1. ## Logarithm help!

Need help with problem.
Log (x-1)-log (x+1)=3-log (x-2)

I know the answer is about 1003.994 but need help figuring out how to get the answer. I got to (×-1)(x-2)=1000x+1000 but don't know where to go from there.

2. ## Re: Logarithm help!

Originally Posted by Nfalconer
Log (x-1)-log (x+1)=3-log (x-2)
Says who?

Havva look:
Wolfram|Alpha: Computational Intelligence)

3. ## Re: Logarithm help!

We are given to solve:

$\displaystyle \log(x-1)-\log(x+1)=3-\log(x-2)$

Rewrite using log rules (and assuming logs are base 10)

$\displaystyle \log\left(\frac{x-1}{x+1}\right)=\log\left(\frac{1000}{x-2}\right)$

This implies:

$\displaystyle \frac{x-1}{x+1}=\frac{1000}{x-2}$

$\displaystyle (x-1)(x-2)=1000(x+1)$

$\displaystyle x^2-3x+2=1000x+1000$

$\displaystyle x^2-1003x-998=0$

Apply the quadratic formula...what do you get?

4. ## Re: Logarithm help!

Originally Posted by DenisB
Says who?

Havva look:
Wolfram|Alpha: Computational Intelligence)
W|A assumes a base of e when none is given.

5. ## Re: Logarithm help!

Originally Posted by Nfalconer
Need help with problem.
Log (x-1)-log (x+1)=3-log (x-2)

I know the answer is about 1003.994 but need help figuring out how to get the answer. I got to (×-1)(x-2)=1000x+1000 but don't know where to go from there.
You are using base 10, see here.

@Nfalconer, I have noticed that you use older norms in mathematics. In today's practice $\log(x)$ is taken to be base $e$ or the natural. In fact $\log_b(x)=\dfrac{\log(x)}{\log(b)}$.

So $\log_{10} (x-1)-\log_{10} (x+1)=3-\log_{10} (x-2)$ translates to $\log_{10}\left(\dfrac{(x-1)(x-2)}{(x-1)}\right)=3$

6. ## Re: Logarithm help!

Originally Posted by MarkFL
W|A assumes a base of e when none is given.
Which is standard practice in mathematics today.

7. ## Re: Logarithm help!

Originally Posted by Plato
Which is standard practice in mathematics today.
Yes, when I was a student, at the elementary level, a log without a base given was assumed to be base 10 and ln() (natural log) was used for base e logs. Then when you got into 3rd year course at the university level, such as course in analysis, log() was used for base e logs.

8. ## Re: Logarithm help!

After using the quadratic formula I did not get 23.844. Please show me how you got that.

9. ## Re: Logarithm help!

Originally Posted by Nfalconer
After using the quadratic formula I did not get 23.844. Please show me how you got that.
That's because he used log base e instead of log base 10 as we pretty much assumed by the problem as written. 23.844 is not correct if we use log base 10.

-Dan

Addendum: Personally I still use ln for log base e. I really don't know why this has been "abandoned."

10. ## Re: Logarithm help!

Originally Posted by Nfalconer
After using the quadratic formula I did not get 23.844. Please show me how you got that.

11. ## Re: Logarithm help!

In some of the computer science courses I took a log without a stated base was assumed to be base 2. That made sense though, obviously.

12. ## Re: Logarithm help!

Originally Posted by Nfalconer
Need help with problem.
Log (x-1)-log (x+1)=3-log (x-2)

I know the answer is about 1003.994 but need help figuring out how to get the answer. I got to (×-1)(x-2)=1000x+1000 but don't know where to go from there.
Here is the Answer based on base e.

$\displaystyle \log\left(\frac{x - 1}{x+1}\right) + \log (x -2)= 3$
$\displaystyle \log\left(\frac{(x - 1)(x-2)}{x+1}\right) = 3$
$\displaystyle \left(\frac{(x-2)(x-1)}{x+1}\right) = \log^{-1}(3)$
$\displaystyle \frac{(x-2)(x-1)}{x+1}=20.08554$
$\displaystyle x^2 - 3x + 2 =20.08554x + 20.08554$
$\displaystyle x^2 - 23.08554x - 18.08554 = 0$

$\displaystyle x =314.1849285458$

13. ## Re: Logarithm help!

Originally Posted by Nfalconer
Need help with problem. Log (x-1)-log (x+1)=3-log (x-2)
I know the answer is about 1003.994 but need help figuring out how to get the answer.
Originally Posted by x3bnm
Here is the Answer based on base e.
$\displaystyle x =314.1849285458$
@x3bnm, isn't it clear to you that the OP clearly implies that it is about $log_{10}~?$

14. ## Re: Logarithm help!

Originally Posted by Plato
@x3bnm, isn't it clear to you that the OP clearly implies that it is about $log_{10}~?$
Sorry for the misunderstanding. But how did he come up with 1003.994? So I was confused. Also, he didn't specifically mention about the base. So I thought why not show him another way. Sorry Plato I didn't mean to offend anybody.

15. ## Re: Logarithm help!

Originally Posted by Nfalconer
Need help with problem.
Log (x-1)-log (x+1)=3-log (x-2)

I know the answer is about 1003.994 but need help figuring out how to get the answer. I got to (×-1)(x-2)=1000x+1000 but don't know where to go from there.

$\displaystyle \log(x-1)-\log(x+1)=3-\log(x-2)$

$\displaystyle \log\left(\frac{x - 1}{x+1}\right) + \log (x -2)= 3$

$\displaystyle \log\left(\frac{(x - 1)(x-2)}{x+1}\right) = 3$

$\displaystyle \left(\frac{(x-2)(x-1)}{x+1}\right) = \log^{-1}(3)$

$\displaystyle \frac{(x-2)(x-1)}{x+1}=1000$

$\displaystyle x^2 - 3x + 2 =1000x + 1000$

$\displaystyle x^2 -1003x -998 = 0$

$\displaystyle x = \frac{1003 \pm \sqrt{1003^{2} - 4(-998)}}{2}$

$\displaystyle x = 1002.003995988$ [Answer]

You can check the quadratic equation below:

Wolfram|Alpha: Computational Intelligence

if you use a calculator the answer will be 1003.994 So you are right.