# How do you solve log4^(2x) = y + 7?

Sep 5, 2015

It turns out this equation is equivalent to $y = 2 \log \left(4\right) x - 7$.

#### Explanation:

the core of this problem is that for any real positive real number $a$ and any real number $b$ we have $\log \left({a}^{b}\right) = b \log \left(a\right)$. Therefore we van rewrite:
$\log \left({4}^{2 x}\right) = y + 7$
as
$2 x \log \left(4\right) = y + 7$
A quick rearrangement gives
$y = 2 \log \left(4\right) x - 7$.

If you want to, you could rewrite $2 \log \left(4\right)$ to $\log \left({4}^{2}\right) = \log \left(16\right)$, or since $4 = {2}^{2}$, $2 \log \left(4\right) = 4 \log \left(2\right)$, but since they are all the same number and you can't write out $\log \left(4\right)$ in decimal notation without rounding off anyway, how you write it down doesn't really matter.