Question

Find the value of $x$ satisfying the equation: ${\log }_5{({2x}^2+3x+2)}^{\frac{1}{4}}={\log }_{25}\left(2x\right)$

Answer 1

$\log \left(2x\right)$ is valid when $x>0$
${\log }_5{({2x}^2+3x+2)}^{\frac{1}{4}}={\log }_{25}\left(2x\right)$

$\Rightarrow \frac{1}{4}{\log }_5({2x}^2+3x+2)=\frac{1}{2}{\log }_5\left(2x\right)[\because \log a^m=m\log a\text{\&}{\log }_{a^m}b=\frac{1}{m}{\log }_{a}b]$

$\Rightarrow {\log }_5({2x}^2+3x+2)={\log }_5(2x{)}^2$

$\Rightarrow {2x}^2+3x+2={\left(2x\right)}^2$

$\Rightarrow -{2x}^2+3x+2=0\Rightarrow x=2asx>0$
Ans: 2