If log_2(log_3(log_4x))=0, log_3(log_4(log_2y))=0, log_4(log_2(log_3z))=0, then x+y+z is equal to:

1971 AMC 12 Problem 21

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21.

\begin{gathered} \log_2(\log_3(\log_4x))=0,\\ \log_3(\log_4(\log_2y))=0,\\ \log_4(\log_2(\log_3z))=0, \end{gathered}

then

x+y+z

is equal to:

50

58

89

111

1296

Undo each logarithm from the outside inward

For the first chain, the successive inner values are

1,3,

and then

x=4^3

Undoing the logarithms from the outside inward gives

\begin{gathered} x=4^3=64,\\ y=2^4=16,\\ z=3^2=9. \end{gathered}

Therefore

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

Therefore, the correct answer is C.