AMC 10 · 2020 · #10

Grade 11 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps
logarithm-propertiesexponentsquadratic-equationszero-product-propertydigit-sum convert-to-algebrawork-backwards ↑ Prerequisites: logarithm-propertiesexponents
📏 Medium solution 💡 3 insights
Problem
A positive integer satisfies one equation: the left side is the base-two logarithm of the base-sixteen logarithm of the number, and the right side is the base-four logarithm of the base-four logarithm of the number. Exactly one positive integer works. Find the sum of its digits.

Pick an answer.

(A)
4
(B)
7
(C)
8
(D)
11
(E)
13
How to solve
Strategy Introduce a Variable

The equation looks tangled because three different bases appear: 2, 4 and 16. But those bases are not really different, since 4 and 16 are both powers of 2. Naming the single quantity x that is the base-2 logarithm of n replaces n everywhere and rewrites each inner logarithm as a simple multiple of x. After that the outer logarithms can be pushed into one common base, the logarithms cancel, and what is left is a quadratic in x that anyone can factor.

1STEP 1

Name the base-2 logarithm x

Give the base-two logarithm a name.

n = 2^x ⟹ log₁₆ n = x/4, log₄ n = x/2
2STEP 2

Put both sides in base 2

Put both sides in the same base.

log₂ (x/4) = log₄ (x/2) = log₂ (x/2)/(log₂ 4) = 1/2log₂ (x/2)
3STEP 3

Clear the half and drop the logarithms

Clear the half and drop the logarithms.

log₂ (x²/16) = log₂ (x/2) ⟹ x²/16 = x/2
4STEP 4

Factor, then throw out the bad root

One root breaks the logarithm and is thrown out.

x² = 8x ⟹ x(x-8) = 0 ⟹ x = 0 (rejected) or x = 8
5STEP 5

Undo the exponent and add the digits

Undoing the exponent and adding digits gives 13.

log₂ n = 8 ⟹ n = 2⁸ = 256, 2 + 5 + 6 = 13
Answer
13
Put n equal to 256 back into the original equation. The base-16 logarithm of 256 is 2 because 16 squared is 256, and the base-2 logarithm of 2 is 1, so the left side is 1. The base-4 logarithm of 256 is 4 because 4 to the fourth power is 256, and the base-4 logarithm of 4 is 1, so the right side is also 1. Both sides equal 1, so 256 really is a solution. The discarded root x equal to 0 would mean n equal to 1, and then the base-16 logarithm of 1 is 0, which makes the outer base-2 logarithm undefined, so no second solution exists and the uniqueness the problem claims holds. The digit sum is 2 plus 5 plus 6, which is 13, matching choice (E).
💡Key takeaway

When one problem mixes logarithms of base 2, 4 and 16, remember they are all powers of 2: name that single exponent and the tangle straightens into a quadratic.

  • Name the base-2 logarithm x
  • Put both sides in base 2
  • Clear the half and drop the logarithms
  • Factor, then throw out the bad root
  • Undo the exponent and add the digits

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