Question

Se x > 0 e x != 1 e logx 7 = - 1/3, calcule log1/x 7^6​

Answer 1

Explicação passo-a-passo:

Lembre-se que:

• $\sf log_{b^m}~a~=\dfrac{1}{m}\cdot log_{b}~a$

• $\sf log_{b}~a^n=n\cdot log_{b}~a$

Assim:

$\sf E=log_{\frac{1}{x}}7^6$

$\sf E=log_{x^{-1}}~7^6$

$\sf E=\dfrac{1}{-1}\cdot log_{x}~7^6$

$\sf E=(-1)\cdot log_{x}~7^6$

$\sf E=6\cdot(-1)\cdot log_{x}~7$

$\sf E=6\cdot(-1)\cdot\Big(-\dfrac{1}{3}\Big)$

$\sf E=\dfrac{6}{3}$

$\sf \red{E=2}$

Answer 2

• Utilize as propriedades básicas dos logaritmos.

logc b = a, então c^a = b.

logc^a b = 1/a . logc b

logx 7 = - 1/3

log1/x 7^6

- 6 . logx 7

- 6 . (- 1/3)

6/3

2

Resposta: 2