## solving for x given: 4e^-.2x + 3e^-.3x = 5

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denicarl
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Joined: Wed Mar 30, 2016 10:56 pm
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### solving for x given: 4e^-.2x + 3e^-.3x = 5

how do I solve for x given: 4e^-.2x + 3e^-.3x = 5
thanks

theshadow
Posts: 136
Joined: Sun Feb 22, 2009 11:12 pm

### Re: solving for x given: 4e^-.2x + 3e^-.3x = 5

how do I solve for x given: 4e^-.2x + 3e^-.3x = 5
The way you've written it means this:

a. $4e^{-0.2}x + 3e^{-0.3}x = 5$

Did you mean this?

b. $4e^{-0.2x} + 3e^{-0.3x} = 5$

Thanks.

denicarl
Posts: 6
Joined: Wed Mar 30, 2016 10:56 pm
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### Re: solving for x given: 4e^-.2x + 3e^-.3x = 5

b-thanks and the 5 should be .5

FWT
Posts: 153
Joined: Sat Feb 28, 2009 8:53 pm

### Re: solving for x given: 4e^-.2x + 3e^-.3x = 5

b-thanks and the 5 should be .5
So you meant this:

$\mbox{Solve }\, 4e^{-0.2x}\, +\, 3e^{-0.3x}\, =\, 0.5$

Try using the steps, etc, explained in your other thread. The equations are pretty much the same so the methods will be pretty much the same too. If you get stuck, please write back showing what you did when working through the steps. Thanks.

maggiemagnet
Posts: 358
Joined: Mon Dec 08, 2008 12:32 am
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### Re: solving for x given: 4e^-.2x + 3e^-.3x = 5

how do I solve for x given: 4e^-.2x + 3e^-.3x = .5
This one is at least as bad as your other one!

$x\, =\, 10 i \, \bigg(2 \pi n\, +\, \pi\, -\, i \log\left[ \, \dfrac{\left(1\,-\, i\, \sqrt{3\,}\right)\, \sqrt[3]{27\,+\,i\, \sqrt{807\,}\,}}{2\, \times\, \sqrt[3]{9\,}}\, +\, \dfrac{4\, \left(1\,+\,i\, \sqrt{3\,}\right)}{\sqrt[3]{3\, \left(27\,+\,i\, \sqrt{807\,}\right)\,}}\right] \bigg)\, \mbox{ for }\, n\, \in\, \mathbb{Z}$

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