If #4^(2n+3) = 8^(n+5)#, what is the value of #n#?

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Answer 1 · 2016-07-31T00:18:46

#n= 9#

Both 4 and 8 are powers of 2.
Write them with 2 as the base.

#4^(2n+3) = 8^(n+5)#

#(2^2)^(2n+3) = (2^3)^(n+5)#

#2^(2(2n+3) )= 2^(3(n+5))#

The bases are equal, so the indices are equal.

#2(2n+3) = 3(n+5)#

#4n +6 = 3n +15#

#n= 9#