Powers: definitions, rules, and conditions on the base
Simplify powers with their domain conditions, distinguish integer from real exponents, and compare powers with a common base.
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
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
In the next fraction, the denominator is explicitly $3\cdot2^t-2^t$, factored as follows:
The denominator is $2^{t+1}>0$, so the cancellation is valid for every real $t$. Finally, set $u=3^x>0$ to obtain
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
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
- Calculate $16^{3/4}$ and $27^{-2/3}$.
- Simplify $\dfrac{x^{3/2}-x^{1/2}}{x^{1/2}}$ and state the condition.
- For $A$ in Example 5, calculate $A$ at $x=13$; why is it no longer 4?
Show hints and answers
- $16^{3/4}=2^3=8$; $27^{-2/3}=1/3^2=1/9$.
- We require $x>0$. Factor the numerator as $x^{1/2}(x-1)$ to obtain $x-1$ on that domain.
- 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
- Logarithms (English lesson)
- Exponential functions (English lesson)
- Logarithmic functions (English lesson)
- Derivatives of exponential and logarithmic functions (English lesson)
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