lg2^x=lg3^y=lg5^z,x*lg2=y*lg3=z*lg5,x=ylg3/lg2,z=y*lg3/lg5,所以2x=2ylg3/lg2,5z=5y*lg3/lg5,比较2lg3/lg2、5lg3/lg5与3的大小.2lg3/lg2=log2(9)与3=log2（8）显然前者大所以2x大于3y；5lg3/lg5=log5(3^5)3=log5(...