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

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Answer 1 · 2015-12-10T17:33:56

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

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

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

Factor without substituting

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

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

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

$x = ln 4$ $x=ln4$ $or$ $" or "$ $x = ln 2$ $x=ln2$

Substitute, then factor

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

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

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

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

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

$x = ln 4$ $x=ln4$ $or$ $" or "$ $x = ln 2$ $x=ln2$

How do you solve e2x−6ex+8=0e^(2x)-6e^x+8=0?