AMC 10 · 2010 · #12

Grade 11 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

logarithm-propertiesexponentslinear-equations-one-var pattern-recognitioneasier-related-problem ↑ Prerequisites: logarithm-properties

📏 Long solution 💡 2 insights

Problem

Five logarithms with matching bases and arguments add to a fixed total. Find the unknown.

Pick an answer.

(A)

(B)

(C)

(D)

256

(E)

1024

How to solve

Strategy Organize Information in More Ways

Tool #15 (Organize Information in More Ways): the five terms look like five different problems only because they are written in five different notations. Every base is a power of 2 and every argument is a power of x, so rewriting all ten numbers that way puts the terms into one common currency where they can be compared. Tool #5 (Look for a Pattern): once rewritten, a pairing jumps out — in each term the exponent on the base and the exponent on x are the same number. That repeated pairing, not the particular exponents 1/2,1,2,3,4, is what the problem is built on. Tool #9 (Solve an Easier Related Problem): rather than run five separate change-of-base computations, prove one small general fact — raising the base and the argument to the same power leaves a logarithm unchanged — and then apply it five times. Proving it also shows exactly why the messy exponents disappear. Tool #4 (Introduce a Variable): naming t=log₂x turns the whole left side into 5t and reduces the problem to a one-step linear equation. Tool #3 (Eliminate Possibilities): the domain restriction rules out non-positive x before any algebra starts, and at the end the same style of check disposes of the four wrong answer choices at once.

1STEP 1

Decide which x are even allowed

The logarithms fix which values are even allowed.

√(x) and log₂x defined⇔ x ＞ 0; √(2), 2, 4, 8, 16 ＞ 0 and ≠ 1

2STEP 2

Write every base and argument as a power

Every base and argument is a power of two.

log₂¹/2x¹/2+log₂¹x¹+log₂²x²+log₂³x³+log₂⁴x⁴=40

3STEP 3

Notice the exponents match in pairs

The two exponents match in every term.

each term = log₂^kx^k, k∈{1/2, 1, 2, 3, 4}

4STEP 4

Prove the matched exponents cancel

Matched exponents cancel completely.

b^t=y ⟹ (b^k)^t=b^tk=(b^t)^k=y^k ⟹ log_b^ky^k=t=log_by

5STEP 5

Collapse to one linear equation

Five equal terms collapse to one linear equation.

5t=40 ⟹ t=8, where t=log₂x

6STEP 6

Undo the logarithm and clear the choices

Undoing the logarithm gives 256, choice (D).

log₂x=8⇔ x=2⁸=256 (D)

Answer

256

Check x=256 against the original equation without using the identity at all, term by term, by asking each logarithm's question directly. (√(2))⁸=2⁴=16=√(256), so the first term is 8. 2⁸=256, so the second is 8. 4⁸=2¹⁶=65536=256², so the third is 8. 8⁸=2²⁴=256³, so the fourth is 8. 16⁸=2³²=256⁴, so the fifth is 8. The total is 5 × 8=40, exactly as required — sufficiency confirmed by hand. Uniqueness holds too, and for a reason independent of the algebra: each of the five bases is greater than 1, so each term is a strictly increasing function of x on (0,∞), and a sum of strictly increasing functions is strictly increasing. A strictly increasing function hits the value 40 at most once, which justifies the problem's phrase "the value of x." The wrong choices are all powers of 2 with exponents 3,4,5,10, and since the left side equals 5log₂x they give 15,20,25,50 — every one of them is what a solver lands on after ending up with the wrong number of copies of log₂x. Dropping a single term, for instance, leaves 4log₂x=40 and points at 1024; the trap is miscounting the copies, not mishandling the exponents.

💡Key takeaway

Each term raises the base and the inside by the same power, and equal raising always cancels, so the whole pile is just five copies of log₂x.

Decide which x are even allowed
Write every base and argument as a power
Notice the exponents match in pairs
Prove the matched exponents cancel
Collapse to one linear equation
Undo the logarithm and clear the choices