Solution
= Solution
Since $1-\log_{10}2=\log_{10}5\ne0$, the logarithm equation is $\log_{10}x=2\log_{10}5=\log_{10}25$. The real logarithm requires $x>0$, and is injective there. Hence
$$
\boxed{x=25.}
$$
= Solution
Since $1-\log_{10}2=\log_{10}5\ne0$, the logarithm equation is $\log_{10}x=2\log_{10}5=\log_{10}25$. The real logarithm requires $x>0$, and is injective there. Hence
$$
\boxed{x=25.}
$$