Question

The sum of value(s) of $x$ satisfying $2-x+3{\log }_{5}2={\log }_5(3^x-5^{2-x})$ is

Answer 1

$2-x+3{\log }_{5}2={\log }_5(3^x-5^{2-x})$
For $\log$ to be defined, we get
$3^x-5^{2-x}>0\Rightarrow \frac{{15}^x-25}{5^x}>0\Rightarrow {15}^x>25\cdots \left(1\right)(\because 5^x>0)$

Now,
$2-x+3{\log }_{5}2={\log }_5(3^x-5^{2-x})\Rightarrow {\log }_5\left(5^{2-x}\right)+{\log }_{5}8={\log }_5(3^x-5^{2-x})\Rightarrow 8\times 5^{2-x}=3^x-5^{2-x}\Rightarrow 9\times 5^{2-x}=3^x\Rightarrow 5^{2-x}=3^{x-2}\Rightarrow 1={15}^{x-2}\Rightarrow x=2$

Verying equation $\left(1\right)$, we get
${15}^2=225>25$

Therefore, the required value of $x$ is $2$.