How do you solve $e^(2x)-6e^x+8=0$ ?

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Answer 1 · 2015-12-10T17:33:56

Use $e^(2x) = (e^x)^2$ to see that this equation is quadratic in $e^x$ .

$e^(2x)-6e^x+8=0$

$(e^x)^2-6e^x+8=0$

Factor without substituting

$(e^x-4)(e^x - 2 ) = 0$

So $e^x-4=0$ $" or "$ $e^x-2=0$

$e^x=4$ $" or "$ $e^x =2$

$x=ln4$ $" or "$ $x=ln2$

Substitute, then factor

Let $u = e^x$ , then we want

$u^2-6u+8=0$

$(u-4)(u-2) = 0$

$u=4$ $" or "$ $u=2$ Recall $u=e^x$ , so

$e^x=4$ $" or "$ $e^x =2$

$x=ln4$ $" or "$ $x=ln2$

How do you solve e^(2x)-6e^x+8=0e^(2x)-6e^x+8=0?