Matemática, perguntado por awdadawdawd, 9 meses atrás

\frac{log_{3}1+log_{10}0,01}{log_{2}\frac{1}{64} .log_{4} \sqrt{8} }
resposta: 4/9
(faça o calculo)

Soluções para a tarefa

Respondido por Usuário anônimo
2

Explicação passo-a-passo:

Lembre-se que:

\sf log_{b}~1=0

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

\sf log_{a}~a=1

\sf log_{b^n}~a=\dfrac{1}{n}\cdot log_{b}~a

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

Assim:

\sf log_{3}~1=0

\sf log_{10}~0,01=log_{10}~10^{-2}

\sf log_{10}~0,01=(-2)\cdot log_{10}~10

\sf log_{10}~0,01=(-2)\cdot1

\sf log_{10}~0,01=-2

\sf log_{2}~\dfrac{1}{64}=log_{2}~2^{-6}

\sf log_{2}~\dfrac{1}{64}=(-6)\cdot log_{2}~2

\sf log_{2}~\dfrac{1}{64}=(-6)\cdot1

\sf log_{2}~\dfrac{1}{64}=-6

\sf log_{4}~\sqrt{8}=log_{2^2}~\sqrt{2^3}

\sf log_{4}~\sqrt{8}=log_{2^2}~2^{\frac{3}{2}}

\sf log_{4}~\sqrt{8}=\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot log_{2}~2

\sf log_{4}~\sqrt{8}=\dfrac{1}{2}\cdot\dfrac{3}{2}\cdot1

\sf log_{4}~\sqrt{8}=\dfrac{3}{4}

Logo:

\sf \dfrac{log_{3}~1+log_{10}~0,01}{log_{2}~\frac{1}{64}\cdot log_{4}~\sqrt{8}}

\sf =\dfrac{0-2}{(-6)\cdot\frac{3}{4}}

\sf =\dfrac{-2}{\frac{18}{4}}

\sf =\dfrac{-2}{-\frac{9}{2}}

\sf =-2\cdot\Big(-\dfrac{2}{9}\Big)

\sf =\red{\dfrac{4}{9}}


sdsadnn: VALEU de novo <3
vlouzin18: ain
Respondido por Makaveli1996
0

Oie, Td Bom?!

 =  \frac{ log_{3}(1)  +  log_{10}(0,01) }{ log_{2}( \frac{1}{64} ) log_{4}( \sqrt{8} )  }

 =  \frac{0 +  log_{10}(10 {}^{ - 2} ) }{ log_{2}(2 {}^{ - 6} ) log_{2 {}^{2} }( 2 {}^{ \frac{3}{2} }  )  }

➛ Calculando os logaritmos separadamente:

I.

 =  log_{10}(10 {}^{ - 2} )

 =  - 2 log_{10}(10)

 =  - 2 \: . \: 1

 =  - 2

II.

 =  log_{2}(2 {}^{ - 6} )

 =  - 6 log_{2}(2)

 =  - 6 \: . \: 1

 =  - 6

III.

 =  log_{2 {}^{2} }(2 {}^{ \frac{3}{2} } )

 =  \frac{ \frac{3}{2} }{2}  \: . \:  log_{2}(2)

 =  \frac{ \frac{3}{2} }{2}  \: . \: 1

 =  \frac{  \frac{3}{2}  }{2}

 =  \frac{3}{4}

• Continuando...

 =  \frac{ - 2}{ - 6 \: . \:  \frac{3}{4} }

 =  \frac{ - 2}{ - 3 \: . \:  \frac{3}{2} }

 =  \frac{ - 2}{  - \frac{9}{2} }

 =  \frac{ - 2  {}^{ \div( - 1)} }{ ( - \frac{9}{2} ) {}^{ \div ( - 1)} }

 =  \frac{2}{ \frac{9}{2} }

 = 2 \div  \frac{9}{2}

 = 2 \: . \:  \frac{2}{9}

 =  \frac{4}{9}

Att. Makaveli1996

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