证明:
(1)
设3^x=4^y=6^z=k
则x=log3(k)=1/logk(3)
y=log4(k)=1/logk(4)
z=log6(k)=1/logk(6)
则1/x+1/(2y)=logk(3)+logk(4)/2
=logk(3)+logk(2)
=logk(6)=1/z
即1/z-1/x=1/2y
(2)由上知,
{3x=3lgk/lg3,
4y=4lgk/lg4,
6z=6lgk/lg6
因k>1,即lgk>0,所以
6/lg6-4/lg4=(lg4^6-lg6^4)/(lg6lg4)>0
--->6z>4y;
4/lg4-3/lg3=(lg3^4-lg4^3)/(lg3lg4)>0
--->4y>3x
因此,3x