SOLUTION: solve: {{{e^(2x) - 2e^x +1=0}}}

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Question 877294: solve:
e%5E%282x%29+-+2e%5Ex+%2B1=0
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Think of this as a "quadratic"...
e%5E%282x%29+-+2e%5Ex+%2B1=0
Let y = e^x
then
y^2 = e^(2x)
.
Now, we can rewrite the equation as:
y%5E2+-+2y+%2B1=0
%28y-1%29%28y-1%29=0
y = {1}
.
Now, we still need to find x:
y = e^x
1 = e^x
ln(1) = x
ln(e^0) = x
0 = x (answer)