• Matéria: Matemática
  • Autor: rafinhaaaaaq
  • Perguntado 5 anos atrás

Considerando log 2=0,3, log3 = 0,5, log5=0,7

a) Log 15
b) Log45
c) Log30
d) Log 1,2
e)log√0,3

Respostas

respondido por: Anônimo
1

Explicação passo-a-passo:

Lembre-se que:

\sf log_{b}~(a\cdot c)=log_{b}~a+log_{b}~c

\sf log_{b}~\Big(\dfrac{a}{c}\Big)=log_{b}~a-log_{b}~c

\sf log_{b}~a^m=m\cdot log_{b}~a

\sf \sqrt[c]{a^b}=a^{\frac{b}{c}}

a)

\sf log~15=log~(3\cdot5)

\sf log~15=log~3+log~5

\sf log~15=0,5+0,7

\sf \red{log~15=1,2}

b)

\sf log~45=log~(3^2\cdot5)

\sf log~45=log~3^2+log~5

\sf log~45=2\cdot log~3+log~5

\sf log~45=2\cdot0,5+0,7

\sf log~45=1+0,7

\sf \red{log~45=1,7}

c)

\sf log~30=log~(2\cdot3\cdot5)

\sf log~30=log~2+log~3+log~5

\sf log~30=0,3+0,5+0,7

\sf \red{log~30=1,5}

d)

\sf log~1,2=log~\Big(\dfrac{12}{10}\Big)

\sf log~1,2=log~\Big(\dfrac{6}{5}\Big)

\sf log~1,2=log~6-log~5

\sf log~1,2=log~(2\cdot3)-log~5

\sf log~1,2=log~2+log~3-log~5

\sf log~1,2=0,3+0,5-0,7

\sf \red{log~1,2=0,1}

e)

\sf log~\sqrt{0,3}=log~\sqrt{\dfrac{3}{10}}

\sf log~\sqrt{0,3}=log~\Big(\dfrac{3}{10}\Big)^{\frac{1}{2}}

\sf log~\sqrt{0,3}=\dfrac{1}{2}\cdot log~\Big(\dfrac{3}{10}\Big)

\sf log~\sqrt{0,3}=\dfrac{1}{2}\cdot(log~3-log~10)

\sf log~\sqrt{0,3}=\dfrac{1}{2}\cdot(0,5-1)

\sf log~\sqrt{0,3}=\dfrac{1}{2}\cdot(-0,5)

\sf log~\sqrt{0,3}=\dfrac{1}{2}\cdot\Big(-\dfrac{1}{2}\Big)

\sf \red{log~\sqrt{0,3}=-\dfrac{1}{4}}

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