Problem 9 · AMC 8 Stretch
Stretch
Algebra & Patterns
list-all-the-casescheck-side-conditions
A number raised to a power equals \(1\) in more ways than people expect. Suppose \(\left(x^2-5x+5\right)^{\,x^2-9x+20}=1\). Find every value of \(x\) that works.
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Answer: x=1,2,3,4,5
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Hint 1 of 4
When is \(A^B=1\)? There are three different ways. Try to list them before solving.
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Hint 2 of 4
Way 1: the base is \(1\) (then \(1\) to any power is \(1\)). Way 2: the exponent is \(0\) (then anything nonzero to the \(0\) power is \(1\)). Way 3: the base is \(-1\) AND the exponent is an even number.
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Hint 3 of 4
Turn each way into a small equation. Base \(=1\): \(x^2-5x+5=1\). Exponent \(=0\): \(x^2-9x+20=0\). Base \(=-1\): \(x^2-5x+5=-1\).
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Approach: Three ways to make a power equal 1, checked case by case
- Way 1: base \(=1\). \(x^2-5x+5=1\) becomes \(x^2-5x+4=(x-1)(x-4)=0\), so \(x=1\) or \(x=4\).
- Way 2: exponent \(=0\) (base not \(0\)). \(x^2-9x+20=(x-4)(x-5)=0\), so \(x=4\) or \(x=5\). At \(x=5\) base \(=5\), at \(x=4\) base \(=1\); both fine.
- Way 3: base \(=-1\) with even exponent. \(x^2-5x+5=-1\) becomes \(x^2-5x+6=(x-2)(x-3)=0\), so \(x=2\) or \(x=3\). Exponent is \(6\) at \(x=2\) and \(2\) at \(x=3\) — both even, so both fine.
- Collecting everything: \(x=1,2,3,4,5\).
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