1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Right from the first step, we have $2^a = kx$; this is by definition of the logarithm. $\endgroup$ – E W H Lee Sep 20 '14 at 4:52
  • $\begingroup$ OMG I am so STUPID. WOOOOOW Thank you!! I'm too tired to be doing math right now, LOL. $\endgroup$ – MarsIsPrettyClose Sep 20 '14 at 4:54
1
$\begingroup$

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:

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.