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July 23, 2014

July 23, 2014

Posted by **Kyle** on Friday, January 11, 2013 at 1:28am.

5*e^(-10x) - 3*e^(-20x) = 2

I'm not sure if I can take natural log of both sides to bring down the exponents.

- Calculus -
**Reiny**, Friday, January 11, 2013 at 8:16amI will assume you want to "solve" for x

let e^(-10x) = y , then e^(-20x) = (e^(-10x)^2 = y^2

then you have

5y - 3y^2 = 2

3y^2 - 5y + 2 = 0

(y-1)(3y-2) = 0

y = 1 or y = 2/3

so e^(-10x) = 1 or e^(-10x) = 2/3

if e^(-10x) = 1

ln e^(-10x) = ln 1 = 0

-10x lne = 0

-10x = 0

x = 0

if e^(-10x) = 2/3

...

-10x = ln2 - ln3

x = -(l2 - ln3)/10 = appr .04055

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