e^(2x)-7e^x+12=0 had two solutions

i got this (7(plus minus)sqrt(7^2-4*e*12))/2e now i replaced e with 2.718 because that is the value of e and I got 1.591 and 0.984. Is this hat I what's suppose to do?

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- Oct 15th 2012, 07:47 AMRD33Equation with 2 solutions
e^(2x)-7e^x+12=0 had two solutions

i got this (7(plus minus)sqrt(7^2-4*e*12))/2e now i replaced e with 2.718 because that is the value of e and I got 1.591 and 0.984. Is this hat I what's suppose to do? - Oct 15th 2012, 07:54 AMMaxJasperRe: Equation with 2 solutions
Roots are:

x1=ln(3)

x2=ln(4) - Oct 15th 2012, 07:59 AMRD33Re: Equation with 2 solutions
- Oct 15th 2012, 08:02 AMMaxJasperRe: Equation with 2 solutions
Let u=e^x then

u^2-7u+12=0

solve for u then plug into u=e^x and find x.

http://mathhelpforum.com/attachment....1&d=1350317240 - Oct 15th 2012, 09:10 AMRD33Re: Equation with 2 solutions
This is what I got when I solved for u:

U= (loge^x(-12)-2)/-7

The log is base e^x

Am I right? - Oct 15th 2012, 10:31 AMSorobanRe: Equation with 2 solutions
Hello, RD33!

Quote:

$\displaystyle \text{Solve for }x\!:\;\;e^{2x}-7e^x+12\:=\:0$

Factor: .$\displaystyle (e^x - 3)(e^x - 4) \:=\:0$

Then: .$\displaystyle \begin{Bmatrix}e^x-3 \:=\:0 & \Rightarrow & e^x \:=\:3 & \Rightarrow & x \:=\:\ln 3 \\ e^x-4 \:=\:0 & \Rightarrow & e^x \:=\:4 & \Rightarrow & x \:=\:\ln4 \end{Bmatrix}$