I want to rewrite $\displaystyle \ln x$ to (something) $\displaystyle \ln 3x$.

Is that possible?

For example, if I want to rewrite $\displaystyle x^2$ to (something) $\displaystyle 2x^2$, it becomes $\displaystyle \frac {1}{2}*2x^2$

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- Jul 23rd 2009, 07:09 PMchengbinlogarithm conversion
I want to rewrite $\displaystyle \ln x$ to (something) $\displaystyle \ln 3x$.

Is that possible?

For example, if I want to rewrite $\displaystyle x^2$ to (something) $\displaystyle 2x^2$, it becomes $\displaystyle \frac {1}{2}*2x^2$ - Jul 23rd 2009, 07:14 PMpickslides
$\displaystyle ln(x) = - ln(3)+ ln(3x) $

- Jul 23rd 2009, 07:17 PMchengbin
- Jul 23rd 2009, 07:24 PMpickslides
I don't know if it can be done with 1 term.

The logic behind my post is simply the addition of logs law, it is:

$\displaystyle ln(a) + ln(b) = ln(a\times b)$

$\displaystyle ln(b) = - ln(a) + ln(a\times b)$

In your case

$\displaystyle ln(x) + ln(3) = ln(3x)$

$\displaystyle ln(x) = - ln(3)+ ln(3x)$ - Jul 23rd 2009, 07:27 PMchengbin
Looks like I was late.

I got how the conversion works on my bed, I got up to tell you I got it and don't bother explaining it. Thanks anyways. - Jul 23rd 2009, 07:33 PMpickslides
Not to worry. Good luck!