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.

With a slight modification, we have
$$5\cdot 2^x=8^{x+2}=(2^3)^{x+2}=2^{3x+6}$$
which means that
$$5=2^{2x+6}\\
\log_2 5=2x+6\\
x=\frac{\log_2 (5)-6}2$$
Transforming to the natural logarithm would look like
$$5=2^{2x+6}\\
\ln 5=(2x+6)\ln 2\\
x=\frac{\ln 5-6\ln 2}{2\ln 2}$$
