AMC 10 · 2008 · #14

Grade 11 algebrageometry-2d

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Multiplicative Rate or Series

logarithm-propertiescircle-circumference convert-to-algebraformula-substitution ↑ Prerequisites: exponentscircle-circumference

📏 Short solution 💡 2 insights

Problem

One circle's radius and circumference are each given as a logarithm of a power. Find the logarithm of one base to the other.

Pick an answer.

(A)

$\frac{1}{4\pi}$

(B)

$\frac{1}{\pi}$

(C)

$\pi$

(D)

$2\pi$

(E)

$10^{2\pi}$

How to solve

Strategy Introduce a Variable

The logarithms look intimidating, but log₁₀{a} and log₁₀{b} are just two ordinary numbers. I name them x and y. The power rule then turns the radius into 2x and the circumference into 4y, so the circle relation C = 2π r becomes a plain linear equation in x and y. The quantity asked for, log_a{b}, is exactly the ratio y/x, so a linear relation between x and y is precisely what I need. Before dividing by x I first check that x cannot be zero, and afterwards I check that a circle like this really exists.

1STEP 1

What a and b must be

Both quantities being positive forces both bases above one.

log₁₀{(a²)} ＞ 0, log₁₀{(b⁴)} ＞ 0, a ＞ 0, a ≠ 1 ⟹ a ＞ 1 and b ＞ 1

2STEP 2

Name the two logarithms

Naming the two logarithms makes each quantity a plain multiple.

x = log₁₀{a}, y = log₁₀{b}, r = 2x, C = 4y

3STEP 3

Glue with the circle formula

The circle formula glues the two together.

C = 2π r ⟹ 4y = 2π(2x) = 4π x ⟹ y = π x

4STEP 4

Work backwards from what is asked

Working backwards from the target gives it directly as π.

t = log_a{b} ⇔ a^t = b ⟹ t x = y ⟹ t = y/x = (π x)/x = π

5STEP 5

Check such a circle actually exists

Such a circle really exists.

a ＞ 1, b = a^π ⟹ r = 2log₁₀{a} ＞ 0, C = 4πlog₁₀{a} = 2π r ✓

6STEP 6

Second route: circumference over diameter

Circumference over diameter confirms π, choice (E).

C/r = 4tlog₁₀{a}/2log₁₀{a} = 2t ⟹ log_a{b} = t = C/2r = C/d = π

Answer

Two independent routes, one algebraic and one structural, both land on π, and a concrete pair (a = 10, b = 10^π, r = 2, C = 4π) satisfies every condition. The answer also passes a dependency test: it should react correctly to changing the exponents. If the circumference were log₁₀{(b²)} instead of log₁₀{(b⁴)}, then 2y = 4π x and the ratio would be 2π — which is exactly trap (D), waiting for anyone who misreads the exponent. Trap (B), 1/π, is log_b{a}, the reciprocal, for anyone who divides the wrong way round. Trap (E), 10²π, has the shape of an antilogarithm; but log_a{b} = log₁₀{b}/log₁₀{a} is a ratio of two logarithms, so it can never come out as a power of 10 here — that choice catches anyone who solves for b instead of for the exponent.

💡Key takeaway

Name the logarithms as plain numbers, and the circle rule C = 2π r turns the mystery ratio log_a{b} into circumference divided by diameter, which is π.

What a and b must be
Name the two logarithms
Glue with the circle formula
Work backwards from what is asked
Check such a circle actually exists
Second route: circumference over diameter