How do you solve $2^x = 5^(x - 2)$ ?

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Answer 1 · 2016-03-04T06:30:08

$x=3.513$

To solve $2^x=5^(x−2)$ , take logs of both the sides, this becomes

$xlog2=(x-2)log5=xlog5-2log5$ and transposing terms this becomes

$xlog5-xlog2=2log5$ or

$x(log5-log2)=2log5$

$x=(2log5)/(log5-log2)=(2xx0.699)/(0.699-0.301)$ or

$x=1.398/0.398=3.513$

How do you solve 2^x = 5^(x - 2) 2^x = 5^(x - 2) ?