Answer
$log_{10}(10x^{2}+20)$
Work Step by Step
We know that $log_{b}xy=log_{b}x+log_{b}y$ (where $x$, $y$, and $b$ are positive real numbers and $b\ne1$).
Therefore, $log_{10}5+log_{10}2+log_{10}(x^{2}+2)=log_{10}(5\times2\times(x^{2}+2))=log_{10}(10\times(x^{2}+2))=log_{10}(10x^{2}+20)$.
