Question 877294
Think of this as a "quadratic"...
{{{e^(2x) - 2e^x +1=0}}}
Let y = e^x
then
y^2 = e^(2x)
.
Now, we can rewrite the equation as:
{{{y^2 - 2y +1=0}}}
{{{(y-1)(y-1)=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)