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Powers: definitions, rules, and conditions on the base

2026-10-22 huynhvietkhanh Comments: 0

Simplify powers with their domain conditions, distinguish integer from real exponents, and compare powers with a common base.

High schoolPowers & Logarithms

Content updated: October 22, 2026

Power rules apply only where the expressions involved are defined. Distinguishing integer exponents from real exponents prevents extending a formula valid for positive bases to cases that have not been defined.

Objectives: calculate and simplify powers; retain conditions arising from denominators or roots; compare powers with the same base.

Prerequisites: multiplication, fractions, nth roots, and absolute values.

On this page: Idea · Method · Examples · Practice · Further learning

1. From integer to real exponents

For an integer $n\ge1$, $a^n$ is the product of $n$ factors equal to $a$; $a$ is the base and $n$ the exponent. For $a\ne0$, define $a^0=1$ and $a^{-n}=1/a^n$. In this lesson, $0^0$ and negative powers of $0$ are undefined.

For $a>0$, $m\in\mathbb Z$, and an integer $n\ge2$, define $a^{m/n}=\sqrt[n]{a^m}$ using the positive root. The value is independent of the representation of the same rational number; for example, $a^{1/2}=a^{2/4}$. A positive base allows consistent use of the rules for every rational exponent.

For $a>0$, a power $a^\alpha$ with an irrational exponent is extended from rational powers by a limit: if $r_n\in\mathbb Q$ and $r_n\to\alpha$, then $a^{r_n}\to a^\alpha$. Existence and independence from the chosen sequence are properties of this continuous extension. Do not use a negative base to define arbitrary real powers.

2. Rules and the direction of comparison

For $a,b>0$ and $u,v\in\mathbb R$, the rules are

$$\begin{aligned}a^ua^v&=a^{u+v},&\frac{a^u}{a^v}&=a^{u-v},\\(a^u)^v&=a^{uv},&(ab)^u&=a^ub^u,\\\left(\frac ab\right)^u&=\frac{a^u}{b^u}.\end{aligned}$$

For integer exponents, many formulas also hold for negative bases different from $0$; do not extend this observation to arbitrary real exponents. For a positive base, $a^x>0$, so division by $a^x$ is allowed. The sum $a^u+a^v$ does not simplify to $a^{u+v}$.

If $a>1$, comparing $a^u,a^v$ preserves the comparison between $u,v$. If $0<a<1$, the direction reverses. For $a=1$, every power is $1$, so no relation between the exponents follows.

3. Eight groups of examples

Example 1. Integer exponents

We have $3\cdot3=3^2$ and $q\cdot q\cdot q\cdot q=q^4$ for every real $q$. Also, $2^{-3}=1/8$ and $2^{-2}/3^2=1/(4\cdot9)=1/36$. Since $3^x>0$, we always have $(3^x)^0=1$ for every real $x$.

Example 2. Simplification using the rules

The following positive-base expressions use real exponents; retain any conditions imposed by variable denominators. We have

$$2^{3x}2^{4x}=2^{7x},\qquad \frac{4x^3y^7}{12xy^5}=\frac{x^2y^2}{3}\quad(xy\ne0).$$
$$\frac{2^{2n}\cdot4^n\cdot2}{16^n}=2^{2n+2n+1-4n}=2.$$
$$\frac{5^{2x-1}\cdot9^{x-2}}{15^{2x-3}}=5^2\cdot3^{-1}=\frac{25}{3}.$$

In the next fraction, the denominator is explicitly $3\cdot2^t-2^t$, factored as follows:

$$\frac{2^t-2^{t-2}}{3\cdot2^t-2^t}=\frac{2^t(1-1/4)}{2^t(3-1)}=\frac38.$$

The denominator is $2^{t+1}>0$, so the cancellation is valid for every real $t$. Finally, set $u=3^x>0$ to obtain

$$\frac{9^x-1}{3^x+1}=\frac{(u-1)(u+1)}{u+1}=u-1=3^x-1.$$

Example 3. Exponent one-half

We have $(0.25)^{1/2}=\sqrt{1/4}=1/2$. The square root is the nonnegative value; do not attach a $\pm$ sign to the value of this expression.

Example 4. Opposite exponents

We have $2x^{1/2}\cdot4x^{-1/2}=8x^0=8$, for $x>0$. The condition comes from the negative one-half exponent; a constant result does not reinstate $x=0$ in the domain.

Example 5. A radical expression independent of x

For $4\le x\le8$, consider $A=\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}$. Set $u=\sqrt{x-4}\in[0;2]$. The two outer radicands equal $(u+2)^2$ and $(u-2)^2$, so both are nonnegative. Therefore

$$A=|u+2|+|u-2|=(u+2)+(2-u)=4.$$

Alternatively, $A^2=2x+2|x-8|=16$ on the specified interval, and $A\ge0$ implies $A=4$. The interval restriction determines the signs in the absolute values.

Example 6. Approximating an irrational exponent

Choose rational numbers $r_1=1$, $r_2=1.4$, $r_3=1.41$, $r_4=1.414,\ldots$ by successively truncating the decimal expansion of $\sqrt2$. Then $r_n\to\sqrt2$, and $3^1,3^{1.4},3^{1.41},3^{1.414},\ldots$ tends to $3^{\sqrt2}$ by the definition of the continuous extension.

Example 7. A base greater than 1

Since $\sqrt2<\sqrt3$ and $2>1$, it follows that $2^{\sqrt2}<2^{\sqrt3}$.

Example 8. A base between 0 and 1

Since $(6/5)^2=36/25<3$, we have $6/5<\sqrt3$. With the base $2/3\in(0;1)$, it follows that $(2/3)^{6/5}>(2/3)^{\sqrt3}$.

4. Practice exercises

  1. Calculate $16^{3/4}$ and $27^{-2/3}$.
  2. Simplify $\dfrac{x^{3/2}-x^{1/2}}{x^{1/2}}$ and state the condition.
  3. For $A$ in Example 5, calculate $A$ at $x=13$; why is it no longer 4?
Show hints and answers
  1. $16^{3/4}=2^3=8$; $27^{-2/3}=1/3^2=1/9$.
  2. We require $x>0$. Factor the numerator as $x^{1/2}(x-1)$ to obtain $x-1$ on that domain.
  3. Here $u=\sqrt{13-4}=3$, so $A=|3+2|+|3-2|=6$. Since $x>8$, the sign of $u-2$ differs from its sign on the original interval.

5. Common mistakes

Retain denominator and root conditions after simplification. Always use $\sqrt{v^2}=|v|$. Distinguish multiplication from a decimal point. Do not use strict monotonicity when the base is 1, or arbitrary real exponents for a negative base.

6. Related lessons

7. Videos and reference material

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