JM24 · Lesson 24
Law of Indices
Interactive tools for powers, scientific notation, and binary — plus comics.
Use → / Enter / click to advance, ← to go back. Each slide builds the idea step by step.
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Part 1: meaning of a power. Part 2: prove the laws — product, quotient, power of a power, roots, then zero and negative indices.
Part 1 · What does a power mean?
1A. Powers: square and cube
Use the arrows. The index counts how many identical factors you multiply.
Square = two identical factors
Cube = three identical factors
General rule
\(x^{n}\) means \(n\) factors of \(x\) multiplied together.
\[ x^{n}=\underbrace{x\times x\times\cdots\times x}_{n\text{ times}} \]
1B. Why \(2x\) is not \(x^{2}\)
Addition changes the coefficient. A power is repeated multiplication.
Square means two identical numbers multiplied
\[ a^{2}=a\times a \]
Number examples
If someone claims \(x\times x=2x\)
Put \(x=3\): then \(x\times x\) is \(3\times 3=9\), but \(2x\) is \(2\times 3=6\). That would force \(3\times 3=6\) — impossible.
\[ \text{If }x\times x=2x\text{ and }x=3,\text{ then }3\times 3=6\text{?} \]
Conclusion
\(3\times 3=9=3^{2}\), so \(x\times x=x^{2}\). Addition is different: \(x+x=2x\).
\[ x\times x=x^{2}\qquad\text{and}\qquad x+x=2x \]
Part 2 · Index laws (add → subtract → multiply → divide → zero → negative)
2. Product (add indices) · \(x^{m}\times x^{n}=x^{m+n}\)
Double-click each block to expand, then count.
Your task
Double-click each block to split it into factors of \(x\).
Answer the question above.
\[ x^{2}\times x^{3}=x^{2+3}=x^{5} \]
3. Quotient (subtract indices) · \(x^{m}\div x^{n}=x^{m-n}\)
Write it as a fraction. Double-click to expand, then cancel matching factors across the line. The denominator must not disappear — when everything cancels, it becomes \(1\).
Cancel across the fraction line
1) Double-click numerator and denominator.
2) Click one top \(x\) and one bottom \(x\) to cancel.
Answer the question above.
\[ \frac{x^{5}}{x^{2}}=x^{5-2}=x^{3} \]
4. Power of a power (multiply indices) · \((x^{m})^{n}=x^{mn}\)
\(x\) to the power of \(n\) means \(n\) factors of \(x\) multiplied together. So \((x^{2})^{3}=x^{2}\times x^{2}\times x^{2}\). Double-click each \(x^{2}\) to split, then count.
Your task
Double-click each \(x^{2}\) block to turn it into \(x\times x\).
Answer the question above.
\[ (x^{2})^{3}=x^{2}\times x^{2}\times x^{2}=x^{6}=x^{2\times 3} \]
5. Roots / fractional indices (divide indices) · \(x^{\frac{n}{m}}=\sqrt[m]{x^{n}}\)
What is a square root?
A square root undoes squaring — it is the inverse of the 2nd power.
\[ \left(\sqrt{a}\right)^{2}=a \]
Higher roots
An \(m\)-th root undoes an \(m\)-th power: \(\left(\sqrt[m]{a}\right)^{m}=a\).
\[ \sqrt[m]{a^{m}}=a \]
Why a fraction?
We want \(r\) with \(\left(x^{r}\right)^{m}=x\). Then \(x^{rm}=x^{1}\), so \(r=\dfrac{1}{m}\).
\[ x^{\frac{1}{m}}=\sqrt[m]{x} \]
General rule
\(x^{\frac{n}{m}}=\left(x^{n}\right)^{\frac{1}{m}}=\sqrt[m]{x^{n}}\) (also equal to \(\left(\sqrt[m]{x}\right)^{n}\)).
\[ x^{\frac{n}{m}}=\sqrt[m]{x^{n}} \]
\[ x^{\frac{n}{m}}=\sqrt[m]{x^{n}} \]
6. Zero exponent · \(x^{0}=1\)
Start from \(0=n-n\)
Any number minus itself is zero. Write the zero index as a difference of equal indices.
\[ 0=n-n \]
Apply the quotient rule
Then \(x^{0}=x^{n-n}=\dfrac{x^{n}}{x^{n}}\). Same number on top and bottom cancels to \(1\) (for \(x\neq 0\)).
\[ x^{0}=x^{n-n}=\dfrac{x^{n}}{x^{n}}=1 \]
Therefore \(0^{0}\) is undefined
Any positive power of \(0\) is \(0\): \(0^{n}=0\). The same pattern would become \(\dfrac{0^{n}}{0^{n}}\) — dividing by zero, which is not allowed.
\[ 0^{0}\text{ is undefined} \]
Summary
Zero exponent is \(1\) only when the base is not zero.
\[ x^{0}=1\quad(x\neq 0),\qquad 0^{0}\text{ undefined} \]
\[ x^{0}=1\quad(x\neq 0),\qquad 0^{0}\text{ undefined} \]
7. Negative exponent · \(x^{-n}=\dfrac{1}{x^{n}}\)
Uses the quotient rule and \(x^{0}=1\).
Proof · via division
Consider \(\dfrac{x^{0}}{x^{n}}\). Left side is \(\dfrac{1}{x^{n}}\) because \(x^{0}=1\).
\[ \dfrac{x^{0}}{x^{n}}=\dfrac{1}{x^{n}} \]
Quotient rule on the indices
Right side: \(x^{0-n}=x^{-n}\). The two sides are equal, so the negative power is a reciprocal.
\[ x^{-n}=\dfrac{1}{x^{n}} \]
Check · via product
\(x^{n}\cdot x^{-n}=x^{n+(-n)}=x^{0}=1\). The number that multiplies \(x^{n}\) to give \(1\) is \(\dfrac{1}{x^{n}}\).
\[ x^{n}\cdot x^{-n}=1 \]
Summary
A negative index means “one over the positive power” (base not zero).
\[ x^{-n}=\dfrac{1}{x^{n}}\quad(x\neq 0) \]
\[ x^{-n}=\frac{1}{x^{n}}\quad(x\neq 0) \]
Write very large or very small numbers as \(a \times 10^{n}\) with \(1 \le a < 10\). Uses the same index laws as the Powers lab — multiply adds indices, divide subtracts.
Part 1 · Standard form
1. What is scientific notation?
Worksheet §1–10: express ordinary numbers in the form \(a \times 10^{n}\).
Standard form
Scientific notation (standard form) writes a number as a coefficient times a power of ten.
\[ a \times 10^{n}\qquad\text{where}\quad 1 \le a < 10 \]
Why \(102\,000\,000 = 1.02 \times 10^{8}\)?
Factor out powers of ten step by step:
\[ 102\,000\,000 = 102 \times 1\,000\,000 = 1.02 \times 100\,000\,000 \]
Powers of ten are named by how many zeros follow the \(1\): \(10^{2}=100\) (two zeros), \(10^{3}=1000\) (three zeros), … so in general \(10^{n}\) is \(1\) followed by \(n\) zeros. Here \(100\,000\,000\) has eight zeros, so \(100\,000\,000=10^{8}\). Therefore
\[ 102\,000\,000 = 1.02 \times 10^{8} \]
Rule: left shift ↔ index \(+1\)
Moving the decimal point one place left multiplies by \(10\), so the power of ten goes up by \(1\). Count how many places you move left — that count is \(n\).
Why \(0.000\,504 = 5.04 \times 10^{-4}\)?
Factor using a fraction of a power of ten:
\[ 0.000\,504 = \dfrac{5.04}{10\,000} = 5.04 \times \dfrac{1}{10\,000} \]
\(10\,000\) has four zeros, so \(10\,000=10^{4}\). A negative power means a reciprocal: \(10^{-4}=\dfrac{1}{10^{4}}=\dfrac{1}{10\,000}\). Therefore
\[ 0.000\,504 = 5.04 \times 10^{-4} \]
Rule: each right shift of the decimal divides by \(10\), so the index goes down by \(1\) (negative for small numbers).
Same idea for any shift
Each left shift is one factor of \(10\) (index \(+1\)); each right shift divides by \(10\) (index \(-1\)). That is \(10^{m}\times 10^{n}=10^{m+n}\).
\[ 2.45 \times 10^{2} \times 10^{6} = 2.45 \times 10^{8} \]
Part 2 · Move the decimal
2. Convert by shifting the decimal point
Convert the number below into standard form.
\(1 \times 10^{0}\)
Your task
Stop when exactly one non-zero digit is before the decimal point (\(1 \le a < 10\)).
Move the decimal until you have one non-zero digit before the point.
\[ 1.02 \times 10^{8} \]
Part 3 · Practice
3. Convert both ways
Random question: ordinary → scientific, or scientific → ordinary.
—
Enter your answer, then Check.
\[ \text{answer} \]
Part 4 · Multiply & divide
4. Use index laws
Combine coefficients, then add or subtract the indices — same rules as \(x^{m} \times x^{n}\) and \(x^{m} \div x^{n}\).
—
Multiply: add indices. Divide: subtract indices.
\[ \text{answer} \]
Part 5 · Compare & order
5. Arrange in order
Drag into ascending order (smallest → largest).
Convert mentally to compare, then drag the cards.
\[ \text{correct order} \]
Guided scientific-notation walkthrough — step through conversion and arithmetic modules. The Scientific notation lab above stays available for practice drills.
Base 2 uses powers of \(2\) as place values — the same indices you met in the Powers lab. Flip bits, match targets, and try worksheet-style challenges.
Part 1 · Base 2 and place value
1. From indices to binary
Each binary digit is a coefficient times a power of \(2\).
Denary expanded form
Binary is base \(2\)
Only digits \(0\) and \(1\)
Each place is either “on” (\(1\)) or “off” (\(0\)) — like light bulbs in a row.
Place value of a digit
\[ \text{In } 10111_{(2)}, \text{ the } 0 \text{ has place value } 2^{3} \]
Part 2 · Bit flip playground
2. Flip the bits
Click a bulb to toggle \(0\leftrightarrow1\). Watch the expanded form and denary value update live.
Divide-by-\(2\) method: stack remainders bottom-up to get binary.
Convert: \(35_{(10)}\)
Binary (read remainders bottom-up):
Click “Divide by 2” for each step.
\[ 35_{(10)} = 100011_{(2)} \]
Part 3 · Challenge
3. Match the denary number
Flip bits until your binary equals the target.
Target: \(35_{(10)}\)
Flip bits to match the target, then Check.
\[ 35_{(10)} = 100011_{(2)} \]
Part 4 · Add, subtract & order
4A. Binary arithmetic (denary answer)
Convert each binary number to denary, then add or subtract.
\(110101_{(2)} + 1101011_{(2)}\)
Answer the question above.
\[ 53 + 107 = 160_{(10)} \]
4B. Order smallest to largest
Drag cards into the row (ascending denary value).
Drag cards into the slots left-to-right (smallest first).
\[ 28 < 55 < 57 \]
