Question

The solution set of $\frac{1}{2}{\log }_3(x+1)-{\log }_9(1-x)={\log }_9(2x+3)$ is

• A. $\left\{\frac{1}{2}\right(\sqrt{5}-1\left)\right\}$
• B. $\left\{\frac{1}{2}\right(\sqrt{5}+1\left)\right\}$
• C. $\{\frac{1}{2},\frac{1}{3}\}$
• D. none of these

Answer 1

The correct option is A $\left\{\frac{1}{2}\right(\sqrt{5}-1\left)\right\}$

$\frac{1}{2}{\log }_3(x+1)-{\log }_9(1-x)={\log }_9(2x+3)$
Above equation is valid when $x+1>0,1-x>0,2x+3>0$
$\Rightarrow -1<x<1$
$\frac{1}{2}{\log }_3(x+1)-{\log }_9(1-x)={\log }_9(2x+3)$
$\Rightarrow {\log }_3(x+1)-2{\log }_9(1-x)=2{\log }_9(2x+3)$
$\Rightarrow {\log }_3(x+1)-{\log }_3(1-x)={\log }_3(2x+3)[\because m{\log }_{a}b={\log }_{a^{1/m}b}]$
$\Rightarrow {\log }_3\left(\frac{x+1}{1-x}\right)={\log }_3(2x+3)[\because \log a-\log b=\log \frac{a}{b}]$
$\Rightarrow \frac{x+1}{1-x}=2x+3$
$\Rightarrow x^2+x-1=0$
$\Rightarrow x=\frac{-1\pm \sqrt{5}}{2}$
Since, $-1<x<1$
Therefore, $x=\frac{\sqrt{5}-1}{2}$
Ans: A