How do you solve: 3^(2-x) = 4^(2x-5) using logs?

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Answer 1 · 2016-02-13T10:05:11

$x=2.1$

As $3^(2-x) = 4^(2x-5)$

$log(3^(2-x))=log(4^(2x-5))$

or $(2-x)log 3=(2x-5)log4$

or $(2-x)/(2x-5)=log4/log3=0.6020-0.4771=0.1249$

Hence $2-x=0.1249(2x-5)=0.2498x-0.6245$

or $1.2498x=2.6245$

or $x=2.6245/1.2498=2.1$