Another logarithm problem

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July 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.
July 5th 2012, 01:55 PM
emakarov

Re: Another logarithm problem

$3e^{-2x} = 10$ Divide both sides by 3

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

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

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

$x = \frac{1}{2}\ln(3/10)$
July 5th 2012, 01:55 PM
Plato

Re: Another logarithm problem

Quote:

Originally Posted by Latsabb

The problem is as follows:

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

Hint: $-2x=\ln(10)-\ln(3)$ , How and Why?
July 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?
July 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 $e^x=y$ then $x=\ln(y)$
July 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.
July 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.
July 6th 2012, 01:33 AM
daigo

Re: Another logarithm problem

All 'e' is is another constant such as $\pi$ and 'ln' is just a different notation for $\log_{e}$ . That's pretty much all you need to know to solve this if you already know basic logs
July 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 $3e^{-2x}= 10$ ? 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)).
July 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 !