AMC 10 · 2022 · #14

Grade 11 algebra
logarithm-propertiessymmetric-polynomialsexponentsquadratic-equations convert-to-algebraidentify-subproblemspattern-recognition ↑ Prerequisites: logarithm-properties
📏 Long solution 💡 3 insights
Problem
A single number is assembled out of base-ten logarithms: cube two of them, add those cubes together, and then add on the product of two others. Find the exact value of the total.

Pick an answer.

(A)
$\frac{3}{2}$
(B)
$\frac{7}{4}$
(C)
2
(D)
$\frac{9}{4}$
(E)
3
How to solve
Strategy Introduce a Variable

The instinct is to reach for a calculator, and the instinct is wrong: each of these four logarithms is irrational, yet all five answer choices are exact rationals. So the problem is not asking for arithmetic — it is asking which combinations of these logarithms are clean even though the pieces are not. Tool #4 (Introduce a Variable) leads, and it leads twice. First it names a=log 5 and b=log 20, because those two never need to be known separately — only their sum matters, and their sum is exactly 2. Then it names t=log 2, because 5, 20, 8 and 0.25 are all 2s and 10s in disguise, so one letter converts the entire problem into ordinary algebra. Tool #7 (Identify Subproblems) does the sorting up front: the two cubes obey one set of rules and the lone product obeys another, so they get handled separately and are only added at the very end. Tool #5 (Look for a Pattern) supplies the identity that makes a³+b³ reachable from a+b alone. Tool #15 (Organize Information in More Ways) does the rewriting in the middle — 5=10/2, 20=2 · 10, 8=2³, 0.25=2⁻² — which is where the problem stops being about logarithms and starts being about a quadratic in one letter.

1STEP 1

Split into two unlike pieces

Split into two unlike pieces.

(log 5)³+(log 20)³_part 1: two cubes + (log 8)(log 0.25)_part 2: one product
2STEP 2

The two logs sum to exactly 2

The two logarithms sum to exactly two.

a=log 5, b=log 20, a+b=log 5+log 20=log(5 · 20)=log 100=2
3STEP 3

Cube the sum, then correct it

Cube the sum, then correct it.

a³+b³=(a+b)³-3ab(a+b)=2³-3ab · 2=8-6ab
4STEP 4

Rewrite everything using log 2

Rewrite everything with one logarithm.

t=log 2: log 5=1-t, log 20=1+t ⟹ ab=(1-t)(1+t)=1-t². log 8=log 2³=3t, log 0.25=log 2⁻²=-2t ⟹ (log 8)(log 0.25)=(3t)(-2t)=-6t²
5STEP 5

Add the parts and watch t vanish

Adding makes the letter vanish, leaving 2.

(2+6t²)+(-6t²)=2 (C)
Answer
2
Check it numerically. With log 2≈ 0.30103: log 5≈ 0.69897 and log 20≈ 1.30103, so the cubes are 0.69897³≈ 0.34149 and 1.30103³≈ 2.20223, summing to about 2.54372. Meanwhile log 8≈ 0.90309 and log 0.25≈ -0.60206, whose product is about -0.54369. The total is about 2.00003, and the small excess is rounding only. That lands on (C) 2 with room to spare — the nearest neighbours, (B) 7/4=1.75 and (D) 9/4=2.25, are each a quarter away, far outside any rounding error. There is also a check that needs no numbers: every answer choice is rational while every logarithm in the problem is irrational, so the irrational parts were always going to have to cancel. Seeing 6t² meet -6t² is that requirement being met, which is evidence the algebra went the intended way. One trap is worth naming. Drop the minus sign on log 0.25=-2t and the last term becomes +6t², giving 2+12t²≈ 3.09 — close enough to (E) 3 to look like the answer. That sign is not a detail; it is the reason the problem works at all.
💡Key takeaway

Before cubing anything, check what the numbers inside the logarithms multiply to: here 5 × 20=100 makes the two logarithms add to exactly 2, and the rest of the expression was built so the leftovers cancel.

  • Split into two unlike pieces
  • The two logs sum to exactly 2
  • Cube the sum, then correct it
  • Rewrite everything using log 2
  • Add the parts and watch t vanish