# Math Help - Can i write this in another way? using Ln instead of Log

1. ## Can i write this in another way? using Ln instead of Log

4/243Log(3x-2)-4/243log(3x+2)

2. Originally Posted by aa751
4/243Log(3x-2)-4/243log(3x-2)
Are my eyes playing tricks on me?

$\frac{4}{243}\log (3x-2)-\frac{4}{243}\log (3x-2)$

If so the answer is 0.

What base is the log here? If you want to use ln you can use the change of base formula. It is $\log_ab = \frac{\ln b}{\ln a}$

3. OH am sorry slides, It is my eyes that isnt working it actually looks like this

4/243Log(3x-2)-4/243log(3x+2)

4. Maybe start by taking out a common factor here

$\frac{4}{243}\log (3x-2)-\frac{4}{243}\log (3x+2)$

$\frac{4}{243}\left(\log (3x-2)-\log (3x+2)\right)$

$\frac{4}{243}\left(\log \frac{(3x-2)}{(3x+2)}\right)$

Depending on what you want to do next you can now use the change of base formula mentioned in post #2. What base is your logarithm?

5. I am not sure =//// I do not really know All i know is that the question wants me to convert that into Ln of f(x)/g(x) I am not sure about the base but I think its 10 It is prolly to the base of 10, that makes most sense right now

6. THe log is to the base 10

7. $\frac{4}{243}\left(\log_{10}\frac{(3x-2)}{(3x+2)}\right)$

$\frac{4}{243}\left(\frac{\ln\frac{(3x-2)}{(3x+2)}}{\ln 10}\right)$

8. Ahh thanks man you are my hero , you saved me

9. Originally Posted by aa751
(4/243)Log(3x-2)-(4/243)log(3x+2) Base of Log= 10 the answer should look like something like this Ln(g(x)/h(x))
difference property of logs ...

$\log(x) - \log(y) = \log\left(\frac{x}{y}\right)$

change of base ...

$\log_b(a) = \frac{\ln{a}}{\ln{b}}$