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

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Answer 1 · 2016-04-11T00:02:21

Since $a = b -> loga = logb$

$log2^x = log5^(x - 2)$

Use the rule $loga^n = nloga$

$xlog2 = (x - 2)log5$

$xlog2 = xlog5 - 2log5$

$xlog2 - xlog5 = -2log5$

$x(log(2/5))= log(1/25)$

$x = (log(1/25))/(log(2/5))$

That's your answer.

Hopefully that helps!

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