2021 AMC 12B Spring Problem 21

Below is the professionally curated solution for Problem 21 of the 2021 AMC 12B Spring, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Spring solutions, or check the answer key.

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Concepts:logarithmbounding to limit cases

Difficulty rating: 2260

21.

Let SS be the sum of all positive real numbers xx for which x22=22x.x^{2^{\sqrt2}}=\sqrt2^{\,2^x}.

Which of the following statements is true?

S<2S\lt\sqrt2

S=2S=\sqrt2

2<S<2\sqrt2\lt S\lt 2

2S<62\le S\lt 6

S6S\ge 6

Solution:

Taking log2,\log_2, the equation becomes 22log2x=2x1.2^{\sqrt2}\log_2 x=2^{x-1}. Substituting x=2x=\sqrt2 gives 2212=221,2^{\sqrt2}\cdot\tfrac12=2^{\sqrt2-1}, which holds, so x=2x=\sqrt2 is a solution.

Let f(x)=2x122log2x.f(x)=2^{x-1}-2^{\sqrt2}\log_2 x. Then f(1)>0,f(1)\gt 0, f(2)=0,f(\sqrt2)=0, f(2)<0,f(2)\lt 0, and f(4)>0,f(4)\gt 0, so there is a second root x0x_0 between 22 and 4.4.

Since ff has no other sign changes, there are exactly two solutions, and S=2+x01.41+3.14.5,S=\sqrt2+x_0\approx 1.41+3.1\approx 4.5, which lies in [2,6).[2,6).

Thus, the correct answer is D.

Problem 21 in Other Years