Question

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

Answer 1

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:

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

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

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

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

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

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

Answer 2

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