Question

x + y + z quando x = log 0,0001 , y = log√2 64 e z = log2 0,125.

Answer 1

x = log 0,0001 >> 10^x = 10^-4, portanto x=-4

y=log√2 64      >>  √2^y = √2^12, portanto y=12

z= log2 0,125  >>  2^z= 125/1000 = 1/8 = 2^-3, >>>>  2^z=2^-3, portanto, z=-3

x+y+z= -4+12-3= 5

Resposta: 5.