Question

Sejam log8 128 = x, log4 64 = y e log2 32 = z, entao x +y +z éigual a : a) 13/2 b) 31/3 c) 13 d) 18 e) 64/3

Answer 1

 

log8 128 = x, log4 64 = y e log2 32 = z, entao x +y +z

 

7/3 + 3 + 5  mmc 3

 

7 + 9 + 15

        3

 

 

31/3

 

letra b   

 

 

 

log8 128 = x,  => 8^x = 128  => 2^3x = 2^7 => 3x = 7  => x=7/3

 

log4 64 = y  => 4^y = 64 => 4^y = 4^3 => y = 3  

 

log2 32 = z =>  2^z = 32 => 2^z = 2^5 => z = 5