How do you solve #2^(x+2) = 8#?

1 Answer
Jan 26, 2016

#x=1#

Explanation:

Given equation is #2^(x+2)=8#

I'm sure you're familiar with the law of exponents and that #a^(m+n)=a^m*a^n#

So, in the main equation, we get #2^(x+2)=2^x2^2=4*2^x#
So, #cancel{4}^1*2^x=cancel{8}^2#
So, #2^x=2#

I'm sure it's obvious from here.