Question

necessito muito esboce o cálculo!!!!!

Exercícios:
Determine o valor das expressões abaixo aplicando as propriedades do
logaritmos
a) log 2 (8 ∙ 4)=

b)log₃ (81/9)=

c) log₂8²=

d) log₂ √16

Answer 1

Resposta:

Logaritmos:

a)

$log_{2}(8 \times 4) = \\ \\ log_{2}(8) + log_{2}(4) = \\ \\ log_{2}( {2}^{3} ) + log_{2}( {2}^{2} ) = \\ \\ 3 \times log_{2}(2) + 2 \times log_{2}(2) = \\ \\ 3 \times 1 + 2 \times 1 = \\ \\ 3 + 2 = \\ \\ 5$

b)

$log_{3}( \frac{81}{9} ) = log_{3}(9) = \\ \\ log_{3}( {3}^{2} ) = \\ \\ 2 \times log_ {3}(3) = \\ \\ 2 \times 1 = \\ \\ 2\\ \\ou\\\\log_{3}(81)\:-\: log_{3}(9) =\\\\ log_{3}({3}^{4} )\:-\:log_{3}({3}^{2} )=\\\\ 4×log_{3}(3)\:-\:2×log_{3}(3)=\\\\4×1\:-\:2×1 =\\\\4\:- 2 = \\\\2$

c)

$log_{2}( {8}^{2} ) = 2 \times log_{2}( {2}^{3} ) = \\ \\ 3\times 2 \times log_{2}(2) = \\ \\ 6 \times 1 = \\ \\ 6$

d)

$log_{2}( \sqrt{16} ) = log_{2}(4) = \\ \\ log_{2}( {2}^{2} ) = \\ \\ 2 \times log_{2}(2) = \\ \\ 2 \times 1 = \\ \\ 2$

Bons Estudos!

Answer 2

Explicação passo-a-passo:

Lembre-se que:

• $\sf log_{b}~(a\cdot b)=log_{b}~a+log_{b}~c$

• $\sf log_{b}~\Big(\dfrac{a}{c}\Big)=log_{b}~a-log_{b}~c$

• $\sf log_{b}~a^m=m\cdot log_{b}~a$

• $\sf log_{a}~a=1$

a)

$\sf log_{2}~(8\cdot4)=log_{2}~8+log_{2}~4$

$\sf log_{2}~(8\cdot4)=log_{2}~2^3+log_{2}~2^2$

$\sf log_{2}~(8\cdot4)=3\cdot log_{2}~2+2\cdot log_{2}~2$

$\sf log_{2}~(8\cdot4)=3\cdot1+2\cdot1$

$\sf log_{2}~(8\cdot4)=3+2$

$\sf \large\red{log_{2}~(8\cdot4)=5}$

b)

$\sf log_{3}~\Big(\dfrac{81}{9}\Big)=log_{3}~81-log_{3}~9$

$\sf log_{3}~\Big(\dfrac{81}{9}\Big)=log_{3}~3^4-log_{3}~3^2$

$\sf log_{3}~\Big(\dfrac{81}{9}\Big)=4\cdot log_{3}~3-2\cdot log_{3}~3$

$\sf log_{3}~\Big(\dfrac{81}{9}\Big)=4\cdot1-2\cdot1$

$\sf log_{3}~\Big(\dfrac{81}{9}\Big)=4-2$

$\sf \large\red{log_{3}~\Big(\dfrac{81}{9}\Big)=2}$

c)

$\sf log_{2}~8^2=2\cdot log_{2}~8$

$\sf log_{2}~8^2=2\cdot log_{2}~2^3$

$\sf log_{2}~8^2=3\cdot 2\cdot log_{2}~2$

$\sf log_{2}~8^2=3\cdot 2\cdot1$

$\sf \large\red{log_{2}~8^2=6}$

d)

$\sf log_{2}~\sqrt{16}=log_{2}~4$

$\sf log_{2}~\sqrt{16}=log_{2}~2^2$

$\sf log_{2}~\sqrt{16}=2\cdot log_{2}~2$

$\sf log_{2}~\sqrt{16}=2\cdot1$

$\sf \large\red{log_{2}~\sqrt{16}=2}$