Matematică
bubuci
2015-11-12 09:43:57
lg2/1+lg3/2+...............+lg10/9=?
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Cont şters
2015-11-12 11:27:38

[latex]lgBig(dfrac{2}{1}Big) + lgBig(dfrac{3}{2}Big) + ...+lgBig(dfrac{10}{9}Big) = \ \ =Big(lg2-lg1Big)+ Big(lg3-lg2Big)+ ...+Big(lg10-lg9Big)= \ \ = lg2+lg3+...+lg10 - lg1-lg2-lg3-...-lg10 = \ \ = lg2+lg3+...+lg10 -(lg1 +lg2+lg3+...+lg10) = \ \ = lg2+lg3+...+lg10 -(0 +lg2+lg3+...+lg10) = \ \ = lg2+lg3+...+lg10 -(lg2+lg3+...+lg10) = \ \ = 0[/latex] [latex]\ $M-am folosit de proprietatea: log_{ig a}b-log_{ig a}c = log_{ig a} Big(dfrac{b}{c}Big) \ \ $La noi baza era 10:quad lg a = log_{ig{10}}a[/latex]

alessia1
2015-11-12 11:28:53

Lg2/1+lg3/2+...............+lg10/9=? [latex]it lgdfrac{2}{1}+lgdfrac{3}{2}+lgdfrac{4}{3}+ ... lgdfrac{10}{9}= lgleft(dfrac{2}{1}cdotdfrac{3}{2}cdotdfrac{4}{3}cdot ... cdotdfrac{10}{9} ight) =lg10=1[/latex]

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