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

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Answer 1 · 2016-02-26T18:50:32

$x = 2 + ln 5 = 3.6094$ $x=2+ln5 = 3.6094$

Take the natural logarithm on both sides and use the laws of logs to obtain

$ln (e^{x - 2}) = ln 5$ $ln(e^(x-2))=ln5$

$therefore (x-2)lne=ln5$

$therefore x=2+ln5$

$=3.6094$

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