Question

If $2\log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}=\log x$, then the value of $x$ will be

• A. $0$
• B. $2$
• C. $1$
• D. $3$

Answer 1

Answer: (B)

$2$

$2\log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}=\log x$ ...(1)

$2\log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}=\log (\frac{15}{18}\times \frac{15}{18})-\log \frac{25}{162}+\log \frac{4}{9}=\log \left(\frac{15\times 15\times 4\times 162}{18\times 18\times 9\times 25}\right)=\log \left(\frac{4\times 162}{18\times 18}\right)=\log 2$...(2)

Equating (1) and (2),

$x=2$