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

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Answer 1 · 2016-02-13T10:05:11

$x = 2.1$ $x=2.1$

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

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

or $(2 - x) log 3 = (2 x - 5) log 4$ $(2-x)log 3=(2x-5)log4$

or $\frac{2 - x}{2 x - 5} = \frac{log 4}{log 3} = 0.6020 - 0.4771 = 0.1249$ $(2-x)/(2x-5)=log4/log3=0.6020-0.4771=0.1249$

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

or $1.2498 x = 2.6245$ $1.2498x=2.6245$

or $x = \frac{2.6245}{1.2498} = 2.1$ $x=2.6245/1.2498=2.1$