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

1 Answer

Jan 16, 2017

#x~=-3.76#

Explanation:

#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#

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