Finding value of x in logarithms? 
Q) Find the value of $x$ in $2 \log x + \log 5 = 2.69897$

So far I got:
$$2 \log x + \log 5 = 2.69897$$
$$\Rightarrow \log x^2 + \log 5 = 2.69897 $$
$$\Rightarrow \log 5x^2 = 2.69897 $$
What should I do next?
Note: In this question $\log(x) \implies \log_{10}(x)$ , it is therefore implied to use $\ln(x)$ to denote natural logarithm

Raise the base of the logarithm to both sides. Then, you get $5 x^2 = b^{2.69897}$ where $b$ is the base of the logarithm (probably $b=10$). Then, solve for $x$ by dividing by $5$ and taking square roots. 
