Problem: 2^2x=k
Maybe someone can tell me what to look for. I am going to fiddle with it a little more.
Either
$\displaystyle \displaystyle \begin{align*} 2^{2x} &= k \\ 2x &= \log_2{k} \\ x &= \frac{1}{2}\log_2{k} \end{align*}$
or
$\displaystyle \displaystyle \begin{align*} 2^{2x} &= k \\ \ln{\left(2^{2x}\right)} &= \ln{k} \\ 2x\ln{2} &= \ln{k} \\ 2x &= \frac{\ln{k}}{\ln{2}} \\ x &= \frac{\ln{k}}{2\ln{2}} \end{align*}$
Start by taking the natural log of both sides of the equation.
Let ln = natural log
2^(2x) = k
ln[2^(2x)] = ln(k)
When we take the natural log of both sides, exponents come down infront of the ln symbol.
So, 2x comes down and it now looks like this:
2xln(2) = ln(k)
At this point, solve the equation for x, dividing both sides by 2(ln2).
x = ln(k)/2(ln2)