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.
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.
Says who?
Answer is ~23.844
Havva look:
Wolfram|Alpha: Computational Intelligence)
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?
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 $\log_{10}\left(\dfrac{(x-1)(x-2)}{(x-1)}\right)=3$
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.
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 $
$\displaystyle \log(x-1)-\log(x+1)=3-\log(x-2)$
$\displaystyle \log\left(\frac{x - 1}{x+1}\right) + \log (x -2)= 3$
$\displaystyle \log\left(\frac{(x - 1)(x-2)}{x+1}\right) = 3$
$\displaystyle \left(\frac{(x-2)(x-1)}{x+1}\right) = \log^{-1}(3)$
$\displaystyle \frac{(x-2)(x-1)}{x+1}=1000$
$\displaystyle x^2 - 3x + 2 =1000x + 1000$
$\displaystyle x^2 -1003x -998 = 0$
$\displaystyle x = \frac{1003 \pm \sqrt{1003^{2} - 4(-998)}}{2}$
$\displaystyle x = 1002.003995988$ [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.