Question

The value of $x$ in $\frac{{\log }_{10}(x-3)}{{\log }_{10}(x^2-21)}=\frac{1}{2}$ is

Answer 1

$\frac{{\log }_{10}(x-3)}{{\log }_{10}(x^2-21)}$ is valid, when $x>3,x^2-21>0,x^2-21\ne 1$
$\Rightarrow x\in (\sqrt{21},\infty )-\left\{\sqrt{22}\right\}$
Now, $\frac{{\log }_{10}(x-3)}{{\log }_{10}(x^2-21)}=\frac{1}{2}$
$\Rightarrow 2{\log }_{10}(x-3)={\log }_{10}(x^2-21)$
$\Rightarrow {(x-3)}^2=x^2-21$
$\Rightarrow x=5$
Ans: 5