AMC 10 · 2006 · #6
Grade 8 arithmeticWhat non-zero real value for x satisfies (7x)14=(14x)7?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Find the one non-zero real number x for which raising 7x to the 14th power gives the same result as raising 14x to the 7th power.
Givens: The equation (7x)^{14} = (14x)^7; x is a real number; x is not zero
Unknowns: The value of x that makes the two sides equal
Understand
Restated: Find the one non-zero real number x for which raising 7x to the 14th power gives the same result as raising 14x to the 7th power.
Givens: The equation (7x)^{14} = (14x)^7; x is a real number; x is not zero
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
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.
Execute — Answer: B
8.EE.A.1 Step 1 Make both exponents 7
- Since 14 = 2 x 7, the left side can be rewritten as a 7th power.
- Group the 14 factors of 7x into seven pairs: (7x)^{14} = ((7x)^2)^7 = (49x^2)^7.
- Now the left side is (49x^2)^7 and the right side is (14x)^7 — both are the 7th power of a base.
💡 Splitting the exponent 14 into 2 times 7 lets both sides wear the same outer power.
8.EE.A.1 Step 2 Equate the bases
- Both sides are now the same thing raised to the 7th power: (49x^2)^7 = (14x)^7.
- Raising to an odd power like 7 is one-to-one for real numbers, so equal 7th powers can only come from equal bases.
- Therefore 49x^2 = 14x.
💡 An odd power never makes two different real numbers collide, so matching 7th powers forces matching bases.
8.EE.C.7 Step 3 Use the non-zero condition
- The equation 49x^2 = 14x is true when x = 0, but the problem asks for a non-zero value, so we may divide both sides by 7x.
- That gives 7x = 2, hence x = 2/7.
- This matches choice (B).
💡 Being told x is non-zero is exactly the permission slip needed to divide away the shared 7x factor.
8.EE.A.1 Since 14 = 2 x 7, the left side can be rewritten as a 7th power. Group the 14 fa 8.EE.A.1 Both sides are now the same thing raised to the 7th power: (49x^2)^7 = (14x)^7. 8.EE.C.7 The equation 49x^2 = 14x is true when x = 0, but the problem asks for a non-zero Review
Reasonableness: Substitute x = 2/7: then 7x = 2 and 14x = 4. The left side is 2^{14} = 16384 and the right side is 4^7 = (2^2)^7 = 2^{14} = 16384. Both sides agree, confirming x = 2/7.
Alternative: Take the 7th root of both sides directly (valid for reals since odd roots are unique): (7x)^{14/7} = 14x, i.e. (7x)^2 = 14x, so 49x^2 = 14x, leading again to 7x = 2 and x = 2/7.
CCSS standards used (min grade 8)
8.EE.A.1Know and apply the properties of integer exponents (Rewriting (7x)^{14} as (49x^2)^7 so both sides share the exponent 7, and concluding that equal 7th powers force equal bases.)8.EE.C.7Solve linear equations in one variable (Dividing 49x^2 = 14x by 7x under the non-zero condition to solve for x.)
⭐ When two powers look mismatched, rewrite one so both share the same exponent — then the insides have to be equal.
⭐ When two powers look mismatched, rewrite one so both share the same exponent — then the insides have to be equal.
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