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设3^x=4^y=6^z=k.x=log3(k)=1/logk(3),y=log4(k)=1/logk(4),z=log6(k)=1/logk(6)
1,1/x+1/(2y)=logk(3)+logk(4)/2=logk(3)+logk(2)=logk(6)=1/z
所以1/x+1/(2y)=1/z
2,3^(1/3)=3^(4/12)=81^(1/12)
4^(1/4)=4^(3/12)=64^(1/12)
6^(1/6)=6^(2/12)=36^(1/12)
3^(1/3)>4^(1/4)>6^(1/6),logk(3)/3>logk(4)/4>logk(6)/6
3/logk(/3)
1,1/x+1/(2y)=logk(3)+logk(4)/2=logk(3)+logk(2)=logk(6)=1/z
所以1/x+1/(2y)=1/z
2,3^(1/3)=3^(4/12)=81^(1/12)
4^(1/4)=4^(3/12)=64^(1/12)
6^(1/6)=6^(2/12)=36^(1/12)
3^(1/3)>4^(1/4)>6^(1/6),logk(3)/3>logk(4)/4>logk(6)/6
3/logk(/3)
1年前
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