How do you solve #27^(2x+1) = 9^(4x)#?

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Answer 1 · 2016-06-10T23:47:24

#x = 3/2#

#27=3^3# and #9 = 3^2#

then

#27^(2x+1) = 9^(4x) equiv (3^3)^{2x+1} = (3^2)^{4x}#

or

#3^{3(2x+1)} = 3^{8x}#

so

#3(2x+1)=8x# Solving for #x# gives #x = 3/2#