Question

If $2\log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}$ is $\log m$, then what is the value of $m$?

• A. $2$
• B. $4$
• C. $-1$
• D. $0$

Answer 1

The correct option is A $2$

Given, $2\log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}=\log m$
$\Rightarrow \log (\frac{15}{18}{)}^2-\log \frac{25}{162}+\log \frac{4}{9}=\log m$ [$\because 2\log a=\log a^2$]
$\Rightarrow \log \frac{225}{324}-\log \frac{25}{162}+\log \frac{4}{9}=\log m$
$\Rightarrow \log \frac{\frac{225}{324}}{\frac{25}{162}}+\log \frac{4}{9}=\log m$ [$\because \log a-\log b=\log \frac{a}{b}$]
$\Rightarrow \log \frac{9}{2}+\log \frac{4}{9}=\log m$
$\Rightarrow \log \frac{9}{2}\times \frac{4}{9}=\log m$ [$\because \log a+\log b=\log ab$]
$\Rightarrow \log 2=\log m$
$\Rightarrow m=2$
Option A is correct.