Matemática, perguntado por Usuário anônimo, 9 meses atrás

Resolva os log em anexo ​

Anexos:

Soluções para a tarefa

Respondido por ShinyComet
5

3. -

a)  \log_{\:125}25=x\Leftrightarrow

\Leftrightarrow125^x=25\Leftrightarrow

\Leftrightarrow\left(5^3\right)^x=5^2\Leftrightarrow

\Leftrightarrow5^{3x}=5^2\Leftrightarrow

\Leftrightarrow 3x=2\Leftrightarrow

\Leftrightarrow x=\dfrac{2}{3}

b)  \log_{\:\frac{1}{4}}32=x\Leftrightarrow

\Leftrightarrow\left(\dfrac{1}{4}\right)^x=32\Leftrightarrow

\Leftrightarrow\left(\dfrac{1}{2^2}\right)^x=2^5\Leftrightarrow

\Leftrightarrow\left(2^{-2}\right)^x=2^5\Leftrightarrow

\Leftrightarrow 2^{-2x}=2^5\Leftrightarrow

\Leftrightarrow-2x=5\Leftrightarrow

\Leftrightarrow x=-\dfrac{5}{2}

c)  \log_{\:9}\dfrac{1}{27}=x\Leftrightarrow

\Leftrightarrow9^x=\dfrac{1}{27}\Leftrightarrow

\Leftrightarrow\left(3^2\right)^x=\dfrac{1}{3^3}\Leftrightarrow

\Leftrightarrow3^{2x}=3^{-3}\Leftrightarrow

\Leftrightarrow2x=-3\Leftrightarrow

\Leftrightarrow x=-\dfrac{3}{2}

4. -

a)  \log_{\:0,25}8=x\Leftrightarrow

\Leftrightarrow\log_{\:\frac{25}{100}}8=x\Leftrightarrow

\Leftrightarrow\log_{\:\frac{1}{4}}8=x\Leftrightarrow

\Leftrightarrow\left(\dfrac{1}{4}\right)^x=8\Leftrightarrow

\Leftrightarrow\left(\dfrac{1}{2^2}\right)^x=2^3\Leftrightarrow

\Leftrightarrow\left(2^{-2}\right)^x=2^3\Leftrightarrow

\Leftrightarrow2^{-2x}=2^3\Leftrightarrow

\Leftrightarrow-2x=3\Leftrightarrow

\Leftrightarrow x=-\dfrac{3}{2}

b)  \log_{\:25}0,008=x\Leftrightarrow

\Leftrightarrow\log_{\:25}\dfrac{8}{1000}=x\Leftrightarrow

\Leftrightarrow\log_{\:25}\dfrac{1}{125}=x\Leftrightarrow

\Leftrightarrow25^x=\dfrac{1}{125}\Leftrightarrow

\Leftrightarrow\left(5^2\right)^x=\dfrac{1}{5^3}\Leftrightarrow

\Leftrightarrow5^{2x}=5^{-3}\Leftrightarrow

\Leftrightarrow2x=-3\Leftrightarrow

\Leftrightarrow x=-\dfrac{3}{2}

c)  \log_{\:0,01}0,001=x\Leftrightarrow

\Leftrightarrow\log_{\:\frac{1}{100}}\dfrac{1}{1000}=x\Leftrightarrow

\Leftrightarrow\left(\dfrac{1}{100}}\right)^x=\dfrac{1}{1000}\Leftrightarrow

\Leftrightarrow\left(\dfrac{1}{10^2}}\right)^x=\dfrac{1}{10^3}\Leftrightarrow

\Leftrightarrow\left(10^{-2}\right)^x=10^{-3}\Leftrightarrow

\Leftrightarrow10^{-2x}=10^{-3}\Leftrightarrow

\Leftrightarrow-2x=-3\Leftrightarrow

\Leftrightarrow x=\dfrac{-3}{-2}\Leftrightarrow

\Leftrightarrow x=\dfrac{3}{2}

Podes ver mais respostas sobre logaritmos e exponenciais em:

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Anexos:

ShinyComet: Obrigado pela "Melhor Resposta" <3
Respondido por Makaveli1996
1

Oie, Td Bom?!

3.

a)

 log_{125}(25)  = x

 log_{5 {}^{3} }(5 {}^{2} )  = x

 \frac{2}{3}  \: . \:  log_{5}(5)  = x

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

 \frac{2}{3}  = x

x =  \frac{2}{3}

b)

 log_{ \frac{1}{4} }(32)  = x

 log_{2 {}^{ - 2} }(2 {}^{5} )  = x

 \frac{5}{ - 2}  \: . \:  log_{2}(2)  = x

 \frac{5}{ - 2}  \: . \: 1 = x

 -  \frac{5}{2}  = x

x =  -  \frac{5}{2}

c)

 log_{9}( \frac{1}{27} )  = x

 log_{3 {}^{2} }(3 {}^{ - 3} )  = x

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

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

 -  \frac{3}{2}  = x

x =  -  \frac{3}{2}

4.

a)

 log_{0,25}(8)  = x

 log_{2 {}^{ - 2} }(2 { }^{3} )  = x

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

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

  - \frac{3}{2}  = x

x =  -  \frac{3}{2}

b)

 log_{25}(0,008)  = x

 log_{5 {}^{2} }(5 {}^{ - 3} )  = x

 \frac{ - 3}{2}  \: . \:  log_{5}(5)  = x

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

 -  \frac{3}{2}  = x

x =  -  \frac{3}{2}

c)

 log_{0,01}(0,001)  = x

 log_{10 {}^{ - 2} }(10 { }^{ - 3} )  = x

 \frac{ - 3}{ - 2}  \: . \:  log_{10}(10)  = x

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

 \frac{3}{2}  = x

x =  \frac{3}{2}

Att. Makaveli1996

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