Simplify, 3 + 2log(base p+q)(1/p^2-q^2)
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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!
Originally Posted by nav92 Simplify, 3 + 2log(base p+q)(1/p^2-q^2) $\displaystyle 3 - 2\log_{p+q}\left[(p+q)(p-q)\right]$ $\displaystyle 3 - 2\log_{p+q}(p+q) - 2\log_{p+q}(p-q)$ $\displaystyle 3 - 2 - 2\log_{p+q}(p-q)$ $\displaystyle 1 - \log(p-q)^2$
hmm, I think he meant log(a*b) where a was P+Q and b was 1/(P^2 - Q^2)
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)
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