AMC12 2021 B

AMC12 2021 B · Q21

AMC12 2021 B · Q21. It mainly tests Logarithms (rare), Manipulating equations.

Let $S$ be the sum of all positive real numbers $x$ for which\[x^{2^{\sqrt2}}=\sqrt2^{2^x}.\]Which of the following statements is true?

设 $S$ 是所有满足 \[x^{2^{\sqrt2}}=\sqrt2^{2^x}\] 的正实数 $x$ 的和。以下哪项陈述正确？

(A) S<\sqrt2 S<\sqrt2

(B) S=\sqrt2 S=\sqrt2

(D) 2\le S<6 2\le S<6

(E) S\ge 6 S\ge 6

Answer

Correct choice: (D)

正确答案：（D）

Solution

Note that \begin{align*} x^{2^{\sqrt{2}}} &= {\sqrt{2}}^{2^x} \\ 2^{\sqrt{2}} \log_2 x &= 2^{x} \log_2 \sqrt{2}. \end{align*} (At this point we see by inspection that $x=\sqrt{2}$ is a solution.) We simplify the RHS, then take the base-$2$ logarithm for both sides: \begin{align*} 2^{\sqrt{2}} \log_2 x &= 2^{x-1} \\ \log_2{\left(2^{\sqrt{2}} \log_2 x\right)} &= x-1 \\ \sqrt{2} + \log_2 \log_2 x &= x-1 \\ \log_2 \log_2 x &= x - 1 - \sqrt{2}. \end{align*} The RHS is a line; the LHS is a concave curve that looks like a logarithm and has $x$ intercept at $(2,0).$ There are at most two solutions, one of which is $\sqrt{2}.$ But note that at $x=2,$ we have $\log_2 \log_2 {2} = 0 > 2 - 1 - \sqrt{2},$ meaning that the log log curve is above the line, so it must intersect the line again at a point $x > 2.$ Now we check $x=4$ and see that $\log_2 \log_2 {4} = 1 < 4 - 1 - \sqrt{2},$ which means at $x=4$ the line is already above the log log curve. Thus, the second solution lies in the interval $(2,4).$ The answer is $\boxed{\textbf{(D) }2\le S<6}.$

注意到 \begin{align*} x^{2^{\sqrt{2}}} &= {\sqrt{2}}^{2^x} \\ 2^{\sqrt{2}} \log_2 x &= 2^{x} \log_2 \sqrt{2}. \end{align*} （此时我们通过观察发现 $x=\sqrt{2}$ 是一个解。）我们化简右边，然后对两边取以 $2$ 为底的对数： \begin{align*} 2^{\sqrt{2}} \log_2 x &= 2^{x-1} \\ \log_2{\left(2^{\sqrt{2}} \log_2 x\right)} &= x-1 \\ \sqrt{2} + \log_2 \log_2 x &= x-1 \\ \log_2 \log_2 x &= x - 1 - \sqrt{2}. \end{align*} 右边是一条直线；左边是一条凹曲线，看起来像对数，在 $(2,0)$ 处有 $x$ 截距。最多有两个解，其中一个是 $\sqrt{2}$。但注意到在 $x=2$ 时，$\log_2 \log_2 {2} = 0 > 2 - 1 - \sqrt{2}$，意味着对数对数曲线在直线上方，因此它必须在 $x > 2$ 的某点再次与直线相交。现在我们检查 $x=4$，发现 $\log_2 \log_2 {4} = 1 < 4 - 1 - \sqrt{2}$，这意味着在 $x=4$ 时直线已经高于对数对数曲线。因此，第二个解位于区间 $(2,4)$ 内。答案是 $\boxed{\textbf{(D) }2\le S<6}$。

Topics