Question

dados os valores de log2 = 0.3 , log3 = 0,48 , log5 = 0,7 e log7 = 0,84. Determine o valor de:

a) log12 e) log200 i) log4,8
b) log49 f) log5[7] j) log7,5
c) log108 g) log3[3] k) log10,5
c) log120 h) log0,03

Os números que estão assim [3], [7] são raiz quadrada

Answer 1

Foi nos dado as informações:

• log (2) = 0,3
• log (3) = 0,48
• log (5) = 0,7
• log (7) = 0,84

Lembrando que quando a base não aparece, assumimos como base 10, ∴ log (a) ⇔ log₁₀ (a).

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Agora vamos determinar o que cada alternativa pede. Mas antes confira algumas das propriedades dos logaritmos usadas aqui:

• logₐ (bᶜ) ⇔ c . logₐ (b)
• logₐ (b.c) ⇔ logₐ (b) + logₐ (c)
• logₐ (b) ⇔ [logₓ (b)]/[logₓ (a)]
• logₐ (a) ⇔ 1
• logₐ (b/c) ⇔ logₐ (b) – logₐ (c)

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Letra A

$\begin{array}{l}\sf log~(12)=log~(2^2\cdot3)\\\\\sf log~(12)=log~(2^2)+log~(3)\\\\\sf log~(12)=2\cdot log~(2)+0,48\\\\\sf log~(2)=2\cdot0,3+0,48\\\\\sf log~(2)=0,6+0,48\\\\\!\boxed{\sf log~(2)\,\approx\,1,08}\end{array}$

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Letra B

$\begin{array}{l}\sf log~(49)=log~(7^2)\\\\\sf log~(49)=2\cdot log~(7)\\\\\sf log~(49)=2\cdot0,84\\\\\!\boxed{\sf log~(49)\,\approx\,1,68}\end{array}$

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Letra C

$\begin{array}{l}\sf log~(108)=log~(2^2\cdot3^3)\\\\\sf log~(108)=log~(2^2)+log~(3^3)\\\\\sf log~(108)=2\cdot log~(2)+3\cdot log~(3)\\\\\sf log~(108)=2\cdot0,3+3\cdot0,48\\\\\sf log~(108)=0,6+1,44\\\\\!\boxed{\sf log~(108)\,\approx\,2,04}\end{array}$

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Letra D

$\begin{array}{l}\sf log~(120)=log~(2^3\cdot3\cdot5)\\\\\sf log~(120)=log~(2^3)+log~(3)+log~(5)\\\\\sf log~(120)=3\cdot log~(2)+0,48+0,7\\\\\sf log~(120)=3+0,3+1,18\\\\\sf log~(120)=0,9+1,18\\\\\!\boxed{\sf log~(120)\,\approx\,2,78}\end{array}$

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Letra E

$\begin{array}{l}\sf log~(200)=log~(2^3\cdot5^2)\\\\\sf log~(200)=log~(2^3)+log~(5^2)\\\\\sf log~(200)=3\cdot log~(2)+2\cdot log~(5)\\\\\sf log~(200)=3\cdot0,3+2\cdot0,7\\\\\sf log~(200)=0,9+1,4\\\\\!\boxed{\sf log~(200)\,\approx\,2,3}\end{array}$

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Letra F

Como foi dito por você, [7] é a raiz quadrada de sete.

$\begin{array}{l}\sf log_5~(\sqrt{7})=log_5~(7^{\frac{1}{2}})\\\\\sf log_5~(\sqrt{7})=\dfrac{1}{2}\cdot log_5~(7)\\\\\sf log_5~(\sqrt{7})=\dfrac{1}{2}\cdot \dfrac{log~(7)}{log~(5)}\\\\\sf log_5~(\sqrt{7})=\dfrac{1}{2}\cdot\dfrac{0,84}{0,7}\\\\\sf log_5~(\sqrt{7})=\dfrac{0,84}{1,4}\\\\\!\boxed{\sf log_5~(\sqrt{7})\,\approx\,0,6}\end{array}$

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Letra G

Como foi dito por você, [3] é a raiz quadrada de três.

$\begin{array}{l}\sf log_3~(\sqrt{3})=log_3~(3^{\frac{1}{2}})\\\\\sf log_3~(\sqrt{3})=\dfrac{1}{2}\cdot log_3~(3)\\\\\sf log_3~(\sqrt{3})=\dfrac{1}{2}\cdot1\\\\\!\boxed{\sf log_3~(\sqrt{3})\,\approx\,0,5}\end{array}$

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Letra H

$\begin{array}{l}\sf log~(0,03)=log~\bigg(\dfrac{3}{100}\bigg)\\\\\sf log~(0,03)=log~(3)-log~(100)\\\\\sf log~(0,03)=0,48-log~(10^2)\\\\\sf log~(0,03)=0,48-2\cdot log~(10)\\\\\sf log~(0,03)=0,48-2\cdot1\\\\\!\boxed{\sf log~(0,03)\,\approx\,-1,52}\end{array}$

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Letra I

$\begin{array}{l}\sf log~(4,8)=log~\bigg(\dfrac{48}{10}\bigg)\\\\\sf log~(4,8)=log~(48)-log~(10)\\\\\sf log~(4,8)=log~(2^4\cdot3)-1\\\\\sf log~(4,8)=log~(2^4)+log~(3)-1\\\\\sf log~(4,8)=4\cdot log~(2)+0,48-1\\\\\sf log~(4,8)=4\cdot0,3+0,48-1\\\\\sf log~(4,8)=1,2+0,48-1\\\\\!\boxed{\sf log~(4,8)\,\approx\,0,68}\end{array}$

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Letra J

$\begin{array}{l}\sf log~(7,5)=log~\bigg(\dfrac{75}{10}\bigg)\\\\\sf log~(7,5)=log~(75)-log~(10)\\\\\sf log~(7,5)=log~(3\cdot5^2)-1\\\\\sf log~(7,5)=log~(3)+log~(5^2)-1\\\\\sf log~(7,5)=0,48+2\cdot log~(5)-1\\\\\sf log~(7,5)=0,48+2\cdot0,7-1\\\\\sf log~(7,5)=0,48+1,4-1\\\\\!\boxed{\sf log~(7,5)\,\approx\,0,88}\end{array}$

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Letra K

$\begin{array}{l}\sf log~(10,5)=log~\bigg(\dfrac{105}{10}\bigg)\\\\\sf log~(10,5)=log~(105)-log~(10)\\\\\sf log~(10,5)=log~(3\cdot5\cdot7)-1\\\\\sf log~(10,5)=log~(3)+log~(5)+log~(7)-1\\\\\sf log~(10,5)=0,48+0,7+0,84-1\\\\\!\boxed{\sf log~(10,5)\,\approx\,1,02}\end{array}$

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Att. Nasgovaskov