How do perform the logarithm on a product term that has a power term?

Question

I have this term $ \prod_{i=0}^N(u_i-x_i-\tau)^{-3/2} $ and on which I need to take the logarithm.

Applying the log make the $\prod$ to $\sum$.

What bothers me is the power $^{-3/2} $.

Will it be

(a) $-3/2\sum_{i=0}^N (u_i-x_i-\tau)$ or will it be

(b) $\sum_{i=0}^N (u_i-x_i-\tau)^{-3/2}$

score 1 · Accepted Answer · answered Sep 22 '18 at 10:24

1

Neither of them: if $u_i-x_i-\tau > 0$ for each $i=0,\dots,N$, then

$$\log \prod_{i=0}^N(u_i-x_i-\tau)^{-3/2} = \sum_{i=0}^N \log (u_i-x_i-\tau)^{-3/2} = -\frac{3}{2}\sum_{i=0}^N \log (u_i-x_i-\tau).$$

answered Sep 22 '18 at 10:24

Gibbs

What if $u_i-x_i-\tau \le 0$? – nashynash Sep 22 '18 at 10:26
If there exists $i$ such that $u_i-x_i-\tau \leq 0$, then $\log (u_i-x_i-\tau)$ is not defined, assuming you are working with real numbers. – Gibbs Sep 22 '18 at 10:28
I believe you meant $\le$ and not $\ge$ – nashynash Sep 22 '18 at 10:29
Yes sure, sorry for the typo. In that case the formula $\log ab = \log a + \log b$ does not apply. – Gibbs Sep 22 '18 at 10:31
Thank you so much. – nashynash Sep 22 '18 at 10:33
Yes, I will. But it says to wait for 3 more minutes and I don't know why. But do not worry, I will. – nashynash Sep 22 '18 at 10:35

1 Answers1