Re: [obm-l] questão logaritmo

To: obm-l@xxxxxxxxxxxxxx
Subject: Re: [obm-l] questão logaritmo
From: saulo nilson <saulo.nilson@xxxxxxxxx>
Date: Tue, 3 Jan 2006 10:12:49 -0200
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(a+b)^2=a^2+b^2+2ab

logo

(a+b)^2=72ab

e a expressao se torna

log72 na base 10

log72= log 8*9 =log(2^3*3^2)=3log2 + 2log3

so que isso esta na base 10, transformando para base 5

3*log2/log10 +2*log3/log10

tudo isto esta na base 5

3*log2/log10 +2*log3/log10=(3log2+2log3)/(log2 + 1)=

=(3m+2n)/(m+1)

abraço, saulo.

On 1/3/06, Bruna Carvalho <bruna.carvalho.pink@gmail.com> wrote:

sendo a^2+b^2=70ab, calcule log((a+b)^2/ab)[10] em função de m=log(2)[5] e n=log(3)[5].