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For how many integers n is \(2^{\mathrm{n}} = \mathrm{n}^2\) ?
Let's start by understanding what we're looking for in everyday language. We have an equation that says "2 raised to the power n equals n squared." This means we need to find integer values where if we take 2 and multiply it by itself n times, we get the same result as multiplying n by itself once.
For example, if n = 4, we'd check: does \(2^4\) equal \(4^2\)? That's asking: does 16 equal 16? Yes it does!
So we're hunting for integer values of n (positive, negative, or zero) that make both sides of the equation \(2^n = n^2\) equal to each other.
Process Skill: TRANSLATE - Converting the mathematical equation into concrete understanding
Let's check different integer values one by one, starting with the easiest cases:
Testing n = 0:
Left side: \(2^0 = 1\) (any number to the power 0 equals 1)
Right side: \(0^2 = 0\)
Since 1 ≠ 0, n = 0 is NOT a solution.
Testing n = 1:
Left side: \(2^1 = 2\)
Right side: \(1^2 = 1\)
Since 2 ≠ 1, n = 1 is NOT a solution.
Testing n = 2:
Left side: \(2^2 = 4\)
Right side: \(2^2 = 4\)
Since 4 = 4, n = 2 IS a solution! ✓
Testing n = 3:
Left side: \(2^3 = 8\)
Right side: \(3^2 = 9\)
Since 8 ≠ 9, n = 3 is NOT a solution.
Testing n = 4:
Left side: \(2^4 = 16\)
Right side: \(4^2 = 16\)
Since 16 = 16, n = 4 IS a solution! ✓
Testing n = 5:
Left side: \(2^5 = 32\)
Right side: \(5^2 = 25\)
Since 32 ≠ 25, n = 5 is NOT a solution.
Testing negative values:
For any negative integer n, \(n^2\) is positive, but \(2^n\) becomes a positive fraction (like 1/2, 1/4, etc.). Since \(n^2\) is always a whole number for integer n, but \(2^n\) gives fractions for negative n, there can't be any negative integer solutions.
Process Skill: CONSIDER ALL CASES - Systematically checking positive, negative, and zero cases
Now let's think about what happens for larger values of n. We need to understand how fast each side of our equation grows:
Exponential growth (\(2^n\)): This grows very rapidly. Each time n increases by 1, we double the result.
Quadratic growth (\(n^2\)): This grows more slowly than exponential growth for large values.
Notice that starting from n = 5, the exponential function \(2^n\) starts pulling ahead of the quadratic function \(n^2\), and this gap only gets bigger. Since \(2^5 = 32 > 25 = 5^2\), and exponential functions grow faster than quadratic functions, there cannot be any more solutions beyond n = 4.
Process Skill: INFER - Drawing conclusions about function behavior for large values
Let's compile our results:
From our systematic testing, we found exactly two integer values that satisfy \(2^n = n^2\):
We confirmed that:
Therefore, there are exactly TWO integers for which \(2^n = n^2\).
The answer is C. Two
We found exactly two integer solutions: n = 2 and n = 4. This matches answer choice C perfectly.
1. Restricting to only positive integers: Students often assume that 'integers' means only positive whole numbers (1, 2, 3, ...) and forget to consider negative integers and zero. This leads them to miss testing important cases like n = 0 and negative values, potentially affecting their final count.
2. Not recognizing the need for systematic testing: Some students may try to solve this algebraically by attempting to manipulate the equation \(2^n = n^2\), not realizing that this type of equation (exponential equals polynomial) typically requires testing specific values rather than algebraic manipulation.
1. Calculation errors with exponents: Students frequently make arithmetic mistakes when computing powers, especially confusing \(2^4 = 16\) with \(4^2 = 16\), or miscalculating values like \(2^3 = 8\) vs \(3^2 = 9\). These calculation errors can lead to identifying wrong solutions or missing correct ones.
2. Incomplete testing of cases: Students may test a few values, find one or two solutions, and then stop prematurely without systematically checking enough cases or without considering the growth rate analysis to confirm no further solutions exist.
3. Incorrect reasoning about negative integers: When testing negative values, students might incorrectly calculate \(2^{-n}\) or make errors in understanding that \(2^{-3} = 1/8\) (a fraction) while \((-3)^2 = 9\) (a positive integer), leading to wrong conclusions about negative solutions.
1. Misunderstanding what to count: Students might correctly find n = 2 and n = 4 as solutions but then get confused about what the question is asking for. They might count the number of equations checked rather than the number of valid integer solutions, or double-count somehow.
2. Second-guessing correct work: After finding exactly two solutions (n = 2 and n = 4), students might doubt their work because it seems 'too simple' and incorrectly convince themselves they missed something, leading them to select 'More than three' or 'One' instead of the correct answer 'Two'.