So here is the problem:
Solved for a in terms of x:
$$a^{x} = 10^{2x + 1}$$
I tried:
$\displaystyle x \cdot \log(a) = (2x+1) \cdot \log\;10 $
$\displaystyle \frac{x}{2x + 1} = \frac{\log\;10} {\log\;a} $
But this is not going in the right direction, the answer according to the book is:
$$ \frac{1} {\log\;a - 2} $$
Excuse the 'power' tag for this question, there is no logarithm tag