3.01a, 3.01b - Fractional powers

Roots as fractional powers

How can raising $16$ to a power produce $\frac18$? The exponent $-\frac34$ contains three instructions: a root, a power and a reciprocal. Begin with the root.

The number $9$ multiplied by itself gives $81$. So taking a square root reverses squaring. In index notation, $(81^{1/2})^2=81^{(1/2)\times2}=81$. This explains the exponent $1/2$: multiplying it by $2$ restores exponent $1$.

The same reasoning gives $a^{1/3}=\sqrt[3]{a}$ and $a^{1/4}=\sqrt[4]{a}$. The denominator tells you how many equal factors rebuild the original number. It does not tell you to divide the original number by that denominator.

Keep the base and change only the root index. Now four equal factors must rebuild $81$.

Our general fractional-power methods use positive bases. Negative bases need care: $\sqrt[3]{-8}=-2$ is real, but $\sqrt{-8}$ is not real. Do not apply an unrestricted fractional-index rule to negative inputs.

Reverse the root with a power to check your answer.

Powers and roots together

The numerator of a fractional index tells you the power; the denominator tells you the root.

Start with $32^{1/5}=2$. Squaring that value gives $(32^{1/5})^2=32^{2/5}=4$. This follows from the power-of-a-power law: multiply the indices.

The two orders give the same result for positive bases. Root first is usually the practical choice when the base is an exact power.

Change only the numerator from $2$ to $3$. The root is unchanged; the final power changes.

A fractional index does not necessarily make a number smaller: $32^{6/5}=2^6=64$. For a base greater than $1$, an exponent above $1$ produces a value above the base.

Keep the root small before taking the power.

Negative fractional indices

A negative index takes the reciprocal of the corresponding positive power; it does not make that value negative.

You already know $a^{-k}=1/a^k$ for an integer $k$ and nonzero $a$. Combine that rule with the root-and-power method, using positive bases throughout.

For a base greater than $1$, the positive power exceeds $1$, so its reciprocal is between $0$ and $1$. If the base is between $0$ and $1$, the reverse happens: for example, $(1/4)^{-1/2}=2$. The sign of the exponent alone does not determine whether the result exceeds $1$.

Change only the numerator of the fractional index. You must now cube the root before taking its reciprocal.

The order “root, power, reciprocal” makes every instruction visible, especially on a non-calculator paper.

Show the root and power before writing the reciprocal.

Estimating unfamiliar powers

Compare unfamiliar roots with known powers, then refine the interval if a particular accuracy is requested.

For positive inputs, square roots increase as their inputs increase. Thus $64<70<81$ gives $8<\sqrt{70}<9$. This locates the root, but to round it to the nearest integer you must compare it with the halfway point $8.5$, not merely choose whichever square number seems closer.

Change only the base from $70$ to $75$. The integer bounds stay the same, but the root crosses the rounding boundary.

The method extends to other powers. Since $2^3<10<3^3$, we know $2<10^{1/3}<3$. Squaring these positive values gives $4<10^{2/3}<9$. To refine this estimate, try a closer cube-root bracket: $2.1^3=9.261<10<10.648=2.2^3$, so $4.41<10^{2/3}<4.84$. A bound on a root becomes a bound on its positive integer power.

Use the midpoint square to justify your rounding.

Quick Recap

• A fractional index's denominator gives the root and its numerator gives the power.
• For positive bases, take the root first to keep the arithmetic small.
• A negative index gives a reciprocal; zero has no negative power.
• Estimate roots with known powers and test a midpoint when rounding.

The opening value is $16^{-3/4}=1/8$: fourth root, cube, reciprocal.