How do you solve #4e^(x - 1) = 64#?

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Answer 1 · 2016-05-18T18:13:16

We have that

#4*e^(x-1)=64#

#(4*e^(x-1))/4=64/4# (Divide by 4 both sides)

#e^(x-1)=16#

#(x-1)*lne=ln16# (Take logarithms in both sides)

#(x-1)=4*ln2# (Remember that #16=2^4#)

#x=1+4*ln2#