1. ## Help with logarithms?

Hello everyone. We've been given an assignment on logarithms and I've managed to answer everything except this one (please see attached picture). I don't know what to do with the 2log9(x-1)log9(x^2-x) part. Do I separate it with a + sign since that's the rule for logarithms being multiplied?

All help is much appreciated! Thank you very much in advance.

2. ## Re: Help with logarithms?

No, it is not true that log(a)log(b) = log(a) + log(b).

I think there is probably a typo in the problem statement. If instead of 2log(x-1)log(x^2-2) it had a minus sign: 2log(x-1)-log(x^2-x) then the problem works out nicely. Do you happen to have the answer for x?

3. ## Re: Help with logarithms?

Originally Posted by ChipB
No, it is not true that log(a)log(b) = log(a) + log(b).

I think there is probably a typo in the problem statement. If instead of 2log(x-1)log(x^2-2) it had a minus sign: 2log(x-1)-log(x^2-x) then the problem works out nicely. Do you happen to have the answer for x?
I agree & had reached the same correction: $2\log(x-1)-\log(x^2-1)=\log\left(\dfrac{(x-1)^2}{(x^2-1)}\right)=\log\left(\dfrac{(x-1)}{(x+1)}\right)$

4. ## Re: Help with logarithms?

Originally Posted by ChipB
No, it is not true that log(a)log(b) = log(a) + log(b).

I think there is probably a typo in the problem statement. If instead of 2log(x-1)log(x^2-x) [greg1313 edit] it had a minus sign: 2log(x-1)-log(x^2-x) then the problem works out nicely.
x = -4/5 would not be a solution to the revised problem with the inserted minus sign, because the logarithm of the second argument
with it substituted makes it undefined. Speaking closer about the base/parent function, f(x) = 2log(x) has a domain of x > 0, while
g(x) = log(x^2) has a domain of x equals all real numbers, except 0.
The hypothetical revised problem has no real solution.

5. ## Re: Help with logarithms?

Hi everyone. Thank you all for the replies. Our teacher said that the revised version is actually supposed to have a positive sign in between the two logarithms.

6. ## Re: Help with logarithms?

Beer soaked ramblings follow.
Originally Posted by kaonashi
Hi everyone. Thank you all for the replies. Our teacher said that the revised version is actually supposed to have a positive sign in between the two logarithms.
Do us all a favor and post the exact problem statement?