Question

Sabendo-se que: logX a=8, logX b=2 e logX c=1 calcular: logx raiz 3 de ab/c

Answer 1

Explicação passo-a-passo:

Equação Logaritmica:

Sabendo que:

Log[x] a = 8 ; Log[x] b = 2 e Log[x] c=1

Então Quanto é:

Log[x] ³√(ab/c)

Para calcular este exercicio , primeiro vamos isolar as variaveis a , b e c .

Log[x] a = 8

x^8 = a

Log[x] b = 2

x^2 = b

Log[x] c = 1

x^1 = c

$\log_{x}\sqrt[3]{\frac{ab}{c}}$

Substituindo ter-se-á:

$\log_{x}\sqrt[3]{\frac{x^8.x^2}{x^1}}$

$\log_{x}\sqrt[3]{\frac{x^{10}}{x^1}}$

$\log_{x}\sqrt[3]{x^{10-1}}$

$\log_{x}\sqrt[3]{x^9}$

$\log_{x}\sqrt[3]{(x^3)^3}$

$\log_{x}x^3$

$\log_{x}x$

$\boxed{\boxed{\mathsf{3.1=3}}}}$$\checkmark$

Espero ter ajudado bastante!)

Answer 2

Explicação passo-a-passo:

Lembre-se que:

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

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

• $\sf log_{b}~a^m=m\cdot log_{b}~a$

• $\sf \sqrt[c]{a^b}=a^{\frac{b}{c}}$

Assim:

$\sf log_{x}~\Big(\dfrac{\sqrt[3]{ab}}{c}\Big)=log_{x}~\Big[\dfrac{(a\cdot b)^{\frac{1}{3}}}{c}\Big]$

$\sf log_{x}~\Big(\dfrac{\sqrt[3]{ab}}{c}\Big)=log_{x}~(a\cdot b)^{\frac{1}{3}}-log_{x}~c$

$\sf log_{x}~\Big(\dfrac{\sqrt[3]{ab}}{c}\Big)=\dfrac{1}{3}\cdot log_{x}~(a\cdot b)-log_{x}~c$

$\sf log_{x}~\Big(\dfrac{\sqrt[3]{ab}}{c}\Big)=\dfrac{1}{3}\cdot(log_{x}~a+log_{x}~b)-log_{x}~c$

$\sf log_{x}~\Big(\dfrac{\sqrt[3]{ab}}{c}\Big)=\dfrac{1}{3}\cdot(8+2)-1$

$\sf log_{x}~\Big(\dfrac{\sqrt[3]{ab}}{c}\Big)=\dfrac{1}{3}\cdot10-1$

$\sf log_{x}~\Big(\dfrac{\sqrt[3]{ab}}{c}\Big)=\dfrac{10}{3}-1$

$\sf log_{x}~\Big(\dfrac{\sqrt[3]{ab}}{c}\Big)=\dfrac{10-3}{3}$

$\sf \red{log_{x}~\Big(\dfrac{\sqrt[3]{ab}}{c}\Big)=\dfrac{7}{3}}$