AMC 10 · 2021 · #21

Grade 11 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In
logarithm-propertiesexponential-functionsign-analysisgraph-reading convert-to-algebraguess-and-checkbound-inequality-then-enumerate ↑ Prerequisites: logarithm-propertiesexponential-function
📏 Long solution 💡 4 insights
Problem
One equation has exponents stacked on top of exponents on both sides. Every positive real that satisfies it counts; add them all together. Instead of computing that sum exactly, decide which of five interval statements about it is true.

Pick an answer.

(A)
$S<\sqrt2$
(B)
$S=\sqrt2$
(C)
$\sqrt2<S<2$
(D)
$2\le S<6$
(E)
$S\ge 6$
How to solve
Strategy Organize Information in More Ways

Head-on the equation is a tower of exponents with nothing to cancel, so Tool #15 (Organize Information in More Ways) sets the route: take log₂ of both sides and the whole thing rearranges into log₂ x = 2^ x-1-√2, one slow log curve against one fast exponential curve. In that form Tool #6 (Guess and Check) instantly finds the planted root x=√2, because that value kills the √2 sitting in the exponent. Tool #1 (Draw a Diagram) then answers the counting question — the log side bends down, the exponential side bends up, so the two curves can meet at most twice and there is exactly one more root to find. Tool #3 (Eliminate Possibilities) finishes without ever computing that root: two clean sign checks cage it between 2 and 4, and that cage is already narrow enough to knock out four of the five choices.

1STEP 1

Take logarithms

Logarithms drop the tower by one level.

2^√2log₂ x = 2^xlog₂√2 = 2^x·1/2 = 2^ x-1 ⟹ log₂ x = 2^ x-1-√2
2STEP 2

Read off the planted root

One root is in plain sight.

x=√2: log₂√2=1/2, 2^ √2-1-√2=2⁻¹=1/2
3STEP 3

Show there are exactly two

Concavity limits it to two roots.

h(x)=log₂ x-2^ x-1-√2, h''(x)=-1/(x²ln 2)-(ln 2)² 2^ x-1-√2 ＜ 0
4STEP 4

Cage the second root

A sign change traps the second root.

h(2)=1-2^ 1-√2 ＞ 0, h(4)=2-2^ 3-√2 ＜ 0 ⟹ 2 ＜ r ＜ 4
5STEP 5

Add and pick the interval

The sum lies in at least two and under six.

S=√2+r, 2+√2 ＜ S ＜ 4+√2 ⟹ 3 ＜ S ＜ 6 ⟹ 2 ≤ S ＜ 6 (D)
Answer
2 ≤ S ＜ 6
Tighten the cage to see how much room the answer has. At x=3 the gap is log₂ 3-2^ 2-√2≈ 1.585-1.501 ＞ 0, and at x=3.2 it is log₂ 3.2-2^ 2.2-√2≈ 1.678-1.724 ＜ 0, so the second root is about 3.14 and S≈√2+3.14≈ 4.55. That sits comfortably inside [2,6), far from both edges, so the crude bracket 3 ＜ S ＜ 6 was never in danger of straddling a boundary. The first root also checks out directly in the untouched original equation: at x=√2 both sides read √2^{ 2^√2}.
💡Key takeaway

When exponents are stacked on both sides, take a logarithm to flatten them, then compare a slow log curve against a fast exponential one — often you only need to cage the crossing, not compute it.

  • Take log base 2 of both sides
  • Read off the planted root
  • Show there are exactly two roots
  • Cage the second root between 2 and 4
  • Add the roots and pick the interval

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