How do you solve $e^-x = 0.04$ ?

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Answer 1 · 2015-10-23T19:01:18

By taking natural logs of both sides:

$e^(-x)=0.04$

Taking natural logs $rArr$

$lne^(-x)=ln(0.04)$

$-xcancel(lne)=-3.219$

$x=3.219$

How do you solve e^-x = 0.04e^-x = 0.04?