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

Oh.
$$2 \log_b x + \log_b 5 = 2.69897=a$$
$$\log_b x=(a-\log_b 5)/2$$
$$x=b^{(a-\log_b 5)/2}$$
And after about a half and a hour you may have $$x=\sqrt{b^{a-\log 5}}=\sqrt{b^a/5}$$ and so on
