laravieira23
Answered

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considere
log 2 = 0,3
log 3 = 0,4
sabendo que quando não há base é 10:
calcule:
[tex]e) log \: \sqrt[4]{12} \\ f) log \: \sqrt[5]{512} \\ g) log \: 0,018 \: \\ [/tex]
[tex]h) log \: \sqrt[5]{7,2} \\ i) log \: 14,4 \\ j) log_{6} \sqrt{15} [/tex]
ME AJUDA POR FAVOR, TO PRECISANDO PRA CONFERIR MINHAS RESPOSTAS to com medo de errar..usei o photomath mas ele só da valores aproximados ​

Sagot :

Nasgovaskov

Considerando log 2 = 0,3, e log 3 = 0,4 , vamos calcular o valor dos logaritmos

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Lembre-se destas propriedades que iremos usar:

  • log a/b <=> log a – log b
  • log a·b <=> log a + log b
  • log a^b <=> b · log a
  • logₐ a <=> 1
  • logₐ b <=> [ logₓ b ]/[ logₓ a] (para mudança de base)

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Obs.: como o próprio enunciado disse, quando a base não aparece, é 10

Letra E)

[tex]\begin{array}{l}\sf =log~\sqrt[\sf4]{\sf12}\\\\\sf log~\sqrt[\sf4]{\sf12}=log~12^{\frac{1}{4}}\\\\\sf log~\sqrt[\sf4]{\sf12}=\dfrac{1}{4}\cdot log~12\end{array}[/tex]

MMC(12) = 2^2.3, assim:

[tex]\begin{array}{l}\sf log~\sqrt[\sf4]{\sf12}=\dfrac{1}{4}\cdot log~(2^2\cdot3)\\\\\sf log~\sqrt[\sf4]{\sf12}=\dfrac{1}{4}\cdot (2\cdot log~2+ log~3)\\\\\sf log~\sqrt[\sf4]{\sf12}=\dfrac{1}{4}\cdot (2\cdot0,3+0,4)\\\\\sf log~\sqrt[\sf4]{\sf12}=\dfrac{1}{4}\cdot (0,6+0,4)\\\\\sf log~\sqrt[\sf4]{\sf12}=\dfrac{1}{4}\cdot 1\\\\\sf log~\sqrt[\sf4]{\sf12}=\dfrac{1}{4}\\\\\!\boxed{\sf log~\sqrt[\sf4]{\sf12}\,\approx\,0,25}\end{array}[/tex]

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Letra F)

[tex]\begin{array}{l}\sf =log~\sqrt[\sf5]{\sf512}\end{array}[/tex]

MMC (512) =2^5 . 2^4, assim:

[tex]\begin{array}{l}\sf log~\sqrt[\sf5]{\sf512}=log~\sqrt[\sf5]{\sf2^5\cdot2^4}\\\\\sf log~\sqrt[\sf5]{\sf512}=log~2\sqrt[\sf5]{\sf2^4}\\\\\sf log~\sqrt[\sf5]{\sf512}=log~2^1\cdot2^{\frac{4}{5}}\\\\\sf log~\sqrt[\sf5]{\sf512}=log~2^{\frac{1\cdot5+4}{5}}\\\\\sf log~\sqrt[\sf5]{\sf512}=log~2^{\frac{9}{5}}\\\\\sf log~\sqrt[\sf5]{\sf512}= \dfrac{9}{5}\cdot log~2\\\\\sf log~\sqrt[\sf5]{\sf512}=\dfrac{9}{5}\cdot 0,3\\\\\sf log~\sqrt[\sf5]{\sf512}=\dfrac{2,7}{5}\end{array}[/tex]

[tex]\begin{array}{l}\!\boxed{\sf log~\sqrt[\sf5]{\sf512}\,\approx\,0,54}\end{array}[/tex]

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Letra G)

[tex]\begin{array}{l}\sf =log~0,018\\\\\sf log~0,018=log~\dfrac{18}{1000}\\\\\sf log~0,018=log~18-log~1000\\\\\sf log~0,018=log~18-log~10^3\\\\\sf log~0,018=log~18-3\cdot log~10\\\\\sf log~0,018=log~18-3\cdot 1\\\\\sf log~0,018=log~18-3\end{array}[/tex]

MMC (18) = 2. 3^2, logo:

[tex]\begin{array}{l}\sf log~0,018=log~(2\cdot3^2)-3\\\\\sf log~0,018=log~2+2\cdot log~3-3\\\\\sf log~0,018=0,3+2\cdot0,4-3\\\\\sf log~0,018=0,3+0,8-3\\\\\!\boxed{\sf log~0,018\,\approx\,-1,9}\end{array}[/tex]

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Letra H)

[tex]\begin{array}{l}\sf =log~\sqrt[\sf5]{\sf7,2}\\\\\sf log~\sqrt[\sf5]{\sf7,2}=log~(7,2)^{\frac{1}{5}}\\\\\sf log~\sqrt[\sf5]{\sf7,2}=\dfrac{1}{5}\cdot log~\bigg(\dfrac{72}{10}\bigg)\\\\\sf log~\sqrt[\sf5]{\sf7,2}=\dfrac{1}{5}\cdot (log~72-log~10) \end{array}[/tex]

MMC (72) = 2^3 . 3^2, assim:

[tex]\begin{array}{l}\sf log~\sqrt[\sf5]{\sf7,2}=\dfrac{1}{5}\cdot (log~2^3\cdot3^2-1)\\\\\sf log~\sqrt[\sf5]{\sf7,2}=\dfrac{1}{5}\cdot (3\cdot log~2+2\cdot log~3-1)\\\\\sf log~\sqrt[\sf5]{\sf7,2}=\dfrac{1}{5}\cdot (3\cdot0,3+2\cdot0,4-1)\\\\\sf log~\sqrt[\sf5]{\sf7,2}=\dfrac{1}{5}\cdot (0,9+0,8-1)\\\\\sf log~\sqrt[\sf5]{\sf7,2}=\dfrac{1}{5}\cdot (0,7)\\\\\sf log~\sqrt[\sf5]{\sf7,2}=\dfrac{0,7}{5}\\\\\!\boxed{\sf log~\sqrt[\sf5]{\sf7,2}\,\approx\,0,14}\end{array}[/tex]

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Letra I

[tex]\begin{array}{l}\sf =log~14,4\\\\\sf log~14,4=log~\dfrac{144}{10}\\\\\sf log~14,4=log~144-log~10\\\\\sf log~14,4=log~144-1\end{array}[/tex]

MMC (144) = 2^4 . 3^2, assim:

[tex]\begin{array}{l}\sf log~14,4=log~(2^4\cdot3^2)-1\\\\\sf log~14,4=4\cdot log~2+2\cdot log~3-1\\\\\sf log~14,4=4\cdot0,3+2\cdot0,4-1\\\\\sf log~14,4=1,2+0,8-1\\\\\!\boxed{\sf log~14,4\,\approx\,1}\end{array}[/tex]

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Letra J)

[tex]\begin{array}{l}\sf =log_6~\sqrt{15}\\\\\sf log_6~\sqrt{15}=log_6~15^{\frac{1}{2}}\\\\\sf log_6~\sqrt{15}=\dfrac{1}{2}\cdot log_6~15 \end{array}[/tex]

[tex]\begin{array}{l}\sf log_6~\sqrt{15}=\dfrac{1}{2}\cdot\dfrac{log~15}{log~6}\\\\\sf log_6~\sqrt{15}=\dfrac{1}{2}\cdot\dfrac{log~3\cdot5}{log~3\cdot2}\\\\\sf log_6~\sqrt{15}=\dfrac{1}{2}\cdot\dfrac{log~3+log~5}{log~3+log~2}\\\\\sf log_6~\sqrt{15}=\dfrac{1}{2}\cdot\dfrac{0,4+log~\dfrac{10}{2}}{0,4+0,3}\\\\\sf log_6~\sqrt{15}=\dfrac{1}{2}\cdot\dfrac{0,4+log~10-log~2}{0,7}\\\\\sf log_6~\sqrt{15}=\dfrac{1}{2}\cdot\dfrac{0,4+1-0,3}{0,7}\\\\\sf log_6~\sqrt{15}=\dfrac{1}{2}\cdot\dfrac{1,1}{0,7}\end{array}[/tex]

[tex]\begin{array}{l}\sf log_6~\sqrt{15}=\dfrac{1,1}{1,4}\\\\\!\boxed{\sf log_6~\sqrt{15}\,\approx\,0,785}\end{array}[/tex]

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Att. Nasgovaskov

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