1. ## Rules of logs?

Why does ln(1/3) = -ln(3)? Is it because (1/3)^-1=3? And then you just put the -1 at the front? If this even makes sense... Thank you in advance.

2. ## Re: Rules of logs?

Originally Posted by Emanresu
Why does ln(1/3) = -ln(3)? Is it because (1/3)^-1=3? And then you just put the -1 at the front? If this even makes sense... Thank you in advance.
Yes that's exactly the reason. One of the rules of logarithms states \displaystyle \displaystyle \begin{align*} \log{\left(m^p\right)} = p\log{m} \end{align*}, so

\displaystyle \displaystyle \begin{align*} \ln{\left(\frac{1}{3}\right)} &= \ln{\left(3^{-1}\right)} \\ &= -1\ln{(3)} \\ &= -\ln{(3)} \end{align*}

3. ## Re: Rules of logs?

You can show that ln(1/3) = -ln(3) in a couple of ways:

1. As you noted, 1/3 = 3^(-1). Since ln(a^b) = b ln(a), if you let a = 3 and b = -1 you get ln(3^-1) = -1 ln(3)

2. Recall that ln(a/b) = ln(a) - ln(b). If you let a = 1 and b = 3 you get ln(1/3) = ln(1)-ln(3) = 0-3 = -3

Hope this helps.

4. ## Re: Rules of logs?

Originally Posted by Emanresu
Why does ln(1/3) = -ln(3)? Is it because (1/3)^-1=3? And then you just put the -1 at the front? If this even makes sense... Thank you in advance.
That is true. But also

$\displaystyle \ln\left(\frac{1}{3}\right)=\ln(1)-\ln(3)=0-ln(3)=-\ln(3)$

5. ## Re: Rules of logs?

Thank you everyone, especially for that second method.