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

1 Answer

Oct 21, 2015

#x=12#

Explanation:

#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#

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