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.

Some of the answers don't mention that $8=2^3$.
$$
5\cdot 2^x = 8^{x+2}
$$
Remember that $8=2^3$.  So
$$
5\cdot 2^x = (2^3)^{x+2}.
$$
$$
5\cdot 2^x = 2^{3(x+2)}
$$
$$
5\cdot 2^x = 2^{3x+6}
$$
$$
5 = 2^{3x+6}\cdot 2^{-x} = 2^{2x+6}
$$
$$
\log_2 5 = 2x+6
$$
$$
-6 + \log_2 5 = 2x
$$
$$
\frac{-6+\log_2 5}{2} = x
$$
