# Logarithms

• October 29th 2010, 12:48 PM
nav92
Logarithms
Simplify, 3 + 2log(base p+q)(1/p^2-q^2)
• October 29th 2010, 01:18 PM
narko
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!
• October 29th 2010, 01:26 PM
skeeter
Quote:

Originally Posted by nav92
Simplify, 3 + 2log(base p+q)(1/p^2-q^2)

$3 - 2\log_{p+q}\left[(p+q)(p-q)\right]$

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

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

$1 - \log(p-q)^2$
• October 29th 2010, 01:32 PM
narko
hmm, I think he meant log(a*b) where a was P+Q and b was 1/(P^2 - Q^2)
• October 29th 2010, 01:35 PM
skeeter
Quote:

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)