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

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

how do I solve for x?

I tried ln 4 +2x+ln5+3x=ln 15 but that doesn't work. Thanks

What you wrote means this:if I have 4e^2x + 5e^3x = 15

how do I solve for x?

a.

Is that right? Or did you mean this?

b.

How did you get to this step?I tried ln 4 +2x+ln5+3x=ln 15 but that doesn't work.

Thanks.

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

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

So you meant this:I meant b

Taking logs of both sides would give you this:and in order to solve say e^(2x) = Y I would take log of both sides so-

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):

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.)

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

- maggiemagnet
**Posts:**358**Joined:**Mon Dec 08, 2008 12:32 am-
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You're right! I think maybe "FWT" hadn't had his/her caffeine yet when s/he posted that!(a^b)(a^c)=(a^(b+c)) so above factoring won't work

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

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

difference of probabilities-actual numbers are

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

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

(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.difference of probabilities-actual numbers are

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

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