题目内容

设x,y,z∈R+,且3x=4y=6z
(1)求证:
1
z
-
1
x
=
1
2y

(2)比较3x,4y,6z的大小.
考点:不等式比较大小,有理数指数幂的化简求值
专题:函数的性质及应用
分析:(1)由于x,y,z∈R+,且3x=4y=6z=k>1,可得x=
lgk
lg3
y=
lgk
lg4
,z=
lgk
lg6
.即可证明;
(2)由于3x=
lgk
lg
33
,4y=
lgk
lg
2
,6z=
lgk
lg
66
.lgk>0,可得
33
=
69
68
=
2
66
>1,即可得出.
解答: (1)证明:∵x,y,z∈R+,且3x=4y=6z=k>1,
∴x=
lgk
lg3
y=
lgk
lg4
,z=
lgk
lg6

1
z
-
1
x
=
lg6
lgk
-
lg3
lgk
=
lg2
lgk

1
2y
=
lg4
2lgk
=
lg2
lgk

1
z
-
1
x
=
1
2y

(2)解:3x=
3lgk
lg3
=
lgk
lg
33

4y=
4lgk
lg4
=
lgk
lg
2

6z=
6lgk
lg6
=
lgk
lg
66

∵k>1,∴lgk>0,
33
=
69
68
=
2
66
>1,
∴3x<4y<6z.
点评:本题考查了指数式与对数的运算性质,考查了推理能力与计算能力,属于中档题.
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