Laws of exponents for real numbers

Start from the $n$th root. For a positive real number $a$ and a positive integer $n$, the $n$th root of $a$, written $\sqrt[n]{a}$, is the positive real number $b$ satisfying $b^n = a$. So $\sqrt[3]{8} = 2$ because $2^3 = 8$, and $\sqrt{9} = 3$ because $3^2 = 9$. Class 9 already uses this idea for square roots and cube roots; the only new step here is writing it with an exponent.

Recall the integer-power laws first. For a positive real base $a$ and integers $p, q$, these hold with no restriction:

where: $a$ is the base (a positive real number), and $p, q$ are the whole-number exponents being combined.

Extend by definition, not by counting. A fraction like $\frac13$ cannot mean "multiply $a$ by itself one-third of a time," so the meaning of $a^{1/n}$ is chosen — chosen to be whatever keeps the laws above true. That choice turns out to be forced, as the derivation below shows, and it gives:

where: $a$ is a positive real number, $n$ is a positive integer (the root taken), and $m$ is a positive integer (the power applied before or after the root — either order gives the same value).

The general rational-exponent laws. With that definition in place, the same four laws now hold for a positive real base $a$ and rational exponents $p, q$:

Nothing about the rules changed between integer exponents and rational exponents — only the type of number allowed as the exponent widened. That is the entire idea of this chapter in one line. This matters because, without exponent notation, every expression involving roots has to be handled with separate root-manipulation rules — multiplying two different roots, or raising a root to a power, would each need its own special-case reasoning. Writing $\sqrt[n]{a}$ as $a^{1/n}$ folds all of that into the four laws already familiar from integer powers, so a single toolkit now covers whole numbers, fractions, and roots alike. This is also why Class 9 introduces rational exponents right after establishing the $n$th root: the notation is a convenience built on a definition, not a new set of facts to memorise separately.

Simplify $\dfrac{(9)^{3/2} \times (27)^{2/3}}{(3)^{2}}$, then extend to a mixed-base product.

Step 1. Express every base as a power of the common prime $3$: $9 = 3^2$ and $27 = 3^3$.

Step 2. Rewrite each factor with the power-of-a-power law $(a^m)^p = a^{mp}$:

Step 3. Substitute back: $\dfrac{3^3 \times 3^2}{3^2}$. Combine the numerator with $a^p \cdot a^q = a^{p+q}$, giving $3^3 \times 3^2 = 3^5$.

Step 4. Apply the quotient law $\dfrac{a^p}{a^q} = a^{p-q}$: $\dfrac{3^5}{3^2} = 3^{5-2} = 3^3 = 27$. The expression simplifies to $27$.

Competitive stretch

Simplify $4^{1/2} \times 8^{1/3}$, where the two bases are not already equal.

Step 1. Rewrite both bases as powers of the prime $2$: $4 = 2^2$, $8 = 2^3$.

Step 2. Apply the power-of-a-power law: $4^{1/2} = (2^2)^{1/2} = 2^{2 \times \frac12} = 2^1$, and $8^{1/3} = (2^3)^{1/3} = 2^{3 \times \frac13} = 2^1$.

Step 3. Both factors now share the base $2$, so combine with $a^p \cdot a^q = a^{p+q}$: $2^1 \times 2^1 = 2^{2} = 4$. So $4^{1/2} \times 8^{1/3} = \mathbf{4}$. The general strategy whenever bases differ is always the same: hunt for a common prime base before touching the product or quotient laws.

• Applying $a^p \cdot a^q = a^{p+q}$ or $\dfrac{a^p}{a^q} = a^{p-q}$ directly to terms with different bases — these laws only combine exponents when the base is identical; rewrite with a common prime base first.
• Confusing $a^{m/n}$ with $\dfrac{a^m}{n}$ — the exponent $\frac{m}{n}$ means "take the $n$th root and raise to the $m$th power," not "divide the power by $n$."
• Mis-simplifying $(a^p)^q$ by adding exponents instead of multiplying — the power-of-a-power rule gives $a^{pq}$, not $a^{p+q}$; that addition rule is reserved for products of powers.
• Forgetting that $a^{1/n} = \sqrt[n]{a}$ in this syllabus requires $a$ to be a positive real number — the rational-exponent laws are stated only for positive real bases.

Ask the tutor: If switching between root notation and fractional-exponent notation still feels shaky, ask the AI tutor to walk through a few more mixed-base simplification problems step by step.