Question

Solution set of equation: $\log x=\frac{1}{2}\log (x+1)$ 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. $\left\{\frac{1}{2}\right(\sqrt{5}+\sqrt{7}\left)\right\}$

Answer 1

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

$\log x=\frac{1}{2}\log (x+1)$
Above equation is valid when $x>0$ and $x+1>0$
$\Rightarrow x>0$
Now given equation is,
$2\log x=\log (x+1)$
$\Rightarrow \log x^2=\log (x+1),[\because m\log a=\log a^m]$
$\Rightarrow x^2=x+1$
$\Rightarrow x=\frac{1\pm \sqrt{5}}{2}$
Since, $x>0$
Therefore, $x=\frac{1+\sqrt{5}}{2}$
Ans: B