I'm new here, looking for some help please.
I've been at this question for 4+ hours, not getting anywhere, haha.
$\log_2 (kx) = a$
Question asks to solve for $x$
So far my best try is
$\log_2 x + \log_2 k = a $
$\log_2 x = a / \log_2 k$
~~~ I feel like this step is very wrong, but I tried it anyways ~~~
$2^{\log_2 x} = 2^{a / \log_2 k}$
$x = 2^{a / \log_2 k}$
Any tip / help would be appreciated, thank you!
Your error is is this statement:
log base 2 x + log base 2 k = a
log base 2 x = a / log base 2 k
For some reason you divided both sides by log base 2 k instead of just subtracting it.
You should have:
$$\log_2x = a - \log_2 k$$
$$x = 2^{(a - \log_2k)}$$ $$x = \frac{2^{(a)}}{2^{(\log_2k)}}$$ $$x = \frac{2^a}k$$