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

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Answer 1 · 2016-07-15T17:30:11

$x = ( ln 25)/ (ln (5/2))$

take logs on each side

$ln 2^x = ln 5^(x - 2)$

then because $ln a^b = b ln a$

$x ln 2 = (x-2) ln 5$

$x (ln 5 - ln 2) = 2 ln 5$

$x = (2 ln 5)/ (ln 5 - ln 2)$

or if you like

$x = ( ln 25)/ (ln (5/2))$

because

$ln 5 - ln 2 = ln (5/2)$

and $2 ln 5 = ln 5^2 = ln 25$

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