# Another logarithm problem

• Jul 5th 2012, 01:46 PM
Latsabb
Another logarithm problem
I keep getting blocked every few pages by something I cant figure out with these logarithm problems, and I apologize for posting up so many questions, but I am really struggling.

The problem is as follows:

3 * e^-2x = 10

Basically, I cant seem to figure out the order, or how to break it down correctly. I typically start off by dividing both sides by 3, but then it goes downhill from there. How do I deal with that e^-2x? I have tried all kind of stuff, and I am not finding a simple solution. The book makes it seem like it should be elementary, since it doesnt even explain it, and it is an example. Maybe it is because it is almost midnight, but I cannot for the life of me see what I am missing. Thanks.
• Jul 5th 2012, 01:55 PM
emakarov
Re: Another logarithm problem
$\displaystyle 3e^{-2x} = 10$ Divide both sides by 3

$\displaystyle e^{-2x}=10/3$ Take ln of both sides, use the fact that $\displaystyle \ln(e^y)=y$ for all y.

$\displaystyle -2x = \ln(10/3)$ Divide by -2

$\displaystyle x = -\frac{1}{2}\ln(10/3)$ You may use the fact that $\displaystyle -\ln(x)=\ln(x^{-1})$ to get

$\displaystyle x = \frac{1}{2}\ln(3/10)$
• Jul 5th 2012, 01:55 PM
Plato
Re: Another logarithm problem
Quote:

Originally Posted by Latsabb
The problem is as follows:

$\displaystyle 3 e^{-2x} = 10$

Hint: $\displaystyle -2x=\ln(10)-\ln(3)$, How and Why?
• Jul 5th 2012, 01:58 PM
Latsabb
Re: Another logarithm problem
Ok, so in general, if I have e raised to something, I can ln both sides, and simply drop the e, and put the powers as constants?
• Jul 5th 2012, 02:09 PM
Plato
Re: Another logarithm problem
Quote:

Originally Posted by Latsabb
Ok, so in general, if I have e raised to something, I can ln both sides, and simply drop the e, and put the powers as constants?

You ought to know that if $\displaystyle e^x=y$ then $\displaystyle x=\ln(y)$
• Jul 5th 2012, 07:53 PM
Latsabb
Re: Another logarithm problem
The book I am working in just introduced e, and has yet to introduce ln.
• Jul 6th 2012, 01:04 AM
emakarov
Re: Another logarithm problem
Quote:

Originally Posted by Latsabb
Ok, so in general, if I have e raised to something, I can ln both sides, and simply drop the e

No, you drop e in one side of the equation and add ln in the other side, as Plato wrote in post #5.

Quote:

Originally Posted by Latsabb
and put the powers as constants?

I am not sure what this means.

Quote:

Originally Posted by Latsabb
The book I am working in just introduced e, and has yet to introduce ln.

I don't see how to solve the original equation without logarithms.
• Jul 6th 2012, 01:33 AM
daigo
Re: Another logarithm problem
All 'e' is is another constant such as $\displaystyle \pi$ and 'ln' is just a different notation for $\displaystyle \log_{e}$. That's pretty much all you need to know to solve this if you already know basic logs
• Jul 6th 2012, 05:34 AM
HallsofIvy
Re: Another logarithm problem
Quote:

Originally Posted by Latsabb
The book I am working in just introduced e, and has yet to introduce ln.

Yet, it asks you to solve $\displaystyle 3e^{-2x}= 10$? Has it dealt with logarithms in general, then? If you have logarithms base 10 ("common" logarithms), you can take the logarithm of both sides to get log(3)- 2x log(e)= 1 and then solve for x: x= (log(3)- 1)/(2log(e)).
• Jul 6th 2012, 10:19 AM
Wilmer
Re: Another logarithm problem
Quote:

Originally Posted by Latsabb
3 * e^-2x = 10

First, you need a set of brackets: 3 * e^(-2x) = 10;
the way you have it means 3 * (e^-2) * x = 10 ; OK?

I'd start this way:
1 / e^(2x) = 10/3

And (as you've been told): log(e) = 1 : tattoo that on your wrist !