How do you solve #2^(2x+5) - 7(2^x)=0# ?

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I know this question is to do with logs, but not sure how to answer the question.

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Answer 1 · 2016-08-16T12:15:20

#x=log (7/32)/log 2=-2.193#, nearly..

Rearranging,

#2#(2x+5)/2^x=2#(2x+5-x)=2^x2^5=32( 2^)x7#>

So, #2^x=7/32#.

Equating logarithms and solving for x,

#x =log(7/32)/log 2#