若log2（log3(log4x)=log3（log4(log2y)=log4（log2(log3z)=0则x+y-z=

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若log2（log3(log4x)=log3（log4(log2y)=log4（log2(log3z)=0则x+y-z=多少
我不会打中括号所以就小括号代替了,

log2（log3(log4x)=0
（log3(log4x)=2^0=1
log4x=3^1=3
x=4^3=64
log3（log4(log2y)=0
log4(log2y)=3^0=1
log2y=4^1=4
y=2^4=16
log4（log2(log3z)=0
log2(log3z)=4^0=1
log3z=2^1=2
z=3^2=9
所以x+y-z=64+16-9=71

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