Mathematics · Article
Exponent laws and power functions
Use integer exponent laws with their nonzero guards, distinguish a reciprocal from a negative value, and read rational powers through principal real roots. Preserve original domains, delimit negative-base conventions, prove selected power-function ranges and interpret scaling with units before comparing a fixed power with a variable-exponent exponential.
Useful background article: Choosing a function for a relationship
Back to MathematicsAfter reading, you can
- Apply integer exponent laws with correct base, quotient and negative-exponent restrictions, and explain parentheses, signs and the lesson's unassigned value for 0^0.
- Interpret reduced rational exponents using principal real roots, determine original real domains, and use general fractional-power laws on positive bases while identifying failures at negative bases.
- Distinguish fixed-power functions from general polynomials and variable-exponent exponentials, prove selected complete domains and ranges, and interpret power scaling within a stated context and its units.
A power has a base and an exponent with different jobs
For a positive integer n, aⁿ means a product of n equal factors a. The base is the quantity repeated, and the exponent counts the factors. Thus 3⁴ = 3·3·3·3 = 81. In (−3)⁴ the entire negative number is the base; in −3⁴ the power is evaluated first and the minus sign is applied afterward. Read the expression's grouping before choosing a law.
The exponent need not remain a positive integer. We extend the definition to zero and negative integers using identities that remain consistent on nonzero bases, then to rational exponents using real roots. Each extension comes with conditions. A formula involving a reciprocal or an even root does not automatically accept every real input just because a positive-integer power does.
A fixed exponent p in a rule x^p makes the input the base. A rule such as 3^x puts the input in the exponent instead. These describe different families of functions. We work with integer and rational fixed exponents here; variable-exponent growth is developed in the next lesson. Keep an exact fraction or radical when its value is not a terminating number, rather than allowing a decimal approximation to replace its definition.
Integer exponent laws follow from products and reciprocals
For positive integers, multiplying powers with the same base joins their factors, so a^m·a^n = a^(m + n). Dividing cancels equal nonzero factors, giving a^m/a^n = a^(m − n) where the original denominator is nonzero. Raising a power to an integer power multiplies exponents. These are different operations: x³·x² joins five factors, while (x³)² joins two groups of three and gives six.
The table uses a clean common setting: a and b are nonzero real numbers, and m and n are any integers. In that setting all displayed expressions are defined, including zero and negative exponents, and the laws hold. With positive integer exponents, some product and power identities also allow zero bases, such as 0³·0² = 0⁵. Check each original expression rather than extending every identity to zero: 0³/0² is still undefined.
The product rule needs the same base; it does not combine 2³·3² into one power by adding exponents. The power of a product distributes to its factors, and the power of a quotient distributes only with a permitted nonzero denominator. Record exclusions before cancelling: x⁴/x² equals x² only on x ≠ 0. The simpler expression can be evaluated at zero, but that does not fill the original quotient's missing input.
| Operation | Identity | How the exponents change |
|---|---|---|
| Same-base product | a^m·a^n = a^(m + n) | Add |
| Same-base quotient | a^m/a^n = a^(m − n) | Subtract numerator exponent minus denominator exponent |
| Power of a power | (a^m)^n = a^(mn) | Multiply |
| Power of a product | (ab)^n = a^n b^n | Apply the same exponent to each factor |
| Power of a quotient | (a/b)^n = a^n/b^n | Apply the same exponent to numerator and denominator |
A negative exponent means a reciprocal
For a ≠ 0, dividing a positive integer power by itself gives 1 and subtracting its exponents gives zero. This motivates the consistent definition a⁰ = 1. It applies to negative nonzero bases as well: (−7)⁰ = 1. In this lesson 0⁰ is unassigned. Do not use the nonzero-base zero-exponent rule to assign it a value. Other settings may specify a convention for a different purpose; none is being added here.
For a ≠ 0 and a positive integer n, define a^(−n) = 1/a^n. For example, 4⁻² = 1/16. A negative exponent does not mean a negative answer. With base −4, an even negative exponent gives (−4)⁻² = 1/16, while an odd one gives (−4)⁻³ = −1/64. The reciprocal retains the sign of the underlying nonzero power.
Zero cannot be raised to a negative integer power because the definition would divide by zero. Likewise, moving a factor across a fraction bar changes the sign of its exponent only when the original factor is nonzero. Simplifying x³·x⁻³ to 1 retains x ≠ 0. Multiplication by zero does not rescue an undefined factor: 0·0⁻¹ is undefined, rather than zero.
| Expression | Meaning | Value |
|---|---|---|
| 4⁻² | 1/4² | 1/16 |
| (−4)⁻² | 1/(−4)² | 1/16 |
| (−4)⁻³ | 1/(−4)³ | −1/64 |
| (−4)⁰ | Nonzero-base zero exponent | 1 |
| 0⁻² | Would require 1/0² | Undefined |
| 0⁰ | No value assigned in this lesson | Unassigned |
Parentheses identify what the power actually applies to
Compare (−5)² = 25 with −5² = −25: the first base includes the minus sign, while the second is the negative of 5². Similarly, (2x)³ = 8x³ because both 2 and x are factors of the base, while 2x³ means only x is cubed. For an odd positive integer power, changing the base's sign changes the output's sign; for an even one, the negative factors pair off and the output is nonnegative.
A power distributes over a product under the law's conditions, but not over addition. For example, (1 + 1)² = 4 whereas 1² + 1² = 2. The square of a sum has the additional cross term: (a + b)² = a² + 2ab + b². Apply a power to the complete grouped expression; do not move it past a plus or minus sign as if that sign were multiplication.
A superscript minus sign in an exponent belongs to that exponent, not to the output. Thus −5⁻² = −1/25, while (−5)⁻² = 1/25. State the original nonzero conditions before rearranging a quotient containing negative powers. An exact substitution at a permitted negative input is a useful sign check, while the law and its domain conditions justify the whole simplification.
A root symbol selects one real value
For y ≥ 0, √y denotes its nonnegative principal square root. Therefore √25 = 5, not ±5. Solving t² = 25 is a different task and has the two solutions 5 and −5. More generally, an even-index real root is defined only for a nonnegative radicand and denotes the nonnegative principal value. For an odd index, every real radicand has one real root, with the radicand's sign: ∛(−27) = −3.
For every real x, √(x²) = |x|. Squaring has removed the sign, and the principal square root does not restore it. At x = −3, √(x²) gives 3 rather than −3. In the opposite order, (√x)² = x is valid only on the original domain x ≥ 0. An equality of simplified expressions does not enlarge an earlier root's domain.
Use root notation to define exact values that need not be rational. The quantity ∛2 is the unique real number whose cube is 2, and √2 is the nonnegative number whose square is 2. A decimal display can approximate those values, but a rounded approximation cannot establish an exact power identity, intercept or preimage. We stay within real roots here; a negative radicand under an even root has no real value in this lesson.
Reduce a rational exponent before reading its real domain
Write a rational exponent as m/n in lowest terms, with integer m and positive integer n. For a positive base x, define x^(m/n) = (√[n]x)^m, taking the positive nth root; when n = 1 this is the integer definition. For positive exponents, zero is also allowed and gives zero. For negative exponents, zero is excluded because taking the power includes a reciprocal.
For a negative base we use a specifically delimited real-root convention: reduced n must be odd, and x^(m/n) means the mth integer power of the unique real nth root of x. Thus (−32)^(1/5) = −2 and (−32)^(2/5) = 4. If the reduced denominator is even, a negative base has no real value under this convention. If m = 0, the reduced exponent is 0/1 and the separate zero-exponent rule still requires a nonzero base.
Reduction is essential. The exponent 2/6 is the same rational number as 1/3, so (−8)^(2/6) means (−8)^(1/3) = −2 under this convention. The different expression √[6]((−8)²) has value 2. Turning an unreduced negative-base exponent into a power-then-even-root expression can therefore change the value. Read the root-first definition with the reduced exponent, and preserve a radical expression's own original domain rather than assigning it a new one through a tempting exponent rewrite.
| Reduced exponent | Real inputs | Reason |
|---|---|---|
| m > 0, n odd | All real x | Odd real root, followed by a positive integer power |
| m > 0, n even | x ≥ 0 | Principal even root |
| m < 0, n odd | x ≠ 0 | Odd root followed by a reciprocal power |
| m < 0, n even | x > 0 | Even root and a nonzero reciprocal denominator |
| m = 0, n = 1 | x ≠ 0 | 0⁰ remains unassigned here |
Use positive bases for the general fractional-power laws
For strictly positive real a and b and rational r and s, the familiar identities hold: a^r·a^s = a^(r + s), a^r/a^s = a^(r − s), (a^r)^s = a^(rs), (ab)^r = a^r b^r and (a/b)^r = a^r/b^r. Positive bases keep every required root and reciprocal defined and positive. The denominator powers cannot be zero. These are the conditions used for general fractional simplifications in this lesson.
The root definition explains the same-base product law. Write r and s with a common positive denominator N and let t be the positive Nth root of a. Then a^r and a^s are integer powers of the same nonzero t, so adding their integer exponents gives a^(r + s). Positive roots likewise make successive rational powers compatible. An individual identity can sometimes hold at zero or at negative inputs, but it needs a separate domain and value check; the blanket fractional-law setting above does not include those bases.
The power-of-power law can fail at a negative base even when both real sides are defined. For instance, ((−1)²)^(1/2) = 1, while multiplying the exponents would give (−1)¹ = −1. Also, (−8)^(2/3) = 4, so ((−8)^(2/3))^(3/2) = 8, whereas (−8)^1 = −8. The principal even root in the later step loses the original sign. Do not extend positive-base fractional laws to negative bases just because the reduced real-root convention supplies values for individual powers.
A power function has one fixed exponent
A power function has the single-term form f(x) = kx^p with fixed coefficient k and fixed exponent p. We use nonzero k and integer or rational p in the examples here, with the real domain justified by the original expression. Multiplying by k rescales the output; it does not repair an undefined power. A zero-output function can be defined separately, but writing 0 times an undefined expression does not assign a value at its missing input.
Positive integer powers are polynomial terms and can accept every real input. The rule 3x² is both a power function and a quadratic polynomial, while 3x² + 1 is a quadratic polynomial with two terms and is not a single pure power. Fractional powers such as √x are not polynomial terms, and negative integer powers such as x⁻² = 1/x² are rational functions on x ≠ 0 rather than polynomials on all real inputs. These family names can overlap when their definitions and domains permit it.
There is a domain detail at exponent zero. Under this lesson's convention, x⁰ equals 1 on x ≠ 0 and has no value at zero. The constant polynomial 1 on all real inputs is an extension with a different domain. Likewise, cancelling a quotient to a power preserves the quotient's original exclusions. Classify the stated function, including its domain, rather than only the appearance of a simplified formula.
A complete range needs a permitted input for every claimed output
The table gives natural real domains and complete ranges for eight coefficient-one rules. For x², every output is nonnegative and any y ≥ 0 is attained at x = √y; the negative preimage is also available when y > 0. For x³, any real y is attained at x = ∛y. For 1/x, no output is zero and every nonzero y is attained at x = 1/y. For 1/x², every output is positive and any y > 0 is attained at x = 1/√y, with its negative also permitted.
The zero-power rule returns 1 at every nonzero input, so its complete range is {1}; at that output the full preimage is the whole nonzero domain, not a short sample list. For √x, every output is nonnegative and any y ≥ 0 is attained at the permitted x = y²; the principal-root condition is why a negative proposed y is rejected. For ∛x, any real y is attained at x = y³.
For x^(2/3) = (∛x)², negative inputs are permitted and outputs are nonnegative. At any y > 0, the equation (∛x)² = y gives ∛x = √y or −√y, hence all preimages are x = y√y and x = −y√y. At y = 0 the sole preimage is 0. This both restricts every output and supplies every proposed nonnegative output, proving the complete range. A domain with additional context restrictions can reduce any of these ranges; the table refers to the specified natural domains.
| Rule | Domain | Range | Complete preimages of a permitted y |
|---|---|---|---|
| x² | All real x | y ≥ 0 | ±√y; zero appears once |
| x³ | All real x | All real y | ∛y |
| 1/x | x ≠ 0 | y ≠ 0 | 1/y |
| 1/x² | x ≠ 0 | y > 0 | ±1/√y |
| x⁰ | x ≠ 0 | {1} | Every nonzero x when y = 1 |
| √x | x ≥ 0 | y ≥ 0 | y² |
| ∛x | All real x | All real y | y³ |
| x^(2/3) | All real x | y ≥ 0 | ±y√y for y > 0; x = 0 for y = 0 |
Scaling an input multiplies a pure power output by a known factor
For positive x and positive scale factor r, a pure power f(x) = kx^p satisfies f(rx) = r^p f(x) for the rational exponents used here, whenever both x and rx belong to the model's domain. Doubling an input multiplies an x² output by 4, an x³ output by 8, an x⁻¹ output by 1/2 and an x⁻² output by 1/4. This is a multiplicative comparison, not a claim of a constant additive change per input unit.
Name the units before calculating. If all dimensions of a box are multiplied by a dimensionless size factor q, each face area scales by q² and volume by q³. A convenient form is V(q) = V₀q³, where V₀ is the prototype volume with cubic-length units and q is a ratio of corresponding lengths in the same units. The powered base is dimensionless, and multiplying it by V₀ supplies the output unit. When variables instead mean numerical measurements in named units, coefficients must carry the units needed by the equation.
The pure scaling identity depends on the pure-power form and its fixed conditions. For x² + 1, doubling x changes only the squared term, so it does not multiply the whole output by four. A shape whose proportions change likewise need not obey the unchanged-shape volume law. A permitted scale factor also cannot authorize an input beyond a maximum size or at an originally excluded zero. State the relationship, its conditions and the permitted input comparison together.
Separate an exact power value from the full rule
The explorer offers the eight rules in the range table: x², x³, 1/x, 1/x², x⁰, √x, ∛x and x^(2/3). Choose an original input and a proposed output independently. The input check asks whether the original power is defined and then gives its exact value. The output check asks whether at least one permitted input produces the proposed y, which need not be your selected input. Read the full range and preimage proof alongside the selected values, including the separate zero-output cases.
Use the guarded exponent laws and the negative-base power-of-power counterexample alongside those checks. A decimal approximation is not an exact radical identity, and a short control list is not the whole domain or preimage set. Keep the convention visible: rational exponents are reduced, negative bases use only an odd reduced root denominator, general fractional laws use positive bases, and x⁰ excludes zero in this lesson.
In a power function such as x³, the exponent is fixed and the base changes. In a real exponential rule such as 3^t, the base 3 is fixed and the exponent changes; standard exponential bases are positive and different from 1. A base of 1 gives a constant rule, and a negative fixed base does not give a real value for every real exponent, as (−3)^(1/2) shows. These distinctions prepare for Exponential growth and decay without requiring logarithms or extending the current real-root calculations to complex numbers.
Check a power function and its possible outputs
Choose a power rule, an input and a proposed output independently. First check the original real domain, then evaluate exactly. To test an output, find inputs and substitute them into the original rule. The complete range proofs cover every real output, including values outside the rational buttons.
f(x) = x²
Original real input domain: x ∈ ℝ. Complete real range: y ∈ ℝ; y ≥ 0.
Evaluate the selected input: x = 2
The input belongs to the original domain. Original function value: 4.
Test the selected proposed output: y = 2
Original equation to solve: x² = 2.
The proposed output is attained. Complete preimage set for this output: {−√2, √2}.
| Candidate input | In the original domain? | Original function value | Equals the proposed output? |
|---|---|---|---|
| −√2 | Yes | 2 | Yes |
| √2 | Yes | 2 | Yes |
Why the stated domain, range and preimages are complete
Every real square is nonnegative. For each real y ≥ 0, its principal square root exists and the inputs −√y and √y both square to y. These are all solutions of x² = y; at y = 0 they are the same input 0. Negative outputs are impossible.
| Input | In the original domain? | Function value |
|---|---|---|
| −8 | Yes | 64 |
| −2 | Yes | 4 |
| −1 | Yes | 1 |
| −1/8 | Yes | 1/64 |
| 0 | Yes | 0 |
| 1/8 | Yes | 1/64 |
| 1 | Yes | 1 |
| 2 | Yes | 4 |
The buttons illustrate the rules with rational inputs and proposed outputs. The full real-domain and range claims follow from the root, sign and reciprocal arguments above; this finite table is not the reason for those claims.
Keep the conditions with the exponent laws
For integer exponents, a nonzero real base makes product, quotient and power-of-a-power laws valid, including negative and zero exponents. Positive integer powers also allow base 0 where no denominator or zero power is introduced. Check that every expression in an identity is defined before using it.
aᵐaⁿ = a^(m+n)
2³·2² = 2⁵. Original function value: 32.
aᵐ/aⁿ = a^(m−n)
2⁴/2². Original function value: 4.
(aᵐ)ⁿ = a^(mn)
(2³)² = 2⁶. Original function value: 64.
aᵐaⁿ = a^(m+n)aᵐ/aⁿ = a^(m−n)(aᵐ)ⁿ = a^(mn)a⁰ = 1a^(−n) = 1/aⁿ
State the real-root convention
Here a rational exponent m/n is first reduced with n > 0. For a positive base, use the positive nth root and raise it to m. A negative base is allowed only when the reduced denominator is odd, using its real odd root. Base 0 is allowed for positive exponents and excluded for negative and zero exponents here. The exponent’s reduced form is part of the convention.
For example, 2/6 reduces to 1/3, so the two expressions below have different meanings. Taking a power first and then a root is a separate operation.
(−8)^(2/6) = (−8)^(1/3) = −2((−8)²)^(1/6) = 2
Use a strictly positive base for the general rational-exponent product, quotient and power-of-a-power laws. Extending a real-root convention to negative bases does not extend every fractional law to them.
A defined counterexample to a fractional power law
((−8)^(2/3))^(3/2) = 4^(3/2) = 8(−8)^((2/3)(3/2)) = (−8)¹ = −8
Both expressions are real and defined, but they have different values. The inner two-thirds power loses the negative sign before the outer principal-root power is taken. Thus multiplying fractional exponents without the positive-base condition can change a value.
The same sign issue appears in a simpler expression: the principal square root of a square gives an absolute value. √((−1)²) = 1 ≠ (−1)¹ = −1.
Square. x = 2: The input belongs to the original domain; Original function value: 4. y = 2: The proposed output is attained; Complete preimage set for this output: {−√2, √2}.
All evaluations and original-equation checks use exact fractions and symbolic real roots. Decimal approximations do not decide domain membership, an output match or whether two preimages are distinct.
Worked examples
A simplification retains both original nonzero inputs
Simplify ((4u⁻²v³)²)/(8u⁻¹v⁴) using positive exponents, state every original real-input restriction and evaluate it at u = −2, v = 3. May the simplified expression be used to assign an original value at v = 0?
- The negative powers u⁻² and u⁻¹ require u ≠ 0. The original denominator 8u⁻¹v⁴ is nonzero only when v ≠ 0 as well. Record both exclusions before simplifying.
- Square the grouped product:
(4u⁻²v³)² = 16u⁻⁴v⁶. The coefficient, the u factor and the v factor all receive the exponent 2; the exponent on u becomes −4, not −2 + 2. - Divide on the recorded domain: the coefficient is 16/8 = 2, the u exponent is −4 − (−1) = −3 and the v exponent is 6 − 4 = 2. Thus the quotient is
2u⁻³v² = 2v²/u³for u ≠ 0 and v ≠ 0. - At u = −2 and v = 3, the value is
2·9/(−8) = −9/4. Directly in the original expression, 4u⁻²v³ = 27 and the denominator is 8·(−1/2)·81 = −324, giving 729/(−324) = −9/4 as a check. - The reduced expression 2v²/u³ could be calculated as zero at v = 0 with u ≠ 0, but the original quotient would have denominator zero there. That new value would extend the function rather than evaluate it. The negative final value at the checked input comes from the negative odd power u³, not merely from a minus sign in an exponent.
The simplified expression is 2v²/u³ on the original domain u ≠ 0 and v ≠ 0. Its value at (−2, 3) is −9/4. Input v = 0 remains excluded despite the reduced formula's apparent value there.
An even root product and an odd-root square have different domains
Find the complete real domains and ranges of F(x) = (x − 4)^(1/2)(x − 4)^(−1/2) and G(x) = (x − 4)^(2/3) under the lesson's reduced real-root convention. Give all preimages for G at an arbitrary permitted output and check G(−4).
- The first factor of F needs x − 4 ≥ 0; its reciprocal-root factor needs x − 4 > 0. Their intersection is x > 4. On that positive-base domain the product law gives
(x − 4)^0 = 1. - Therefore F has domain x > 4 and range {1}; every x > 4 is a preimage of 1. At x = 4 the reciprocal root is undefined, so the simplification to 1 cannot supply an original value there.
- The reduced denominator 3 in G's exponent is odd, so the cube root of x − 4 exists for every real x. G is the square of that cube root; its outputs are nonnegative, including zero at x = 4.
- For any y > 0, solving
(∛(x − 4))² = ygives cube root values √y and −√y. Cubing yields all preimagesx = 4 + y√yandx = 4 − y√y, both real and permitted. At y = 0 the sole preimage is x = 4. Thus the complete range of G is y ≥ 0. - At x = −4, the shifted base is −8, whose cube root is −2, so G(−4) = 4. For the proposed output y = 2, the two exact irrational inputs are 4 + 2√2 and 4 − 2√2; their shifted cube roots are √2 and −√2, respectively, and both squares give 2.
F has domain x > 4 and range {1}, with every domain input producing 1. G has domain all real x and range y ≥ 0; at y > 0 its complete preimages are 4 ± y√y, and at y = 0 its sole preimage is 4. G(−4) = 4.
A size ratio keeps square and cubic units clear
A prototype rectangular box has dimensions 2 centimetres by 1 centimetre by 1 centimetre. Similar boxes multiply every dimension by a dimensionless factor q with 1 ≤ q ≤ 3. Find volume and the area of a face corresponding to the prototype's 2-by-1-centimetre face, prove the complete volume range and compare q = 1 with q = 2.
- The scaled dimensions are 2q, q and q centimetres. Multiply all three:
V(q) = 2q³cubic centimetres. The named face has sides 2q and q centimetres, soA(q) = 2q²square centimetres. The other kind of face has area q², so identifying the face matters. - The factor q is dimensionless because it is the ratio of corresponding side lengths measured in the same unit. The coefficients 2 in V and A represent the prototype volume 2 cubic centimetres and the prototype face area 2 square centimetres, respectively.
- For 1 ≤ q ≤ 3, cubing positive inputs gives 1 ≤ q³ ≤ 27, so 2 ≤ V(q) ≤ 54 cubic centimetres. Conversely, for any proposed numerical volume v in [2, 54], choose
q = ∛(v/2), which lies in [1, 3] and gives V(q) = v. This proves the complete range, including both endpoints. - At q = 1 the volume is 2 cubic centimetres and the named face area is 2 square centimetres. At q = 2 the volume is 16 cubic centimetres and the face area is 8 square centimetres. Doubling all dimensions multiplies volume by 8 and face area by 4.
- The comparison is valid because both sizes are permitted and the proportions are unchanged. It does not authorize q = 0 or q > 3 in this model, and it does not describe a box in which only one dimension is doubled. The power law is tied to the stated simultaneous scaling.
V(q) = 2q³ cubic centimetres and the specified face area is A(q) = 2q² square centimetres, on 1 ≤ q ≤ 3. The complete volume range is [2, 54] cubic centimetres. From q = 1 to 2, volume rises from 2 to 16 and that face area from 2 to 8, giving factors 8 and 4.
Try it yourself
Evaluate (−5)², −5², (−5)⁻², −5⁻² and (−5)⁻³. Explain which bases include the minus sign and whether a negative exponent necessarily gives a negative value. State the status of 0⁰ and 0⁻² in this lesson.
Show the answer and reasoning
The five values are 25, −25, 1/25, −1/25 and −1/125, respectively. A negative exponent does not necessarily give a negative value. Here 0⁰ is unassigned and 0⁻² is undefined.
Parentheses make −5 the base in the first, third and fifth expressions. Without them the minus is applied after the power of 5. Negative exponents take reciprocals of the nonzero corresponding positive powers; the even negative-base power is positive and the odd one negative. The zero-exponent identity requires a nonzero base, while a negative power of zero would divide by zero.
Simplify x³·x⁻⁵ and x⁴/x⁴, stating the original real domain of each. Do the final formulas permit you to insert an original value at zero? Compare with the valid product 0³·0².
Show the answer and reasoning
x³·x⁻⁵ = 1/x² on x ≠ 0, and x⁴/x⁴ = 1 on x ≠ 0. Neither original expression has a value at zero. The separate positive-power product 0³·0² is defined and equals 0⁵ = 0.
The first original expression includes a negative power, and the second has a denominator x⁴, so both exclude zero. Adding or subtracting exponents on those domains gives −2 and 0. The constant reduced result does not remove a denominator exclusion. Positive integer powers of zero are defined, which is why the last product has different input conditions.
Under the reduced real-root convention, evaluate (−32)^(3/5), (−32)^(−2/5) and 81^(−3/4). Does (−16)^(3/4) have a real value? Explain the root index, sign and reciprocal conditions.
Show the answer and reasoning
The three real values are −8, 1/4 and 1/27. The expression (−16)^(3/4) has no real value under this convention.
The fifth root of −32 is −2: cubing it gives −8, while its negative second power gives 1/(−2)² = 1/4. The principal fourth root of 81 is 3, so its negative third power is 1/27. A negative base with reduced even root denominator 4 is outside the real domain; the positive-power numerator does not repair that missing root.
Compare H(x) = √(x²) with J(x) = (√x)². Give each original real domain and correct simplified rule, and use x = −3 to explain why applying the general fractional power-of-power law without a positive-base guard would fail.
Show the answer and reasoning
H has all real inputs and H(x) = |x|. J has domain x ≥ 0 and J(x) = x there. At x = −3, H(−3) = 3 rather than −3; J is undefined.
The square inside H is always nonnegative, and the principal outer root selects its nonnegative magnitude. Multiplying the exponents to claim (x²)^(1/2) = x would wrongly retain a negative sign. J takes the even root first, so it excludes negative inputs and its later square does not enlarge the original domain.
For F(x) = (x + 1)^(−1/2), give the complete real domain and range, evaluate F(3) and supply a permitted input for every proposed output in the range. Is zero an attained output?
Show the answer and reasoning
Domain x > −1; range y > 0. F(3) = 1/2. For any y > 0, the permitted input x = 1/y² − 1 gives F(x) = y; zero is not attained.
The rule is 1/√(x + 1), so the even-root radicand must be positive to keep the denominator nonzero. All outputs are positive. For a positive y, the proposed input has x + 1 = 1/y² > 0, with principal root 1/y, so the reciprocal is y. A finite reciprocal of a positive denominator cannot equal zero.
Classify 3x², 3x² + 1, 3x⁻² and 3^x as power, polynomial, quadratic or exponential where appropriate. State the original domain of 3x⁻², and explain why the rule x⁰ under this lesson's convention differs from the constant polynomial 1 on all real inputs.
Show the answer and reasoning
3x² is a power function and a quadratic polynomial. 3x² + 1 is a quadratic polynomial, not one pure power. 3x⁻² is a power function and a rational function on x ≠ 0, not a polynomial on all real inputs. 3^x is exponential. The rule x⁰ equals 1 only on x ≠ 0 here; the constant polynomial also defines a value at zero.
A power rule has a fixed exponent on the changing base; a polynomial uses nonnegative integer powers and accepts all real inputs in its natural domain. A genuine quadratic has degree two, and adding a constant need not preserve pure-power scaling. The exponential has fixed positive base 3 and a changing exponent. The zero-power convention preserves the missing input zero, so the two constant-looking rules have different domains.
For the dimensionless rule f(x) = 12x⁻² restricted to 1 ≤ x ≤ 4, prove its complete range, find f(2) and compare f(2x) with f(x) when both inputs are in the domain. Which original inputs permit that doubling comparison?
Show the answer and reasoning
The complete range is 3/4 ≤ y ≤ 12; f(2) = 3. When both inputs are permitted, f(2x) = f(x)/4. Such original inputs are 1 ≤ x ≤ 2.
On the domain, 1 ≤ x² ≤ 16 and the positive reciprocal gives 3/4 ≤ 12/x² ≤ 12. For any y in those bounds, choose x = √(12/y); then 1 ≤ x ≤ 4 and f(x) = y, proving the whole range. Doubling multiplies x² by four and hence divides its reciprocal by four. Combining 1 ≤ x ≤ 4 with 1 ≤ 2x ≤ 4 gives the comparison domain 1 ≤ x ≤ 2.
What to carry forward
- Identify the complete base before applying a power. Parentheses, a leading minus sign and a minus in the exponent play different roles.
- For nonzero real bases and integer exponents, add exponents in same-base products, subtract in quotients and multiply in powers of powers. Preserve every original denominator or negative-power exclusion.
- A nonzero base raised to zero gives 1; 0⁰ remains unassigned here. A negative exponent means a reciprocal and does not by itself make the output negative.
- Principal even roots require nonnegative radicands and select nonnegative values; odd real roots accept every real radicand. In particular, √(x²) = |x| rather than x on all real inputs.
- Reduce rational exponents with positive denominator. Negative bases use only odd reduced denominators under the stated real-root convention; general fractional exponent laws use strictly positive bases.
- A complete power-function range requires both an output restriction and a permitted input for every proposed output. Check zero cases, original domains and all preimages rather than inferring completeness from samples.
- A pure power has fixed exponent and scales by a known multiplicative factor when both inputs are permitted. Retain units and context, and distinguish a changing base x^p from a fixed-base exponential b^x.
References
Sources for the concepts and teaching approaches in this article.
- College Algebra 2eOpenStax, Rice UniversityTextbook chapterOpenStax · College Algebra 2e · 1.2 Exponents and Scientific NotationOpenStax · College Algebra 2e · 1.3 Radicals and Rational ExponentsOpenStax · College Algebra 2e · 5.2 Power Functions and Polynomial FunctionsOpenStax · College Algebra 2e · 6.1 Exponential Functions