Problem 21 · 2016 Math Kangaroo
Stretch
Algebra & Patterns
casework
How many different real solutions does the following equation have?
\((x^2 - 4x + 5)^{\,x^2 + x - 30} = 1\)
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Answer: C — 3
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Hint 1 of 3
A power equals 1 in only three ways.
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Hint 2 of 3
Either the exponent is 0, or the base is 1, or the base is -1 with an even exponent.
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Hint 3 of 3
Notice the base \(x^2-4x+5 = (x-2)^2+1\) is always at least 1, which kills one case.
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Approach: three cases for base^exponent = 1
- The base \(x^2-4x+5 = (x-2)^2+1 \ge 1\), so it can never be \(-1\); only two cases survive.
- Base \(= 1\): \((x-2)^2+1 = 1\) gives \(x = 2\).
- Exponent \(= 0\): \(x^2+x-30 = 0\) gives \(x = 5\) or \(x = -6\), and the base is nonzero at both.
- The distinct real solutions are \(2, 5, -6\): that is 3 of them (C).
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