How do you solve $e^{2 x} - (4 e^{x}) + 3 = 0$ $e^(2x)-(4e^x)+3=0$ ?

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Answer 1 · 2018-06-20T09:37:31

$x = 0 and x = ln (3)$ $x=0 and x=ln(3)$

Let $u = e^{x}$ $u=e^x$ . Then the equation will become $u^{2} - 4 u + 3 = 0$ $u^2-4u + 3=0$ .

This equation can be solved by factoring

$(u - 3) (u - 1) = 0$ $(u -3)(u-1) =0$
#u = 3 or 1

Now we see that

$e^{x} = 3 and e^{x} = 1$ $e^x = 3 and e^x = 1$
$x = ln 3 and x = 0$ $x = ln3 and x= 0$

Hopefully this helps!

Answer 2 · 2018-06-20T09:41:08

$x = 0 or ln 3 \approx 1.09861$ $x = 0 or ln3 approx 1.09861$

$e^{2 x} - 4 e^{x} + 3 = 0$ $e^(2x)-4e^x+3=0$

Let $ϕ = e^{x} \to x = ln ϕ$ $phi = e^x -> x=lnphi$

$:. phi^2 - 4phi +3=0$

$(phi-3)(phi-1)=0$

Hence, $phi = 1 or 3$

$phi =1 -> x=ln1 =0$

$phi =3 -> x= ln3 approx 1.09861$

How do you solve e^(2x)-(4e^x)+3=0e2x−(4ex)+3=0e^(2x)-(4e^x)+3=0?