text{adalah variabel}$.
• $a_1,a_2\text{koefisien}x$
• $b_1,b_2\text{koefisien}y$.
• $c_1,c_2\text{adalah konstanta}$.
• $a_1,a_2,b_1,b_2,c_1,\text{dan}{c}_2\text{adalah bilangan riil}$.

$\text{2. Sistem Persamaan Linear Tiga Variabel}$

$\begin{array}{l}\\ \underset{―}{\mathbf{\text{Bentuk Umum}}}& :\\ & \{\begin{array}{l}{a}_{1}x+b_{1}y+c_{1}z=d_1\\ a_{2}x+b_{2}y+c_{2}z=d_2\\ a_{3}x+b_{3}y+c_{3}z=d_3\end{array}\\ \\ \mathbf{\text{Keterangan}}& \bullet a_1,a_2,a_3,\\ & b_1,b_2,b_3,\\ & c_1,c_2,c_3,\\ & d_1,d_2,d_3\\ & \text{semuanya adalah bilangan real}\end{array}$

$\text{B. Penyelesaian Sistem persamaan Linear}$

Menentukan penyelesaian atau himpunan penyelesaian (HP) dari sistem persamaan linear baik yang terdiri dari dua variabel ataupun tiga variabel adalah menentukan pasangan koordinat yang memenuhi sistem persamaan tersebut di bilangan riil. Adapun cara menyelesaikan sistem persamaan linear ini

• Metode Substitusi
• Metode Eliminasi
• Metode Eliminasi-Substitusi 
• Metode Determinan Matrik
• Metode Invers Matrik (Matrik Persegi minimal ordo 2x2)

$\text{1. Metode Eliminasi-Substitusi}$

Adapun langkah-langkah dalam penyelesaian model tipe ini (Metode Substitusi dan Metode Eliminasi mengikuti karena prosesnya terangkum di langkah gabungan ini) adalah:

• buatlah dua buah kelompok persamaan yang memungkinkan dapat disederhanakan (kalau bisa ambil yang termudah dan sederhana menurut Anda)
• Salah satu variabel dihilangkan dengan cara menyamakan koefisien variabel yang bersangkutan kemudian mengeliminasikan dengan persamaan linear yang dipilih pada langkat pertama tadi.
• Nilai variabel yang didapatkan disubstitusikan ke dalam salah satu persamaan pada langkah pertama tadi juga.
• Jika diperlukan lagi, prinsipnya kembali pada poin pertama tadi

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

$\begin{array}{l}\\ 1.& \text{Tentukanlah penyelesaian SPLDV dari}\\ & \{\begin{array}{l}2x-y=7\\ x-y=-1\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Mis}& \text{alkan}\\ & \{\begin{array}{ll}2x-y=7& .....(1)\\ x-y=-1& .....(2)\end{array}\\ & \begin{array}{rl}& \text{dari persamaan}(2)\text{didapatkan}\\ & x=y-1.\text{Bentuk ini kemudian}\\ & \text{kita substitusikan ke}\\ & \text{persamaan}(1).\end{array}\\ & \begin{array}{rl}2x-y& =7\\ 2(y-1)-y& =7\\ 2y-2-y& =7\\ y& =9.....(3)\\ \text{Selanjutnya}& \text{nilainya kita}\\ \text{substitusikan ke}& \text{persamaan}(2)\\ x& =y-1\\ x& =9-1\\ x& =8\end{array}\\ & \begin{array}{rl}& \text{Sehingga},\\ & \{\begin{array}{l}x=8\\ y=9\end{array}\\ & \text{Jadi, HP}=\{(8,9)\}\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah penyelesaian SPLDV dari}\\ & \{\begin{array}{l}2x-y+z=-4\\ 2x-y-2z=-3\\ x+3y-z=0\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Per}& \text{hatikan misal}\\ & \{\begin{array}{l}2x-y+z=-4..........(1)\\ 2x-y-2z=-3........(2)\\ x+3y-z=0.............(3)\end{array}\\ & \begin{array}{rl}& \text{dari persamaan}(2)\text{didapatkan}\\ & 2x-y=2z-3.\text{Bentuk ini}\\ & \text{kita substitusikan ke}\\ & \text{persamaan}(1).\end{array}\\ & \begin{array}{rl}2x-y+z& =-4\\ (2z-3)+z& =-4\\ 3z& =-1\\ z& =-\frac{1}{3}.....(4)\\ \text{Selanjutnya}& \text{nilai tersebut kita}\\ \text{substitusikan ke}& \text{pers.}(2)\text{dan}(3)\end{array}\\ & \begin{array}{rl}& \text{Selanjutnya}\\ & \{\begin{array}{l}6x-3y=-11.....(2)\\ x+3y=-\frac{1}{3}.....(3)\end{array}\end{array}\\ & \begin{array}{rl}& \text{dengan cara seperti}\\ & \text{poin 1.a kita akan}\\ & \text{mendapatkan nilai}\\ x& =-\frac{34}{21}\text{dan}\\ y& =\frac{3}{7}\end{array}\\ & \begin{array}{rl}& \text{Sehingga},\\ & \{\begin{array}{l}x=-\frac{34}{21}\\ y=\frac{3}{7}\\ z=-\frac{1}{3}\end{array}\\ & \text{HP}=\left\{(-\frac{34}{21},\frac{3}{7},-\frac{1}{3})\right\}\end{array}\end{array}\end{array}$