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

1 Answer
Oct 27, 2016

Answer:

#x=5.097141119#

#x=5.1#

Explanation:

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#