How do you solve this logarithmic equation?

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1 Answer
Cesareo R.
Aug 15, 2017

See below.

Explanation:

#e^x+e^(2x)-e^(3x)=0# or

#e^x(1+e^x-e^(2x))=0# but #e^x ne 0 forall x in RR# and #e^(2x)=(e^x)^2# so

#1+e^x-(e^x)^2=0# and now solving for #e^x# we get

#e^x = 1/2(1pm sqrt5)-> e^x = 1/2(1+sqrt(5))# and finally

#x = log_e( 1/2(1+sqrt(5))) approx 0.481212#

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