Question

Mackenzie 1996 Se (2^(x).k^(y+1).5(t+3)).(2^(x-1).k^(y).5^(t+1)) = 150, então k vale:
quero os calculos

Answer 1

$2 ^{x}.k ^{y+1} .5 ^{t+3} .2 ^{x-1}.k[tex] ^{y}$ k ^y [/tex].5 ^{t+1} =150 \\ 2 ^{2x-1} .k ^{2y+1} .5 ^{2t+4} =2.3.5 ^{2} \\ k=3[/tex]