Matematică
cosminbeti
2015-11-11 02:49:39
Trebuie schimbate bazele?
Răspunsuri la întrebare
ursuboris123
2015-11-11 04:30:00

log(2)8*√2=log(2)2³*2^(1/2)=log(2)2^(3+1/2)=log(2)2^7/2=7/2 log(3)3*√3=log(3)3*3^(1/2)=log(3)3^(1+1/2)=log(3)3^3/2=3/2 7/2-3/2=4/2=2

robert18
2015-11-11 04:31:15

[latex]log_{ig2}8sqrt2-log_{ig3}3sqrt3= \ \ = log_{ig2}8+log_{ig2}sqrt2- ig(log_{ig3}3+log_{ig3} sqrt3ig) = \ \ = 3+log_{ig2}2^{dfrac{1}{2}}-Big(1+log_{ig3}3^{dfrac{1}{2}}Big) = \ \ = 3+dfrac{1}{2}cdot log_{ig2}2-Big(1+dfrac{1}{2}cdot log_{ig3}3Big) = \ \ = 3+dfrac{1}{2}-Big(1+dfrac{1}{2}Big) = \ \ = dfrac{7}{2} - dfrac{3}{2}} = \ \ = dfrac{4}{2} = \ \ =2[/latex]

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