How do you solve for x?: $log_2(2^(x+3))=15$ ?

Question

1702 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-21T07:19:38

$x=12$

$log_2(2^(x+3))=15=>$ if $log_a(x)=yhArrx=a^y$ , so:

$2^(x+3)=2^15=>$ when the same bases to different exponents are

equal then the exponents must be equal, then:

$x+3=15=>$ subtract 3 from both sides:

$x=12$

How do you solve for x?: log_2(2^(x+3))=15log_2(2^(x+3))=15?