Multiplying and Dividing with Standard Form

The key is that when multiplying powers of 10, we add the exponents — we do not multiply them. The calculation separates into two parts: the coefficients (the numbers at the front, \( a \) in \( a \times 10^n \), so \( 3 \times 2 = 6 \)) and the powers of 10 (\( 10^4 \times 10^3 = 10^{4+3} = 10^7 \)). A student who multiplies the powers would get \( 10^{12} \), which is far too large.

We can verify by writing the numbers in full: \( 30\,000 \times 2\,000 = 60\,000\,000 = 6 \times 10^7 \). The answer has 7 digits after the 6, confirming the power of 7, not 12.

When dividing standard form numbers, we divide the coefficients (\( 8 \div 2 = 4 \)) and subtract the powers (\( 10^{5 \; – \; 2} = 10^3 \)). A student who adds the powers instead of subtracting would get \( 4 \times 10^7 \), which is far too large.

We can check: \( 800\,000 \div 200 = 4\,000 = 4 \times 10^3 \). Division makes numbers smaller, so the power of 10 should decrease — subtracting the exponents achieves this.

Multiplying coefficients gives \( 7 \times 5 = 35 \) and adding powers gives \( 10^{3+4} = 10^7 \), so the initial result is \( 35 \times 10^7 \). But 35 is not between 1 and 10, so this is not in standard form. We write \( 35 = 3.5 \times 10^1 \), giving \( 3.5 \times 10^1 \times 10^7 = 3.5 \times 10^8 \). The adjustment adds 1 to the power.

A student who leaves the answer as \( 35 \times 10^7 \) has performed the arithmetic correctly but has not given a final answer in standard form. In standard form, the coefficient must satisfy \( 1 \leq a < 10 \).

Dividing the coefficients gives \( 6 \div 3 = 2 \). Subtracting the powers gives \( 10^{2 \; – \; 5} = 10^{-3} \). It is crucial to subtract in the correct order: dividend minus divisor (top minus bottom), which gives \( 2 \; – \; 5 = -3 \). A student who subtracts “the smaller from the larger” would get \( 10^3 \), giving \( 2 \times 10^3 = 2000 \) — a million times too big.

We can verify: \( 600 \div 300\,000 = 0.002 = 2 \times 10^{-3} \). Since we are dividing a smaller number by a larger one, the result must be less than 1, confirming a negative power of 10.

Dividing the coefficients gives \( 2 \div 8 = 0.25 \). Subtracting the powers gives \( 10^3 \). This leaves \( 0.25 \times 10^3 \). To adjust this to standard form, we must multiply the \( 0.25 \) by 10 to get 2.5.

To keep the number’s overall value balanced, if we multiply the coefficient by 10, we must divide the \( 10^3 \) by 10, dropping the power to \( 10^2 \). The correct answer is \( 2.5 \times 10^2 \).

Example: \( (5 \times 10^3) \times (3 \times 10^2) = 15 \times 10^5 = 1.5 \times 10^6 \)

Another: \( (4 \times 10^4) \times (8 \times 10^1) = 32 \times 10^5 = 3.2 \times 10^6 \)

Creative: \( (2.5 \times 10^{-2}) \times (6 \times 10^5) = 15 \times 10^3 = 1.5 \times 10^4 \) — works with negative exponents and decimal coefficients too.

Trap: \( (2 \times 10^4) \times (4 \times 10^2) = 8 \times 10^6 \) — A student might rush, see 2 and 4, and think \( 2+4=6 \) triggers an adjustment, confusing the addition of powers with the multiplication of coefficients. But \( 2 \times 4 = 8 \), which is already between 1 and 10. No adjustment is needed, so this does not satisfy the condition.

Example: \( (4 \times 10^3) \times (3 \times 10^4) = 12 \times 10^7 = 1.2 \times 10^8 \)

Another: \( (6 \times 10^5) \times (2 \times 10^2) = 12 \times 10^7 = 1.2 \times 10^8 \)

Creative: \( (1.5 \times 10^4) \times (8 \times 10^3) = 12 \times 10^7 = 1.2 \times 10^8 \) — uses a decimal coefficient and requires spotting that \( 1.5 \times 8 = 12 \).

Trap: \( (4 \times 10^4) \times (3 \times 10^4) = 12 \times 10^8 = 1.2 \times 10^9 \) — a student sees the “1.2” and thinks the answer must be \( 1.2 \times 10^8 \), but the powers add to 8, and the coefficient adjustment adds another 1, making the final power 9.

Example: \( (3 \times 10^2) \div (6 \times 10^5) = 0.5 \times 10^{-3} = 5 \times 10^{-4} \)

Another: \( (2 \times 10^3) \div (8 \times 10^6) = 0.25 \times 10^{-3} = 2.5 \times 10^{-4} \)

Creative: \( (9 \times 10^{-1}) \div (3 \times 10^2) = 3 \times 10^{-3} \) — involves a number already less than 1.

Trap: \( (6 \times 10^{-2}) \div (2 \times 10^{-5}) = 3 \times 10^3 \) — a student sees the negative exponents in both numbers and assumes the answer must have a negative exponent too. But \( -2 \; – \; (-5) = -2 + 5 = 3 \), so the result has a positive exponent.

Example: \( (3 \times 10^5) \div (6 \times 10^2) = 0.5 \times 10^3 = 5 \times 10^2 \)

Another: \( (2 \times 10^4) \div (5 \times 10^1) = 0.4 \times 10^3 = 4 \times 10^2 \)

Creative: \( (1 \times 10^{-1}) \div (4 \times 10^3) = 0.25 \times 10^{-4} = 2.5 \times 10^{-5} \) — involves negative exponents and produces a very small answer.

Trap: \( (9 \times 10^5) \div (3 \times 10^2) = 3 \times 10^3 \) — a student might offer this thinking all divisions need adjustment, but \( 9 \div 3 = 3 \), which is already between 1 and 10. No adjustment is needed.

When multiplying powers of 10, you add the exponents, not multiply them. This is the index law: \( 10^a \times 10^b = 10^{a+b} \). For example, \( 10^3 \times 10^2 = 10^{3+2} = 10^5 \), not \( 10^{3 \times 2} = 10^6 \). You can verify by writing out the tens: \( 1000 \times 100 = 100\,000 = 10^5 \).

A student might test \( 10^2 \times 10^2 \) and notice that both \( 2 + 2 \) and \( 2 \times 2 \) give 4. This is a coincidence, not evidence the rule works — try \( 10^3 \times 10^2 \) and the answers diverge (5 vs 6). The correct operation is always addition.

It depends on the product of the coefficients. If the coefficient product is between 1 and 10, no adjustment is needed — for example, \( (3 \times 10^2) \times (2 \times 10^4) = 6 \times 10^6 \), which is already in standard form since 6 is between 1 and 10.

However, if the product of the coefficients is 10 or more, the result is not in standard form — for example, \( (5 \times 10^3) \times (4 \times 10^2) = 20 \times 10^5 \), and \( 20 \times 10^5 \) must be rewritten as \( 2 \times 10^6 \).

This is true when the coefficient product stays between 1 and 10. For example, \( (3 \times 10^2) \times (2 \times 10^3) = 6 \times 10^5 \), and \( 2 + 3 = 5 \) matches the final power.

But it’s false when the coefficients multiply to 10 or more, because adjusting the coefficient adds 1 to the power. For example, \( (5 \times 10^3) \times (4 \times 10^2) \): adding the powers gives \( 3 + 2 = 5 \), but the answer is \( 2 \times 10^6 \) — the final power is 6, not 5. The “extra” 1 comes from rewriting 20 as \( 2 \times 10^1 \).

The power in the answer depends on the relative sizes of the original powers. If the dividend has a larger power, the result is positive — for example, \( (8 \times 10^5) \div (2 \times 10^2) = 4 \times 10^3 \), where \( 5 \; – \; 2 = 3 > 0 \).

But if the dividend has a smaller power, the result is negative — for example, \( (6 \times 10^2) \div (3 \times 10^5) = 2 \times 10^{-3} \), where \( 2 \; – \; 5 = -3 < 0 \). Students who always "subtract the smaller from the larger" will never get a negative power, which prevents them from working with very small numbers.

Because both numbers are less than 10, the maximum possible coefficient result is just under 10 (e.g., \( 9.9 \div 1 = 9.9 \)). The coefficient can drop below 1 (e.g., \( 2 \div 8 = 0.25 \)), which means you must multiply the coefficient by 10 and subtract 1 from the power.

You will never need to add 1 to the power in division; adjusting upwards is exclusive to multiplication.

This is the “multiply the powers” misconception — one of the most common errors in standard form. The student correctly multiplied the coefficients (\( 3 \times 2 = 6 \)) but then multiplied the exponents (\( 5 \times 4 = 20 \)) instead of adding them (\( 5 + 4 = 9 \)).

The index law states \( 10^a \times 10^b = 10^{a+b} \), not \( 10^{a \times b} \). The correct answer is \( 6 \times 10^9 \). You can see the error by checking the magnitude: \( 6 \times 10^{20} \) is over 100 billion times larger than the correct answer.

The answer is correct, but the reasoning is dangerously flawed. The student’s rule — “subtract the smaller power from the bigger power” — only works when the dividend has the larger power. It fails when the divisor has the larger power.

For example, applying this rule to \( (4 \times 10^2) \div (8 \times 10^5) \), the student would again compute \( 5 \; – \; 2 = 3 \) and get \( 0.5 \times 10^3 = 5 \times 10^2 \). But the correct calculation is \( 2 \; – \; 5 = -3 \), giving \( 0.5 \times 10^{-3} = 5 \times 10^{-4} \) — a completely different answer. The correct rule is always dividend power minus divisor power.

The student has correctly multiplied the coefficients and correctly added the powers, but has stopped too early. This is the “forgetting to adjust” misconception — the student has not converted the result into proper standard form.

In standard form, the coefficient must be between 1 and 10 (i.e., \( 1 \leq a < 10 \)). Since 30 is not in this range, we rewrite: \( 30 = 3 \times 10^1 \), so \( 30 \times 10^5 = 3 \times 10^1 \times 10^5 = 3 \times 10^6 \). The correct answer is \( 3 \times 10^6 \).

The student has divided the coefficients correctly but has subtracted the powers in the wrong order. They computed \( 5 \; – \; 3 = 2 \) instead of the correct \( 3 \; – \; 5 = -2 \). When dividing, you always subtract the divisor’s power from the dividend’s power: the number you’re dividing comes first.

The correct answer is \( 3 \times 10^{-2} = 0.03 \). We can check: \( 6000 \div 200\,000 = 0.03 \). The student’s answer of \( 3 \times 10^2 = 300 \) is 10,000 times too large — a clear sign that the power is wrong.