How do you solve $15e^-x=645$ ?

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Answer 1 · 2017-01-16T17:08:30

$x~=-3.76$

$e^-x=645/15=43$

We take logs of both sides:

$loge^-x=log43~=3.76$

But $loge^-x$ , by definition, is the power to which we raise the base $e$ to get $e^-x$ , so here $-x~=3.76$ .

And $x~=-3.76$

How do you solve 15e^-x=64515e^-x=645?