Powers

These look similar because they use the same two numbers, but the positions matter. \( 2^5 \) means \( 2 \times 2 \times 2 \times 2 \times 2 = 32 \), while \( 5^2 \) means \( 5 \times 5 = 25 \). So \( 2^5 = 32 \) and \( 5^2 = 25 \) — they are not equal.

Squaring is like forming a shape (area), while powers of 2 are like branching (doubling).

This shows that powers are not commutative — unlike addition and multiplication, swapping the base and the exponent changes the result. In fact, \( a^b = b^a \) is only true for certain special pairs (such as \( 2^4 = 4^2 = 16 \)), not in general.

At first glance, 10 seems like a much larger base than 2, so students might expect \( 10^2 \) to be larger. But \( 10^2 = 100 \), while \( 2^{10} = 1024 \). A large exponent with a small base can far exceed a small exponent with a large base.

This happens because \( 2^{10} \) involves multiplying 2 by itself ten times: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024. Each step doubles the previous value, so the result grows far faster than students might expect. The misconception that “a bigger base always gives a bigger result” ignores the crucial role of the exponent.

The expression \( 2^4 \) means “multiply 2 by itself 4 times”: \( 2 \times 2 \times 2 \times 2 = 16 \). It does not mean “2 multiplied by 4,” which would give 8. Powers (repeated multiplication) and ordinary multiplication are different operations entirely.

A good way to see why they must be different: \( 2 \times 4 = 8 \), but \( 2^4 = 16 \). The gap grows even wider with larger exponents — for example, \( 2 \times 10 = 20 \) but \( 2^{10} = 1024 \). Confusing power notation with multiplication leads to answers that are far too small.

\( 1^{100} \) means \( 1 \times 1 \times 1 \times \ldots \times 1 \) (one hundred times). Since multiplying by 1 never changes the value, the result stays at 1 no matter how many times you multiply. So \( 1^{100} = 1 \).

Students who think “raising to a big power must give a big number” are confusing the effect of the exponent on different bases. When the base is 2 or more, a larger exponent does make the result grow — but 1 is special. \( 1^n = 1 \) for every positive integer \( n \), regardless of how large \( n \) is.

Example: \( 2^4 = 16 \) and \( 4^2 = 16 \)

Another: \( 2^6 = 64 \) and \( 4^3 = 64 \)

Creative: \( 2^3 = 8 \) and \( 8^1 = 8 \). In fact, any number to the power of 1 equals itself, so you can always pair \( a^b \) with the result raised to the power 1.

Trap: \( 2^3 \) and \( 3^2 \) — a student might think that swapping the base and exponent always gives the same answer (the “powers are commutative” misconception). But \( 2^3 = 8 \) and \( 3^2 = 9 \), which are not equal.

Example: 64, since \( 8^2 = 64 \) and \( 4^3 = 64 \)

Another: 1, since \( 1^2 = 1 \) and \( 1^3 = 1 \)

Creative: 729, since \( 27^2 = 729 \) and \( 9^3 = 729 \). Numbers that are both square and cube are always sixth powers — here, \( 729 = 3^6 \).

Trap: 9 — it is a square number (\( 3^2 = 9 \)) but not a cube number (\( \sqrt[3]{9} \approx 2.08 \)). A student might think any square number is automatically a cube number too, or confuse “cube number” with “multiple of 3” (9 happens to be a multiple of 3, which could reinforce the confusion that multiples of 3 and cube numbers are the same thing).

Example: \( 3^2 = 9 \)

Another: \( 5^3 = 125 \)

Creative: \( 1^{99} = 1 \) — since 1 is odd and any power of 1 is still 1, the result is always odd regardless of the exponent.

Trap: \( 6^3 = 216 \) — a student might think that because the exponent (3) is odd, the answer must be odd. But the parity of the result is determined by the base, not the exponent: since 6 is even, \( 6^n \) is even for every positive integer \( n \). To get an odd result, you need an odd base — the exponent is irrelevant to parity.

Example: \( n = 5 \), since \( 2^5 = 32 \) and \( 5^2 = 25 \), and \( 32 > 25 \)

Another: \( n = 10 \), since \( 2^{10} = 1024 \) and \( 10^2 = 100 \), and \( 1024 > 100 \)

Creative: \( n = 1 \), since \( 2^1 = 2 \) and \( 1^2 = 1 \), and \( 2 > 1 \). Students often start testing from \( n = 2 \) upwards and miss this one.

Trap: \( n = 4 \), since \( 2^4 = 16 \) and \( 4^2 = 16 \), so \( 2^n = n^2 \) — they are equal, not \( 2^n > n^2 \). A student might fail to evaluate both expressions properly and assume \( 16 > 16 \). (Note: \( n = 3 \) also fails since \( 2^3 = 8 < 9 = 3^2 \).)

For any base of 2 or more, this is true: for example, \( 3^2 = 9 \), which is larger than the base of 3.

However, this fails for 1 (\( 1^2 = 1 \)) and for numbers between 0 and 1. For example, \( 0.5^2 = 0.25 \). Here, the result is smaller than the base (\( 0.25 < 0.5 \)). This is a crucial concept when dealing with probability or geometric sequences. TRUE case: \( 5^2 = 25 > 5 \). FALSE case: \( 0.5^2 = 0.25 < 0.5 \).

Students may assume that since the base is 3, every power of 3 must end in 3. But the units digits of powers of 3 cycle in a repeating pattern: \( 3^1 = 3 \), \( 3^2 = 9 \), \( 3^3 = 27 \), \( 3^4 = 81 \), \( 3^5 = 243 \). The units digits cycle: 3, 9, 7, 1, 3, 9, 7, 1, …

So the units digit is 3 only when the exponent leaves remainder 1 when divided by 4 (i.e., exponents 1, 5, 9, 13, …). For other exponents, the units digit is 9, 7, or 1. TRUE case: \( 3^1 = 3 \) (units digit 3). FALSE case: \( 3^2 = 9 \) (units digit 9, not 3).

Factors usually come in pairs: if \( a \times b = n \), then both \( a \) and \( b \) are factors. This makes students think every number has an even number of factors. But for square numbers, the square root pairs with itself — for example, \( 9 = 3 \times 3 \), so 3 only gets counted once, not twice.

Check: 1 has 1 factor (odd). 4 has factors 1, 2, 4 — that’s 3 (odd). 9 has factors 1, 3, 9 — that’s 3 (odd). 16 has factors 1, 2, 4, 8, 16 — that’s 5 (odd). 36 has factors 1, 2, 3, 4, 6, 9, 12, 18, 36 — that’s 9 (odd). This works because every square number has exactly one factor that pairs with itself (the square root), creating the odd count.

Students might think that as powers of 2 get very large, they could eventually end in 0 (perhaps by analogy with powers of 10, which always end in 0). But the units digits of powers of 2 follow a strict cycle: \( 2^1 = 2 \), \( 2^2 = 4 \), \( 2^3 = 8 \), \( 2^4 = 16 \), \( 2^5 = 32 \). The units digits cycle: 2, 4, 8, 6, 2, 4, 8, 6, …

Zero never appears in this cycle. For a number to end in 0, it must be divisible by both 2 and 5. A power of 2 is never divisible by 5 (since 5 is not a factor of 2), so no power of 2 can ever end in 0.

The student has confused exponentiation with multiplication — treating \( 3^4 \) as \( 3 \times 4 \) rather than as 3 multiplied by itself 4 times. The expression \( 3^4 \) means \( 3 \times 3 \times 3 \times 3 = 81 \), not \( 3 \times 4 = 12 \).

This is one of the most common errors with power notation. The exponent tells you how many times to multiply the base by itself — it is not a second number to multiply by. A useful check: \( 3 \times 4 \) would just be written as \( 3 \times 4 \), so the power notation \( 3^4 \) must mean something different.

The student has reached the correct answer (\( 3^2 = 9 > 8 = 2^3 \)), but their reasoning relies on the flawed belief that the “bigger base always gives a bigger result.” This happens to work here, but fails in general.

Counterexample: compare \( 2^5 \) and \( 5^2 \). By the student’s logic, \( 5^2 \) should be larger because \( 5 > 2 \). But \( 2^5 = 32 \) and \( 5^2 = 25 \), so the smaller base actually gives the larger result. The size of the answer depends on both the base and the exponent — neither one alone determines the outcome. The correct way to compare is to evaluate both expressions.

The student has confused “squaring” with “doubling.” These sound similar and both involve the number 2, but they are completely different operations.

Squaring means multiplying the number by itself: \( 7^2 = 7 \times 7 = 49 \). Doubling means multiplying by 2: \( 7 \times 2 = 14 \). A good way to reinforce the difference: squaring 10 gives 100, but doubling 10 gives only 20 — these are very different results.

The student has confused the index law for multiplying powers with the index law for raising a power to a power. When multiplying powers with the same base, you add the exponents, not multiply them: \( 2^3 \times 2^2 = 2^{3+2} = 2^5 = 32 \). The student’s answer of \( 2^6 = 64 \) is incorrect.

We can verify by expanding: \( 2^3 \times 2^2 = (2 \times 2 \times 2) \times (2 \times 2) = 2 \times 2 \times 2 \times 2 \times 2 = 2^5 \). There are \( 3 + 2 = 5 \) twos being multiplied, not \( 3 \times 2 = 6 \). The rule where you multiply exponents applies to a different situation: \( (2^3)^2 = 2^{3 \times 2} = 2^6 \).