8^(y - 2) = 2^y
a.y = 1
b.y = 3
c.y = -3
d.y = -1

Question

8^(y - 2) = 2^y
 a.y = 1 
 b.y = 3 
 c.y = -3 
 d.y = -1

larseighner · Answer

Can you convert $ 8^{y-2}$ to a power of 2?  That is $ \large2^?$ ?

anonymous · Answer

im lost on what to do

larseighner · Answer

Okay.  Let me make it bigger

$$ \Large8^{y-2} $$

is on the left.

$$ \Large 2^y $$

is on the right.

You can solve this problem if you can express both terms as the same base number to some power.

So how can you convert

$$ \Large8^{y-2} $$ to

$$ \Large2^{something} $$

anonymous · Answer

do you divide it by 4?

larseighner · Answer

Here's a hint

$$ \Large 8 = 2^3 $$

anonymous · Answer

so the answer would be 3?

larseighner · Answer

$$\Large8^{y-2} $$

is what you are trying to convert.

I just told you 

$$\Large 8 = 2^3$$

So what can you do?