# Simplifying / Solving for $x$

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!

• Right from the first step, we have $2^a = kx$; this is by definition of the logarithm. – E W H Lee Sep 20 '14 at 4:52
• OMG I am so STUPID. WOOOOOW Thank you!! I'm too tired to be doing math right now, LOL. – MarsIsPrettyClose Sep 20 '14 at 4:54

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:

$$\log_2x = a - \log_2 k$$

$$x = 2^{(a - \log_2k)}$$ $$x = \frac{2^{(a)}}{2^{(\log_2k)}}$$ $$x = \frac{2^a}k$$

• Okay, so it does work, I just messed up on that subtraction part. Thank you so much! – MarsIsPrettyClose Sep 20 '14 at 4:59