Solve for $x$ in $3^x=5^2$ by logarithm. Was solving using properties of logarithm but got stuck at the equation $x\log 3=\log5+\log5$
Hint: $\log 3$ and $\log 5$ in your equation are just constants. How would you solve for $x$ if they were replaced by say, $1$ and $2$?
$$3^x=5^2\implies \log3^x=\log5^2\implies x\log3=2\log5\implies x=\frac{2\log5}{\log3}$$
I hope that helps :)