AMC12 2003 A

AMC12 2003 A · Q24

AMC12 2003 A · Q24. It mainly tests Logarithms (rare), Manipulating equations.

If $a\geq b > 1,$ what is the largest possible value of $\log_{a}(a/b) + \log_{b}(b/a)?$

若 $a\geq b > 1,$ 则 $\log_{a}(a/b) + \log_{b}(b/a)$ 的最大可能值是多少？

(A) -2 -2

(B) 0 0

(D) 3 3

(E) 4 4

Answer

Correct choice: (B)

正确答案：（B）

Solution

Using logarithmic rules, we see that \[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\] \[=2-\left(\log_{a}b+\frac {1}{\log_{a}b}\right)\] Since $a$ and $b$ are both greater than $1$, using AM-GM gives that the term in parentheses must be at least $2$, so the largest possible values is $2-2=0 \Rightarrow \boxed{\textbf{B}}.$ Note that the maximum occurs when $a=b$.

利用对数运算规则，有 \[\log_{a}a-\log_{a}b+\log_{b}b-\log_{b}a = 2-(\log_{a}b+\log_{b}a)\] \[=2-\left(\log_{a}b+\frac {1}{\log_{a}b}\right)\] 由于 $a$ 与 $b$ 都大于 $1$，用 AM-GM 可知括号内的量至少为 $2$，因此最大值为 $2-2=0 \Rightarrow \boxed{\textbf{B}}.$ 注意最大值在 $a=b$ 时取得。

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