设3^x=4^y=6^z=k.(k>1)
x=log3(k)
y=log4(k)
z=log6(k)
1/z-1/x=logk(6)-logk(3)=logk(2)
1/(2y)=1/2*logk(4)=logk(2)
故:1/z-1/x=1/(2y)
设3^x=4^y=6^z=k.(k>1)
x=log3(k)
y=log4(k)
z=log6(k)
1/z-1/x=logk(6)-logk(3)=logk(2)
1/(2y)=1/2*logk(4)=logk(2)
故:1/z-1/x=1/(2y)