Question

Solve for $x$, ${\log }_{6}9-{\log }_{9}27+{\log }_{8}x={\log }_{64}x-{\log }_{6}4$

• A. $\frac{7}{8}$
• B. $\frac{11}{8}$
• C. $\frac{1}{8}$
• D. $\frac{5}{8}$

Answer 1

The correct option is C $\frac{1}{8}$

${\log }_{6}9-{\log }_{9}27+{\log }_{8}x={\log }_{64}x-{\log }_{6}4$
${\log }_{8}x-{\log }_{64}x={\log }_{9}27-{\log }_{6}4-{\log }_{6}9$
${\log }_{8}x-{\log }_{8^2}x={\log }_{3^2}27-{\log }_{6}36$
${\log }_{8}x-\frac{1}{2}{\log }_{8}x=\frac{1}{2}{\log }_{3}27-2$
$\Rightarrow \frac{1}{2}{\log }_{8}x=\frac{3}{2}-2$
$\Rightarrow {\log }_{8}x=-1$
$\Rightarrow 8^{-1}=x$
or $x=\frac{1}{8}$
Given eqn is defined for $x>0$
Hence, $x=\frac{1}{8}$is solution of given equation