AMC 10 · 2006 · #6
Grade 8 arithmeticPick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The two sides use different exponents (14 and 7), so they can't be compared directly. Re-expressing the 14th power as a 7th power of a new base makes both sides the 7th power of something, and then the two bases must match. That turns a scary-looking exponent equation into a short algebra problem.
Make both exponents 7
Since 14 = 2 × 7, pair up the factors of 7x: (7x)¹⁴ = ((7x)²)⁷ = (49x²)⁷, so the left side is now a 7th power too.
Splitting the exponent 14 into 2 times 7 lets both sides wear the same outer power.
8.EE.A.1Look For A PatternEquate the bases
Both sides read (49x²)⁷ = (14x)⁷, and an odd power is one-to-one on the reals, so the bases must match: 49x² = 14x.
An odd power never makes two different real numbers collide, so matching 7th powers forces matching bases.
An odd power never sends two different real numbers to the same place, so matching powers force matching bases.
▸ Why?
An exponent counts how many times a factor is used, and an odd count keeps the sign of the base.
▸ Why?
So two equal powers of that kind must have equal bases, with no second possibility.
Use the non-zero condition
Since x is not zero, divide 49x² = 14x by the shared factor 7x: 7x = 2, so x = 2/7 — choice (B).
Being told x is non-zero is exactly the permission slip needed to divide away the shared 7x factor.
8.EE.C.7Eliminate PossibilitiesWhen two powers look mismatched, rewrite one so both share the same exponent — then the insides have to be equal.
- Make both exponents 7
- Equate the bases
- Use the non-zero condition