Question

Considere f x = log 2 (na base x) e g x = log 5 (na base x) , calcule o valor de [f(1/128) - g(125)] / 2

Answer 1

Explicação passo-a-passo:

Temos as seguintes funções:

$f(x)=\log_{x}2$

$g(x)=\log_{x}5$

Calculando $f(\frac{1}{128} )$:

$f(\frac{1}{128} )=\log_{\frac{1}{128} }2=a\\\\\bigg(\dfrac{1}{128}\bigg) ^{a}=2\\\\(2^{-7})^{a} =2^{1} \\\\2^{-7a} =2^{1} \\\\\text Igualando\ os\ expoentes:\\\\-7a=1\\\\a=\dfrac{1}{-7} \\\\\boxed{f\bigg(\frac{1}{128}\bigg )=-\dfrac{1}{7} }$

Calculando $g(125 )$:

$g(125)=\log_{125}5=b\\\\125^{b}=5\\\\(5^{3})^{b}=5^{1} \\\\5^{3b}=5^{1}\\\\\text Igualando\ os\ expoentes:\\\\3b=1\\\\b=\dfrac{1}{3}\\\\\boxed{ g(125)=\dfrac{1}{3}}$

Calculando o valor da expressão do problema:

$\dfrac{f(\frac{1}{128} )-g(125)}{2}=\dfrac{\dfrac{-1}{7}-\dfrac{1}{3} }{2} \\\\\dfrac{f(\frac{1}{128} )-g(125)}{2}=\dfrac{\dfrac{-3}{21}-\dfrac{7}{21} }{2} \\\\\dfrac{f(\frac{1}{128} )-g(125)}{2}=\dfrac{\dfrac{-10}{21} }{2} \\\\\dfrac{f(\frac{1}{128} )-g(125)}{2}=\dfrac{-10 }{21.2} \\\\\boxed{\boxed{\dfrac{f(\frac{1}{128} )-g(125)}{2}=\dfrac{-5 }{21} }}$

Answer 2

Explicação passo-a-passo:

Estudo das funções

Dada as funções :

$\sf{ f(x)~=~\log_{x}2 ~~e~g(x) = \log_{x}5 }$

Determinar: $\sf{ \dfrac{ f\Big(\frac{1}{128}\Big) - g(125) }{2} }$

• Primeiro vamos achar a $\sf{f\Big(\dfrac{1}{128}\Big) }$.

$\iff \sf{ f\Big( \dfrac{1}{128}\Big)~=~ \log_{\frac{1}{128}}2 }$

$\iff \sf{ f\Big( \dfrac{1}{128}\Big)~=~ \log_{2^{-7}}2 }$

$\iff \sf{ f\Big( \dfrac{1}{128}\Big)~=~ -\dfrac{1}{7}\log_{2}2 ~=~ -\dfrac{1}{7}*1 }$

$\iff ~~~ \boxed{ \sf{ f\Big( \dfrac{1}{128}\Big)~=~ -\dfrac{1}{7} } }$

• Agora vamos achar a g(125)

$\iff \sf{ g(125) ~=~ \log_{125}5 }$

$\iff \sf{ g(125) ~=~ \log_{5^3}5 }$

$\iff \sf{ g(125) ~=~ \dfrac{1}{3}\log_{5}5 ~=~\dfrac{1}{3}*1 }$

$\iff ~~~\boxed{ \sf{ g(125)~=~ \dfrac{1}{3} } }$

• Então vamos ter que :

$\iff \sf{ \dfrac{ f\Big(\frac{1}{128}\Big) - g(125) }{2} ~=~ \dfrac{ -\frac{1}{7} - \frac{1}{3} }{2} }$

$\iff \sf{ \dfrac{ f\Big(\frac{1}{128}\Big) - g(125) }{2}~=~ \dfrac{ \frac{-3 - 7 }{7*3} }{2} }$

$\iff \sf{ \dfrac{ f\Big(\frac{1}{128}\Big) - g(125) }{2}~=~ \dfrac{ -\frac{10}{21} }{2}~=~-\dfrac{10}{21}*\dfrac{1}{2} }$

$\green{ \iff \boxed{ \sf{ \dfrac{ f\Big(\frac{1}{128}\Big) - g(125) }{2}~=~ - \dfrac{5}{21} } \sf{ \longleftarrow Resposta } } }$

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Espero ter ajudado bastante!)

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