Laws of Indices

When we multiply powers with the same base, we add the indices because we are combining groups of repeated multiplication. \( 2^3 \) means \( 2 \times 2 \times 2 \) and \( 2^4 \) means \( 2 \times 2 \times 2 \times 2 \). Writing them together gives seven 2s multiplied: \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7 \).

A common error is to multiply the indices to get \( 2^{12} \), or to multiply the bases to get \( 4^7 \). Both are wrong. We add the indices (\( 3 + 4 = 7 \)) because multiplication combines the groups — it doesn’t multiply how many are in each group.

Consider the pattern of dividing by 5 each time: \( 5^3 = 125 \), \( 5^2 = 25 \), \( 5^1 = 5 \). Each step divides by 5, so the next step gives \( 5^0 = 5 \div 5 = 1 \). The pattern demands the answer is 1, not 0.

Alternatively, using the division law: \( 5^3 \div 5^3 = 5^{3-3} = 5^0 \). But we also know \( 5^3 \div 5^3 = 125 \div 125 = 1 \). So \( 5^0 \) must equal 1. Students often assume “to the power of zero means zero of something” and write 0, but the index tells us about repeated multiplication, not about the value directly.

\( (3^2)^3 \) means \( 3^2 \times 3^2 \times 3^2 \). Using the multiplication law, we add the indices: \( 2 + 2 + 2 = 6 \), giving \( 3^6 \). We can verify: \( 3^2 = 9 \), and \( 9^3 = 729 \). Also, \( 3^6 = 729 \). Both agree.

The common error is to add the indices (\( 2 + 3 = 5 \)) to get \( 3^5 = 243 \), confusing the power of a power law with the multiplication law. When raising a power to another power, we multiply the indices (\( 2 \times 3 = 6 \)), because we are repeating the group of repeated multiplications.

Extending the pattern of dividing by 4: \( 4^2 = 16 \), \( 4^1 = 4 \), \( 4^0 = 1 \), \( 4^{-1} = \frac{1}{4} \), \( 4^{-2} = \frac{1}{16} \). Each step divides by 4, so \( 4^{-2} = 1 \div 4 \div 4 = \frac{1}{16} \).

A negative index does not mean a negative answer. The rule \( a^{-n} = \frac{1}{a^n} \) gives \( 4^{-2} = \frac{1}{4^2} = \frac{1}{16} \), which is a small positive number. Students who write \( -16 \) are confusing the negative sign in the index with a negative sign in front of the result.

By applying the multiplication law of indices: \( 9^{1/2} \times 9^{1/2} = 9^{1/2 + 1/2} = 9^1 = 9 \). We need a number which, when multiplied by itself, equals 9.

Since we know that \( 3 \times 3 = 9 \), it follows that the value of \( 9^{1/2} \) must be 3. A power of \( \frac{1}{2} \) is mathematically equivalent to taking the square root. Students will often incorrectly halve the base to get 4.5, failing to recognize that fractional indices represent roots.

Example: \( a^3 \times a^5 = a^8 \)

Another: \( 2^4 \times 2^6 = 2^{10} \)

Creative: \( a^{-3} \times a^7 = a^4 \) — this works with negative indices too; \( -3 + 7 = 4 \).

Trap: \( a^3 \times b^5 \) — this looks like a multiplication of powers, and a student might write \( ab^8 \) by adding the indices. But the bases are different (\( a \) and \( b \)), so the multiplication law of indices does not apply. This expression cannot be simplified.

Example: \( a^3 \div a^3 = a^0 = 1 \)

Another: \( 7^0 = 1 \)

Creative: \( \frac{3^7 \times 3^2}{3^9} = \frac{3^9}{3^9} = 3^0 = 1 \) — combining the multiplication and division laws in a single expression.

Trap: \( a^1 \) — a student might think “to the power of 1 means it equals 1.” The expression \( a^1 \) simplifies to just \( a \). This only equals 1 if \( a \) itself is 1. The power that always guarantees a result of 1 (for any non-zero base) is \( a^0 \).

Example: \( a^2 \div a^5 = a^{-3} \)

Another: \( a^3 \times a^{-7} = a^{-4} \)

Creative: \( (a^{-2})^3 = a^{-6} \) — using the power of a power law with a negative starting index.

Trap: \( \frac{x^2}{x^{-3}} \) — many students will see the 2 and the -3, do simple addition or subtraction in their head (\( 2 – 3 = -1 \)), and think this gives a negative index. But dividing by a negative index means subtracting a negative: \( 2 – (-3) = 5 \), resulting in a positive index of \( x^5 \).

Example: \( 2^{-3} = \frac{1}{8} \)

Another: \( 8^{-1} = \frac{1}{8} \)

Creative: \( \frac{2^2}{2^5} = 2^{-3} = \frac{1}{8} \) — using the division law to arrive at a negative index, then converting to a fraction.

Trap: \( \frac{1}{2^{-3}} \) — a student might think “there’s a \( 2^{-3} \) in the denominator, and \( 2^{-3} = \frac{1}{8} \), so this must be \( \frac{1}{8} \) too.” But \( \frac{1}{2^{-3}} = 2^3 = 8 \), not \( \frac{1}{8} \). Dividing by a number with a negative index flips it to a positive index.

This is the multiplication law of indices: \( a^m \times a^n = a^{m+n} \) for any base \( a \) and any indices \( m \) and \( n \). It works because multiplying powers combines groups of repeated multiplication. For example, \( a^3 \times a^2 = (a \times a \times a) \times (a \times a) = a^5 \), and \( 3 + 2 = 5 \).

This applies even with negative or zero indices. For example, \( a^4 \times a^{-1} = a^3 \) (since \( 4 + (-1) = 3 \)), and \( a^5 \times a^0 = a^5 \) (since \( 5 + 0 = 5 \)). Students might think there are exceptions for special cases, but the law always holds provided the bases are the same.

The correct power of a power law is \( (a^m)^n = a^{mn} \), not \( a^{m+n} \). So this statement is only true when \( mn = m + n \).

True case: \( m = 2, n = 2 \): \( (a^2)^2 = a^4 \) and \( a^{2+2} = a^4 \). Both give \( a^4 \), so the statement holds. False case: \( m = 3, n = 2 \): \( (a^3)^2 = a^6 \) but \( a^{3+2} = a^5 \). These are different, so the statement fails. This is a very common confusion — students mix up when to add and when to multiply the indices.

For any non-zero value of \( a \), \( a^0 = 1 \), not 0. This can be shown by the pattern \( a^3, a^2, a^1, a^0 \) where each step divides by \( a \), so \( a^0 = a^1 \div a = 1 \). Or by the division law: \( a^n \div a^n = a^{n-n} = a^0 \), and any number divided by itself is 1.

When \( a = 0 \), the expression \( 0^0 \) is undefined (not equal to 0 or 1). So there is no value of \( a \) for which \( a^0 = 0 \). Students often assume “the zero index means nothing” or “zero of something,” but the index 0 arises from dividing a power by itself, which always gives 1.

A negative index means taking the reciprocal, not making the number negative. \( x^{-2} = \frac{1}{x^2} \), whereas \( -x^2 \) is the negative of \( x \) squared. For example, if \( x = 3 \), \( 3^{-2} = \frac{1}{9} \), but \( -3^2 = -9 \). They are never equal for real numbers.

This targets a very common misconception where students confuse the negative sign in the index with a negative sign applied to the entire term.

The “multiply the indices” misconception. The student has confused the multiplication law with the power of a power law. When multiplying powers with the same base, you add the indices, not multiply them. \( a^4 \times a^3 = a^{4+3} = a^7 \). You multiply indices only when raising a power to another power: \( (a^4)^3 = a^{12} \).

The student’s answer of \( a^{12} \) would be correct for \( (a^4)^3 \), not for \( a^4 \times a^3 \).

The “divide the indices” misconception. The student has the correct answer (\( a^2 \)), but only by coincidence. The division law says you subtract the indices: \( a^4 \div a^2 = a^{4-2} = a^2 \). Here, \( 4 \; – \; 2 = 2 \) and \( 4 \div 2 = 2 \) happen to give the same result, so the error is hidden.

But try \( a^5 \div a^2 \): the correct answer is \( a^{5-2} = a^3 \), while dividing the indices gives \( a^{5 \div 2} = a^{2.5} \), which is wrong. Getting the right answer by the wrong method is dangerous because the student won’t realise their mistake until the numbers stop being convenient.

The “add the indices” misconception (power of a power). The student has confused the power of a power law with the multiplication law. For \( (a^m)^n \), you multiply the indices: \( (a^3)^2 = a^{3 \times 2} = a^6 \), not \( a^{3+2} = a^5 \). You add indices when multiplying powers: \( a^3 \times a^2 = a^5 \).

The student’s answer of \( a^5 \) would be correct for \( a^3 \times a^2 \), not for \( (a^3)^2 \). A useful check: \( (a^3)^2 = a^3 \times a^3 = a^6 \).

The “only apply the power to the variable” misconception. When a product is raised to a power, the power applies to every factor inside the bracket. \( (3a)^2 = 3^2 \times a^2 = 9a^2 \), not \( 3a^2 \). The student treated the 3 as if it were outside the bracket.

The bracket means the entire expression \( 3a \) is being squared: \( (3a)^2 = 3a \times 3a = 9a^2 \). This error is equivalent to thinking \( (3a)^2 \) is the same as \( 3 \times a^2 \), when in fact it equals \( 3^2 \times a^2 \).

The “multiply the indices” misconception combined with coefficients. The student correctly multiplied the coefficients (\( 3 \times 4 = 12 \)) but incorrectly multiplied the indices. The coefficients are regular numbers being multiplied, but the indices represent repeated multiplication and must be added according to the multiplication law (\( 2 + 5 = 7 \)).

The correct answer is \( 12x^7 \).