## solving for x, given exponentials: 4e^2x + 5e^3x = 15

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denicarl
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### solving for x, given exponentials: 4e^2x + 5e^3x = 15

if I have 4e^2x + 5e^3x = 15
how do I solve for x?
I tried ln 4 +2x+ln5+3x=ln 15 but that doesn't work. Thanks

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

### Re: solving for x, given exponentials: 4e^2x + 5e^3x = 15

if I have 4e^2x + 5e^3x = 15
how do I solve for x?
What you wrote means this:

a. $4e^2x + 5e^3 x = 15$

Is that right? Or did you mean this?

b. $4e^{2x} + 5e^{3x} = 15$
I tried ln 4 +2x+ln5+3x=ln 15 but that doesn't work.
How did you get to this step?

Thanks.

denicarl
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### Re: solving for x, given exponentials: 4e^2x + 5e^3x = 15

I meant b and in order to solve say e^(2x) = Y I would take log of both sides so-
if I have a sum of exponentials on one side can I use same methodology-the actual numbers in the problem I was trying to solve are probabilities

e^(-.2x) - e^(-.12x) = .5 how do I solve for x

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

### Re: solving for x, given exponentials: 4e^2x + 5e^3x = 15

I meant b
So you meant this:

$\mbox{Solve }\, 4e^{2x}\, +\, 5e^{3x}\, =\, 15\, \mbox{ for the value of }\, x.$
and in order to solve say e^(2x) = Y I would take log of both sides so-
Taking logs of both sides would give you this:

$\ln\left(\,4e^{2x}\, +\, 5e^{3x}\,\right) =\, \ln(15)$

You can't take a log apart in the way you did. This won't work. (You can learn about log rules here.)

You might want to try using exponent rules (lesson):

$a^{bc}\, =\, (a^b)\, (a^c)$

$4e^{2x}\, +\, 5e^{3x}\, =\, (4)(e^2)(e^x)\, +\, (5)(e^3)(e^x)\, =\, (4e^2\, +\, 5e^3)\,e^x\,=\, 15$

The parenthetical is just a number. Divide through. Once you have e^x equal to just a number, you can take logs. (To learn how to solve exponential equations, try this lesson.)

denicarl
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### Re: solving for x, given exponentials: 4e^2x + 5e^3x = 15

(a^b)(a^c)=(a^(b+c)) so above factoring won't work

maggiemagnet
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### Re: solving for x, given exponentials: 4e^2x + 5e^3x = 15

(a^b)(a^c)=(a^(b+c)) so above factoring won't work
You're right! I think maybe "FWT" hadn't had his/her caffeine yet when s/he posted that!

According to Wolfram's "Alpha" solver, this doesn't solve simply. They give this:

$x\, =\, i \, \left(2 \pi n\, -\, i \log\bigg[\dfrac{1}{10125}\, \left(\sqrt[3]{ \dfrac{3074546109375}{2}\, -\, \dfrac{13839609375\, \sqrt{49345}}{2} \,}\, \right. \right. \\ \\ \left. \left. \mbox{ }\, +\, 675 \sqrt[3]{ \dfrac{1}{2}\, \left\{9997\,+\,45\, \sqrt{ 49345\,}\right\}\,}\, -\, \dfrac{4}{15}\bigg]\right) \mbox{ for }\, n\, \in\, \mathbb{Z}$

Where did you start from, that you ended up with this equation? Thank you!

denicarl
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### Re: solving for x, given exponentials: 4e^2x + 5e^3x = 15

difference of probabilities-actual numbers are

(1-e^(-.2x))-(1-e^(-.16x))=.5

e^(-.16x)-e^(-.2x)=.5 solve for x

anonmeans
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Joined: Sat Jan 24, 2009 7:18 pm
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### Re: solving for x, given exponentials: 4e^2x + 5e^3x = 15

difference of probabilities-actual numbers are

(1-e^(-.2x))-(1-e^(-.16x))=.5

e^(-.16x)-e^(-.2x)=.5 solve for x
Wolfram Alpha still gives a complicated answer. Assuming your equation is right, then maybe the homework is expecting a numerical approximation instead of an algebraic solution.