Question

1º)Sabendo que log3(7x – 1) = 4 e que log2(y2 -36) = 6, calcule x e Y​

Answer 1

Resposta:

C.E: 7x - 1 > 0 e y² - 36 > 0

$log_ab=c\implies b=a^c\\\\7x-1=3^4\\\\7x=81+1\\\\7x=82\\\\x=\frac{82}{7} \\\\y^2-36=2^6\\\\y^2=64+36\\\\y^2=100\\\\y=10~~ou~~y=-10$