Question

6) Considere log 2 0,3010 , log 3 0,4771 e calcule usando as propriedades dos logaritmos:
a) log 8 b) log 72
c) log√2 d) log 5
e) log 3000 f) log 0,0001

Answer 1

Temos que: log 2 = 0,3010 e log 3 = 0,4771

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Algumas propriedades dos logaritmos:

• log a^b <=> b.log a
• log a.b <=> log a + log b
• log a/b <=> log a – log b
• logₐ a <=> 1

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Letra A)

$\begin{array}{l}\sf =\log~8\\\\\sf \log~8=\log~2^3\\\\\sf \log~8=3\cdot\log~2\\\\\sf \log~8=3\cdot0,3010\\\\\!\boxed{\sf \log~8\,\approx\, 0,903}\end{array}$

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Letra B)

$\begin{array}{l}\sf =\log~72\end{array}$

MMC (72) = 2.2.2.3.3 = 2³ . 3², assim:

$\begin{array}{l}\sf \log~72=\log~(2^3\cdot3^2)\\\\\sf \log~72=\log~2^3+\log~3^2\\\\\sf \log~72=3\cdot\log~2+2\cdot\log~3\\\\\sf \log~72=3\cdot0,3010+2\cdot0,4771\\\\\sf \log~72=0,903+0,9542\\\\\!\boxed{\sf \log~72\,\approx\,1,8572}\end{array}$

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Letra C)

$\begin{array}{l}\sf =\log~\sqrt{2}\\\\\sf \log~\sqrt{2}=\log~2^{\frac{1}{2}}\\\\\sf \log~\sqrt{2}= \dfrac{1}{2}\cdot\log~2\\\\\sf \log~\sqrt{2}=\dfrac{1}{2}\cdot0,3010\\\\\sf \log~\sqrt{2}=\dfrac{0,3010}{2}\\\\\!\boxed{\sf \log~\sqrt{2}\,\approx\,0,1505}\end{array}$

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Letra D)

$\begin{array}{l}\sf =log~5\\\\\sf \log~5=\log~\dfrac{10}{2}\\\\\sf \log~5=\log~10-\log~2\end{array}$

Lembre que, como a base não aparece, é um logaritimo decimal, portanto a base é 10. Assim log₁₀ 10 = 1:

$\begin{array}{l}\sf \log~5=1-0,3010\\\\\!\boxed{\sf \log~5\,\approx\,0,699}\end{array}$

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Letra E)

$\begin{array}{l}\sf =\log~3000\\\\\sf \log~3000=\log~(1000\cdot3)\\\\\sf \log~3000=\log~1000+\log~3\\\\\sf \log~3000=\log~10^3+0,4771\\\\\sf \log~3000=3\cdot\log~10+0,4771\\\\\sf \log~3000= 3\cdot1+0,4771\\\\\sf \log~3000=3+0,4771\\\\\!\boxed{\sf \log~3000\,\approx\,3,4771}\end{array}$

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Letra F)

$\begin{array}{l}\sf =\log~0,0001\\\\\sf \log~0,0001=\log~\dfrac{1}{10000}\\\\\sf \log~0,0001=\log~\dfrac{1}{10^4}\\\\\sf \log~0,0001=\log~(10^4)^{-1}\\\\\sf \log~0,0001=\log~10^{-4}\\\\\sf \log~0,0001=-\:4\cdot\log~10\\\\\sf \log~0,0001=-\:4\cdot1\\\\\!\boxed{\sf \log~0,0001=-\:4}\end{array}$

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Att. Nasgovaskov