How do you solve #e^(2x) = 2e^x + 1 = 0#?

1 Answer
Daniel L.
May 19, 2018

See explanation.

Explanation:

If you substitute the expression #e^x# with a new variable you get a quadratic equation:

#e^(2x)-2e^x+1=0#

#t^2-2t+1=0#

#(t-1)^2=0#

#t=1#

Now we have to return to #x# by solving:

#e^x=1=>e^x=e^0=>x=0#

Answer: The equation has one solution #x=0#.

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