Hanifah Blog: logaritma

Jika $\displaystyle ^{x+y}\log{2}=a$ dan $\displaystyle ^{x-y}\log{8}=b$ , dengan 0 < y < x, maka $^4\log(x^2-y^2)=....$

$\dfrac{a+3b}{ab}$
$\dfrac{a+b}{2ab}$
$\dfrac{a+b}{4ab}$
$\dfrac{3a+b}{2ab}$
$\dfrac{3a+b}{4ab}$

Jawab :

$\displaystyle \begin{aligned} ^4\log(x^2-y^2)&=\:{^2}^2\log\left\{(x+y)(x-y)\right\}\\ &=\tfrac{1}{2}\cdot \left\{\;^2\!\log(x+y)+^2\!\log(x-y)\right\}&~~~~~...\:(1)\end{aligned}$

Diketahui juga :

$\displaystyle \begin{aligned} ^{x+y}\log{2}=a\:\Leftrightarrow\: ^2\log(x+y)=\frac{1}{a} &~~~~~...\:(2) \end{aligned}$

dan

$\displaystyle \begin{aligned} ^{x-y}\log{8}=b&\:\Leftrightarrow\: ^8\log(x-y)=\frac{1}{b}\\ &\:\Leftrightarrow\: {^2}^3\log(x+y)=\frac{1}{b}\\ &\:\Leftrightarrow\: \frac{1}{3}\cdot{^2}\log(x-y)=\frac{1}{b}\\ &\:\Leftrightarrow\: ^2\log(x-y)=\frac{3}{b} &~~~~~...\:(3) \end{aligned}$