Question

(PUC-SP) Se [log na base 2(log na base 3 log na base 4 x) = log na base 3(log na base 4 log na base 2 y) = log na base 4(log na base 2 log na base 3 z) = 0], então x + y + z é:


A. 50

B. 58

C. 89

D. 111

E. 1296

Answer 1

Resposta:

Separando em três equações temos:

$1)log_{2}( log_{3}( log_{4}(x) ) ) = 0 \\ 2)log_{3}( log_{4}( log_{2}(y) ) ) = 0 \\ 3)log_{4}( log_{2}( log_{3}(z) ) ) = 0$

Resolvendo cada uma:

1)

$log_{2}( log_{3}( log_{4}(x) ) ) = 0 \\ log_{3}( log_{4}(x) ) = {2}^{0} \\ log_{3}( log_{4}(x) ) = 1 \\ log_{4}(x) = {3}^{1} \\ log_{4}(x) = 3 \\ x = {4}^{3} \\ x = 64$

2)

$log_{3}( log_{4}( log_{2}(y) ) ) = 0 \\ log_{4}( log_{2}(y) ) = {3}^{0} \\ log_{4}( log_{2}(y) ) = 1 \\ log_{2}(y) = {4}^{1} \\ log_{2}(y) = 4 \\ y= {2}^{4} \\ y = 16$

3)

$log_{4}( log_{2}( log_{3}(z) ) ) = 0 \\ log_{2}( log_{3}(z) ) = {4}^{0} \\ log_{2}( log_{3}(z) ) = 1 \\ log_{3}(z) = {2}^{1} \\ log_{3}(z) = 2 \\ z= {3}^{2} \\ z = 9$

Logo x+y+z=64+16+9=89

Alternativa C)

Answer 2

Resposta: c 89 ok porque base4 lon na base  x

Explicação passo a passo:

estudei