AMC 10 · 2015 · #8

Grade 11 algebra
logarithm-propertiesexponents convert-to-algebraidentify-subproblems ↑ Prerequisites: exponentslogarithm-properties
📏 Medium solution 💡 2 insights
Problem
A number is raised to a logarithm and the result is rooted. Find the value.

Pick an answer.

(A)
5
(B)
$\sqrt[4]{2015}$
(C)
625
(D)
2015
(E)
$\sqrt[4]{5^{2015}}$
How to solve
Strategy Introduce a Variable

The only awkward object here is log₅ 2015, and the fastest way to defuse a logarithm is to name it: set x = log₅ 2015, which converts the whole log into the single plain equation 5^x = 2015 (Introduce a Variable). After that the obstacle is cosmetic — the power is written in base 625 while the useful fact is written in base 5 — so I rewrite 625 as 5⁴ to put everything over one base (Organize Information in More Ways). Once the bases agree, the outer 1/4 power exactly undoes the 4, and the stored equation 5^x = 2015 finishes it. Finally I bracket the value between two powers of 5 and check the five choices against that window (Eliminate Possibilities), which confirms the answer by a route that does not reuse the algebra.

1STEP 1

Name the logarithm

The logarithm gets a name.

x = log₅ 2015 ⟺ 5^x = 2015
2STEP 2

Put everything over base 5

Everything rewrites over one base.

625^x = (5⁴)^x = 5⁴x
3STEP 3

The fourth root undoes the 4

The root undoes the power exactly.

(5⁴x)¹/4 = 5^ 4x · 1/4 = 5^x
4STEP 4

Cash in the stored equation

The stored equation gives 2015.

(625^log₅ 2015)¹/4 = (5⁴x)¹/4 = 5^x = 2015
5STEP 5

Trap the value between powers of 5

Bracketing the value confirms it, choice (D).

5⁴ = 625 < 2015 < 3125 = 5⁵ ⟹ 4 < x < 5 ⟹ 625 < 5^x < 3125
Answer
2015
A numerical pass agrees. log₅ 2015 ≈ 4.727, so the inner power is 625⁴.727 = 5¹8.91 ≈ 1.65 × 10¹³, and its fourth root is about 2.0 × 10³ — matching 2015 to the precision of the estimate. The structural check is even cleaner: the outer 1/4 power exactly cancels the 4 hidden inside 625 = 5⁴, so the expression is really just 5^log₅ 2015, and an exponential undoing its own logarithm must return the number inside. It is worth noticing that x = log₅ 2015 is irrational — if x = p/q then 5^p = 2015^q, impossible since 13 divides 2015 but not any power of 5 — so the whole-number exponent rules are not enough here; the real-exponent power rule for a positive base is what carries the argument. Choice (D) 2015 is consistent with every check.
💡Key takeaway

Write 625 as 5⁴ and the outer fourth root cancels that 4, leaving 5^log₅ 2015 — and a base-5 power sitting on a base-5 logarithm just hands the number back.

  • Name the logarithm
  • Put everything over base 5
  • The fourth root undoes the 4
  • Cash in the stored equation
  • Trap the value between powers of 5