Problem 5 · AMC 8 Stretch
Stretch
Algebra & Patterns
Number Theory
reduce-and-expandpattern-recognitionlogical-reasoning
Insisting the rule \(x^m\cdot x^n=x^{m+n}\) keeps working tells you what new exponents must mean. Multiplying \(x^{1/2}\) by itself gives \(x^{1/2}\cdot x^{1/2}=x^{1}=x\). So \(x^{1/2}\) equals which familiar expression in \(x\)? (Write it using a square root, e.g. sqrt(x).)
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Answer: the square root of x
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Hint 1 of 4
Don't try to picture a 'negative bag' or a 'half bag.' Instead demand that 'when you multiply, you add the exponents' still works.
Still stuck? Show hint 2 →
Hint 2 of 4
Multiply \(x^{1/2}\) by itself: the rule gives \(x^{1/2}\cdot x^{1/2}=x^{1/2+1/2}=x^{1}=x\).
Still stuck? Show hint 3 →
Hint 3 of 4
A number that, times itself, gives \(x\) is a square root of \(x\).
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Approach: Extend a pattern by requiring the exponent rule to keep holding
- Throw away the 'bag of x's' picture for new exponents and REQUIRE \(x^m\cdot x^n=x^{m+n}\) to keep holding; the rule then forces the definitions.
- Multiply \(x^{1/2}\) by itself: \(x^{1/2}\cdot x^{1/2}=x^{\frac12+\frac12}=x^{1}=x\).
- So \(x^{1/2}\) is a number that times itself gives \(x\) — that is exactly the square root: \(x^{1/2}=\sqrt{x}\).
- The same trick handles negatives: \(x^{-3}\cdot x^{3}=x^{0}=1\), so \(x^{-3}=\frac{1}{x^3}\). We keep the useful rule and let it define the new cases.
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