How do you solve #4^(x+4) = 8^x#?

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How do you solve #4^(x+4) = 8^x#?

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Answer 1 · 2015-07-23T18:34:36

I found: #x=8#

Explanation:

You can write it as:
#(2^2)^(x+4)=(2^3)^x#
#2^(2(x+4))=2^(3x)# take the #log_2# on both sides to get rid of the #2# (where you have #cancel(log_2)(cancel(2)^x)=x#) and get:
#2(x+4)=3x#
#2x+8=3x#
#x=8#

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How do you solve #4^(x+4) = 8^x#?

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