Respostas

2014-08-22T21:13:40-03:00
E aí mano,

a)~~log(a\cdot b)=loga\cdot b~~~(~~).\\\\
b)~~log(a+b)=loga+logb~~~(~~).\\\\
c)~~logm\cdot a=m\cdot loga~~~(~~)\\\\
d)~~loga^m=logm\cdot a~~~(~~)\\\\
\boxed{e)~~loga^m=m\cdot loga~~~(X)}.\\.

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