Question

(Função Logarítmica) Se (2 raiz de 2)^x= 64, o valor de log de base 1/8 x é:


resposta: -2/3

Answer 1

$(2\sqrt2)^x=64\\\\ (2\cdot2^{\frac1 2})^x=2^6\\\\ (2^{1+\frac1 2})^x=2^6\\\\ (2^{\frac3 2})^x=2^6\\\\ 2^{\frac3 2 x}=2^6$

Como as bases são iguais, podemos igualar os expoentes:

$\dfrac{3}{2}x=6\\\\ 3x=2\cdot6\\\\ 3x=12\\\\ x=\dfrac{12}{3}\\\\ \boxed{x=4}$

Agora vamos calcular o log de x na base 1/8, que chamaremos de y:

$y=\log_{\frac1 8}x\\\\ y=\log_{\frac 1 8}4\\\\ \left(\dfrac{1}{8}\right)^y=4\\\\ (8^{-1})^y=2^2\\\\ (2^3)^{-y}=2^2\\\\ 2^{-3y}=2^2$

Como as bases são iguais, podemos igualar os expoentes:

$-3y=2\\\\ 3y=-2\\\\ \boxed{y=-\dfrac{2}{3}}$