Question

log (2.4.8.64) base 2

Answer 1

Log(4096) base 2 = X 

4096=2^x 
2^12=2^x
x=12  

Log de  4096 base 2 = 12 

Answer 2

$log_{2}(2.4.8.64) = log_{2}2+ log_{2}4+ log_{2}8+ log_{2}64 \\ \\ log_{2}2+ log_{2}2^{2}+log_{2}2^{3}+log_{2}2^{6} \\ \\ log_{2}2+ 2.log 2+ 3.log_{2}2+ 6log_{2}2 \\ \\ 1 + 2.1 + 3.1 + 6.1 \\ \\ 1 + 2 + 3 + 6 = 12$

Espero ter ajudado.