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

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Answer 1 · 2016-04-11T00:02:21

Since $a = b \to log a = log b$ $a = b -> loga = logb$

${log 2}^{x} = {log 5}^{x - 2}$ $log2^x = log5^(x - 2)$

Use the rule ${log a}^{n} = n log a$ $loga^n = nloga$

$x log 2 = (x - 2) log 5$ $xlog2 = (x - 2)log5$

$x log 2 = x log 5 - 2 log 5$ $xlog2 = xlog5 - 2log5$

$x log 2 - x log 5 = - 2 log 5$ $xlog2 - xlog5 = -2log5$

$x (log (\frac{2}{5})) = log (\frac{1}{25})$ $x(log(2/5))= log(1/25)$

$x = \frac{log (\frac{1}{25})}{log (\frac{2}{5})}$ $x = (log(1/25))/(log(2/5))$

That's your answer.

Hopefully that helps!

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