1. ## Logarithms

Simplify, 3 + 2log(base p+q)(1/p^2-q^2)

2. You can factorise
1/(P^2 - Q^2) to 1/(P + Q)(P - Q)

Then you get 3 + log(1/(P-Q). You can make this 3+ log1 - log(P-Q) = 3 + 0 - log(P-Q)

Oh: And I refuse 2 use that math text app.. I don't understand it, some crazy foreign stuff imo!

3. Originally Posted by nav92
Simplify, 3 + 2log(base p+q)(1/p^2-q^2)
$3 - 2\log_{p+q}\left[(p+q)(p-q)\right]$

$3 - 2\log_{p+q}(p+q) - 2\log_{p+q}(p-q)$

$3 - 2 - 2\log_{p+q}(p-q)$

$1 - \log(p-q)^2$

4. hmm, I think he meant log(a*b) where a was P+Q and b was 1/(P^2 - Q^2)

5. Originally Posted by narko
hmm, I think he meant log(a*b) where a was P+Q and b was 1/(P^2 - Q^2)
note the word "base" by (p+q)