Question

log7 49=y e log4 ⅛=z então z y é:
a)-4/3
b)2/3
c)-3
d)-2/3
e)-3/2

Answer 1

$log7^{49} =y$
$7^{y} =49$
$7^y=7^2$
$y=2$

$log4 ^{ \frac{1}{8} }=z$
$4^z= \frac{1}{8}$
$(2^2)^z = 8^{-1}$
$2^{2z} =(2^3) ^{-1}$
$2z=-3$
$z= \frac{-3}{2}$

$z.y = \frac{-3}{2} .2 = \frac{-6}{2} =-3$

$Resposta:c)$