题目内容
已知2a=3b=k(k≠1),且2a+b=ab,则实数k的值为( )
分析:由2a=3b=k(k≠1),知a=log2k,b=log3k,故
=logk2,
=logk3,由2a+b=ab,知
+
=2logk3+logk2=logk18=1,由此能求出k.
| 1 |
| a |
| 1 |
| b |
| 2 |
| b |
| 1 |
| a |
解答:解:∵2a=3b=k(k≠1),
∴a=log2k,b=log3k,
∴
=logk2,
=logk3,
∵2a+b=ab,
∴
+
=2logk3+logk2
=logk9+logk2
=logk18=1,
∴k=18.
故选D.
∴a=log2k,b=log3k,
∴
| 1 |
| a |
| 1 |
| b |
∵2a+b=ab,
∴
| 2 |
| b |
| 1 |
| a |
=logk9+logk2
=logk18=1,
∴k=18.
故选D.
点评:本题考查指数式和对数式的相互转化,是基础题,解题时要认真审题,仔细解答,注意对数性质的灵活运用.
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