1
$\begingroup$

I've been trying to solve this problem for a while now and can't seem to figure it out.

If $3x - y = 12$, what is the value of $\frac{8^x}{2^y}$?

The answer should be $2^{12}$ but I'm not sure how to get there. Here's what I begin with.

$$y= 3x - 12$$ $$\frac{8^x}{2^{3x-12}}$$

And then I get stuck. I don't know how to simplify the equation when both the base and the exponents are different. Any help would be much appreciated.

$\endgroup$
2
$\begingroup$

Always try to get your exponentials with the same base. Note that we have

$$8=2^3$$

Which means

$$\frac{8^x}{2^{3x-12}}=\frac{2^{3x}}{2^{3x-12}}$$

Can you take it from here?

$\endgroup$
2
$\begingroup$

$8=2^3$ S0, the numerator is $2^{3x}$

$2^{3x} \over 2^{3x-12}$=$2^{12}$

$\endgroup$

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.