How do you solve #14^(x+1) = 36#?

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Answer 1 · 2016-04-19T02:07:19

Take the logarithm of each side. A logarithm rule lets us take the exponent outside of the logarithm.

#14^{x+1}=36#

#ln(14^{x+1})=ln(36)#

#(x+1)*ln(14)=ln(36)#

#(x+1)=ln(36)/{ln(14)}#

#x=ln(36)/{ln(14)}-1#

#x\approx 0.357878#

Substitute #x# into the problem to check the answer.