What is the quadratic formula of $e^(2x) - 2e^x = 1$ ?

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Answer 1 · 2015-10-31T21:33:40

Recognise this as quadratic in $e^x$ and hence solve using the quadratic formula to find:

$x = ln(1+sqrt(2))$

This is an equation that is quadratic in $e^x$ , rewritable as:

$(e^x)^2-2(e^x)-1 = 0$

If we substitute $t = e^x$ , we get:

$t^2-2t-1 = 0$

which is in the form $at^2+bt+c = 0$ , with $a=1$ , $b=-2$ and $c=-1$ .

This has roots given by the quadratic formula:

$t = (-b+-sqrt(b^2-4ac))/(2a) = (2+-sqrt(4+4))/2 = 1+-sqrt(2)$

Now $1-sqrt(2) < 0$ is not a possible value of $e^x$ for Real values of $x$ .

So $e^x = 1 + sqrt(2)$ and $x = ln(1+sqrt(2))$

What is the quadratic formula of e^(2x) - 2e^x = 1e^(2x) - 2e^x = 1?