# Thread: converting into differnce of logarithms.

1. ## converting into differnce of logarithms.

ln((x+6)(x-8))/(x-7)^2)^5/3 , x > 8

2. ## Re: converting into differnce of logarithms.

You have too many closing parentheses.

3. ## Re: converting into differnce of logarithms.

As emakarov said, the problem has unmatched parentheses- it is impossible to tell if the entire logarithm is to the 5/3 power, or just the argument of the logarithm, or just the denominator.

But to answer your question generally, log(A/B)= log(A)- log(B). It is also true that log(AB)= log(A)+ log(B) and that log(A^n)= nlog(A). You may need to use those.

4. ## Re: converting into differnce of logarithms.

I noticed that, :/, I don't know how to enter formulas to where they actually look like they are written.

5. ## Re: converting into differnce of logarithms.

its suppose to be, ln( ((x+6)(x-8))/(x-7)^2)^5/3 , x > 8

6. ## Re: converting into differnce of logarithms.

Originally Posted by blondedude092
its suppose to be, ln( ((x+6)(x-8))/(x-7)^2)^5/3 , x > 8
$\frac{5}{3}\ln\left[\frac{(x+6)(x-8)}{(x-7)^2}\right]$

$\frac{5}{3}\left[\ln(x+6) + \ln(x-8) - 2\ln(x-7)\right]$