Problem with logarithms Problem:
Solve:
$$\frac{1}{2^x} = \frac{5}{8^{x+2}}$$
My attempt:
$$\frac{1}{2^x} = \frac{5}{8^{x+2}}$$
$$\Rightarrow 5 \cdot 2^x = 8^{x+2}$$
$$\Rightarrow 2^{\log_2 5+x} = 8^{x+2} $$
$$\Rightarrow (\log_2 5 + x)(\log_a 2) = (x+2)(\log_a 8)$$
And then just keep going like this, but I'm obviously wrong already as the answer is:
$$ x = \frac{\ln 5 - 9 \ln 2}{2 \ln 2}$$
What am I doing wrong?
EDIT: If someone could enlarge the latex for me that would be great.

$2^{\log_2 5+x} = 8^{x+2}$ implies that $2^{\log_2 5+x} = 2^{3x+6}$ and so $\log_2 5+x=3x+6$. We can conclude that $x=\frac{\log_2 5-6}{2}$.
