How do you solve #7*e^(x-3)=57#?

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Answer 1 · 2016-10-27T02:13:26

#x=5.097141119#

#x=5.1#

We start by rearranging to have #e^(x-3)# by its self.

#7*e^(x-3)=57#

#e^(x-3)=57/7#

the inverse function of e^x is ln(x) thus,

#ln(e^(x-3))=x-3#

so,

#x-3=ln(57/7)#

#x=ln(57/7)+3#

#x=ln(57)-ln(7)+3#

#x=5.097141119#

#x=5.1#

so to check,

#7*e^(5.097141119-3)#

#=57#