log2^(log3^x)=log3^(log4^y)=0则x+y=

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log2^(log3^x)=log3^(log4^y)=0则x+y=log2^(log3^x)=log3^(log4^y)=0则x+y=log2^(log3^x)=log3^(log4^y)=0则x+

log2^(log3^x)=log3^(log4^y)=0则x+y=
log2^(log3^x)=log3^(log4^y)=0则x+y=

log2^(log3^x)=log3^(log4^y)=0则x+y=


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好像弄错了,log2^(log3^x)中2是底数吧?如果是这样应该这么算:
log2^(log3^x)=0,则log3^x=1,x=3
log3^(log4^y)=0,则log4^y=1,y=4
x+y=7