Question

The number of solution(s) of the equation $\frac{1}{2}{\log }_{\sqrt{3}}\left(\frac{x+1}{x+5}\right)+{\log }_9{(x+5)}^2=1$ is/are

• A. $0$
• B. $1$
• C. $2$
• D. infinite

Answer 1

Answer: (B)

$1$

Given equation

$\frac{1}{2}{\log }_{\sqrt{3}}\left(\frac{x+1}{x+5}\right)+{\log }_9{(x+5)}^2=1$

$\frac{1}{2}\left(\frac{1}{1/2}\right){\log }_3\left(\frac{x+1}{x+5}\right)+\frac{1}{2}{\log }_3{(x+5)}^2={\log }_{3}3(\because {\log }_{a^n}b=\frac{1}{n}{\log }_{a}b)$

$\Rightarrow \frac{2}{2}{\log }_3\left(\frac{x+1}{x+5}\right)+\frac{1}{2}.2{\log }_3{(x+5)}^{}={\log }_{3}3$

$\Rightarrow {\log }_3\{\frac{x+1}{x+5}.(x+5)\}={\log }_{3}3(\because \log m+\log n=\log mn;{\log }_{n}n)$

$\Rightarrow$ $(x+1)=3$

$\Rightarrow x=2$

So, only one solution is possible.