Question

(UNITAU-SP) Sabendo-se que A= log 3^raiz quadrada de 7 elevado a 7+ log0,001, é correto afirmar que:
A) A=7
B) A= 3
C) A= 1
D) A= 0
E) A= -1

Answer 1

Explicação passo-a-passo:

• $\sf log_{\sqrt[3]{7}}~7=x$

$\sf (\sqrt[3]{7})^x=7$

$\sf (7^{\frac{1}{3}})^x=7^1$

$\sf 7^{\frac{x}{3}}=7^1$

Igualando os expoentes:

$\sf \dfrac{x}{3}=1$

$\sf x=3\cdot1$

$\sf \red{x=3}$

• $\sf log~0,001=y$

$\sf 10^y=0,001$

$\sf 10^y=\dfrac{1}{1000}$

$\sf 10^y=\dfrac{1}{10^3}$

$\sf 10^y=10^{-3}$

Igualando os expoentes:

$\sf \red{y=-3}$

Logo:

$\sf A=log_{\sqrt[3]{7}}~7+log~0,001$

$\sf A=3-3$

$\sf \red{A=0}$

Letra D

Answer 2

Oie, Td Bom?!

■ Resposta: Opção D.

$A = log_{ \sqrt[3]{7} }(7) + log_{10}(0.001)$

$A = log_{7 {}^{ \frac{1}{3} } }(7) + log_{10}( \frac{1}{1000} )$

$A = log_{7 {}^{ \frac{1}{3} } }(7) + log_{10}( \frac{1}{10 {}^{3} } )$

$A = log_{7 {}^{ \frac{1}{3} } }(7) + log_{10}(10 {}^{ - 3} )$

$A = \frac{1}{ \frac{1}{3} } \: . \: log_{7}(7) - 3 log_{10}(10)$

$A = \frac{1}{ \frac{1}{3} } \: . \: 1 - 3 \: . \: 1$

$A = 3 - 3$

$A = 0$

Att. Makaveli1996