How do you solve $e^{2 x} - 5 e^{x} + 6 = 0$ $e^(2x)-5e^x+6=0$ ?

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Answer 1 · 2016-04-21T17:53:50

$x = ln 3 or x = ln 2$ $x=ln3 or x=ln2$

Let $u = e^{x}$ $u=e^x$

Then we have

$u^{2} - 5 u + 6 = 0$ $u^2-5u+6=0$

$(u - 3) (u - 2) = 0$ $(u-3)(u-2)=0$

$u - 3 = 0 or u - 2 = 0$ $u-3=0 or u-2=0$

$e^{x} - 3 = 0 or e^{x} - 2 = 0$ $e^x-3=0 or e^x-2=0$

$e^{x} = 3 or e^{x} = 2$ $e^x=3 or e^x=2$

${ln e}^{x} = ln 3 or {ln e}^{x} = ln 2$ $lne^x=ln3 or lne^x = ln 2$

$x ln e = ln 3 or x ln e = ln 2$ $x ln e=ln3 or xlne=ln2$

$x = ln 3 or x = ln 2$ $x=ln3 or x=ln2$

How do you solve e^(2x)-5e^x+6=0e2x−5ex+6=0e^(2x)-5e^x+6=0?