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Operations with Variable Exponents

43 questions

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Question 1Question →
Assume xx and yy are non-zero real numbers. When the algebraic expression (2x−3y4)34x5y−2\frac{(2x^{-3}y^4)^3}{4x^5y^{-2}} is fully simplified to the form cxaybcx^ay^b, what is the integer value of bb?
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Answer: 14

Answer

14
The expression simplifies to 2x−14y142x^{-14}y^{14}. Because the question asks for the exponent of yy, which is denoted by bb in the target expression cxaybcx^ay^b, the correct value is 14.

Step-by-Step Solution

1
Apply the power of a product rule to the numerator.
(2x−3y4)3=23(x−3)3(y4)3=8x−9y12(2x^{-3}y^4)^3 = 2^3(x^{-3})^3(y^4)^3 = 8x^{-9}y^{12}
When an algebraic term with coefficients and variables is raised to a power, the outer exponent must be distributed to each individual factor inside the parentheses.
2
Set up the division by applying the quotient rule for each base.
8x−9y124x5y−2=(84)⋅(x−9x5)⋅(y12y−2)\frac{8x^{-9}y^{12}}{4x^5y^{-2}} = \left(\frac{8}{4}\right) \cdot \left(\frac{x^{-9}}{x^5}\right) \cdot \left(\frac{y^{12}}{y^{-2}}\right)
Separating the expression by identical bases allows the independent application of the quotient rule for exponents.
3
Evaluate the exponent for the yy term to find the value of bb.
y12÷y−2=y12−(−2)=y12+2=y14y^{12} \div y^{-2} = y^{12 - (-2)} = y^{12 + 2} = y^{14}
When dividing exponential terms with the same base, subtract the denominator's exponent from the numerator's exponent. Care must be taken with the double negative.
4
Identify the requested integer value.
b=14b = 14
The question specifically asks for the integer value of bb in the final simplified form cxaybcx^ay^b, which represents the exponent of the yy term.

Key Concept

Applying the power of a power and quotient of powers rules for variable exponents.
Question 2Question →

Consider the algebraic expression below, where x≠0x \neq 0 and y≠0y \neq 0. Which of the following is its fully simplified form?

(3x−2y32xy−2)−2×(9x24y4)−1 \left( \frac{3x^{-2}y^3}{2xy^{-2}} \right)^{-2} \times \left( \frac{9x^2}{4y^4} \right)^{-1}
Show answer & explanation

Answer: 16x481y6 \frac{16x^4}{81y^6}

Answer

The expression fully simplifies to 16x481y6\frac{16x^4}{81y^6}.
The correct expression is derived by sequentially simplifying the contents of the parentheses using the quotient rule, applying the outer negative exponents to both the numerical coefficients and variables, and then multiplying the resulting terms by adding the exponents of like bases.

Step-by-Step Solution

1
Simplify the fraction inside the first set of parentheses using the quotient rule for exponents (am/an=am−na^m / a^n = a^{m-n}).
3x−2y32xy−2=32x−2−1y3−(−2)=32x−3y5 \frac{3x^{-2}y^3}{2xy^{-2}} = \frac{3}{2} x^{-2-1} y^{3-(-2)} = \frac{3}{2} x^{-3} y^5
Simplifying inside the parentheses first follows the order of operations and makes subsequent calculations more manageable.
2
Apply the outer negative exponent (−2-2) to all components of the first simplified term, including the numerical coefficient.
(32x−3y5)−2=(32)−2(x−3)−2(y5)−2=49x6y−10 \left( \frac{3}{2} x^{-3} y^5 \right)^{-2} = \left( \frac{3}{2} \right)^{-2} (x^{-3})^{-2} (y^5)^{-2} = \frac{4}{9} x^6 y^{-10}
The power of a product rule states that (ab)n=anbn(ab)^n = a^n b^n, and applying a negative power to a fraction flips it: (a/b)−n=(b/a)n(a/b)^{-n} = (b/a)^n.
3
Apply the outer negative exponent (−1-1) to the second term.
(9x24y4)−1=4y49x2=49x−2y4 \left( \frac{9x^2}{4y^4} \right)^{-1} = \frac{4y^4}{9x^2} = \frac{4}{9} x^{-2} y^4
A negative exponent of −1-1 simply takes the reciprocal (flips the fraction).
4
Multiply the two resulting expressions by grouping coefficients and using the product rule for exponents (am×an=am+na^m \times a^n = a^{m+n}).
(49x6y−10)×(49x−2y4)=(49×49)(x6×x−2)(y−10×y4)=1681x4y−6 \left( \frac{4}{9} x^6 y^{-10} \right) \times \left( \frac{4}{9} x^{-2} y^4 \right) = \left( \frac{4}{9} \times \frac{4}{9} \right) (x^6 \times x^{-2}) (y^{-10} \times y^4) = \frac{16}{81} x^4 y^{-6}
When multiplying terms with the same base, their exponents are added together.
5
Rewrite the final expression using only positive exponents.
1681x4y−6=16x481y6 \frac{16}{81} x^4 y^{-6} = \frac{16x^4}{81y^6}
Standard mathematical notation requires fully simplified algebraic expressions to be written without negative exponents.

Key Concept

Simplifying rational expressions using the laws of exponents, including the quotient rule, power rule, product rule, and handling negative exponents alongside numerical coefficients.
Question 3Question →

Connect the unsimplified exponential expressions in the left column to their corresponding reduced forms in the right column, assuming xx and yy are non-zero constants.

Click a left item, then click its matching right item

Items

(2x−3y2)34x−5y−1 \frac{(2x^{-3}y^2)^3}{4x^{-5}y^{-1}}
(3x4y−2)−2⋅(9x0y5) (3x^4y^{-2})^{-2} \cdot (9x^0y^5)
(15x3y−45x−2y3)−1 \left( \frac{15x^3y^{-4}}{5x^{-2}y^3} \right)^{-1}
6x−1y4⋅2x5y−23x8y−3 \frac{6x^{-1}y^4 \cdot 2x^5y^{-2}}{3x^8y^{-3}}

Matches

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Answer

Each unsimplified expression matches to a unique reduced form by carefully applying the power, product, and quotient rules of exponents.
Each expression is correctly simplified by systematically applying the product, quotient, and power rules of exponents, paying special attention to numerical coefficients and negative signs.

Step-by-Step Solution

1
Simplify the first expression by expanding the numerator's power and applying the quotient rule.
(2x−3y2)34x−5y−1=8x−9y64x−5y−1=2x−4y7=2y7x4 \frac{(2x^{-3}y^2)^3}{4x^{-5}y^{-1}} = \frac{8x^{-9}y^6}{4x^{-5}y^{-1}} = 2x^{-4}y^7 = \frac{2y^7}{x^4}
The power of a product rule applies exponents to coefficients, while the quotient rule requires subtracting denominator exponents from numerator exponents.
2
Simplify the second expression by resolving the negative outer exponent and multiplying the coefficients and variables.
(3x4y−2)−2⋅(9x0y5)=(3−2x−8y4)⋅(9⋅1⋅y5)=19x−8y4⋅9y5=x−8y9=y9x8 (3x^4y^{-2})^{-2} \cdot (9x^0y^5) = (3^{-2}x^{-8}y^4) \cdot (9 \cdot 1 \cdot y^5) = \frac{1}{9}x^{-8}y^4 \cdot 9y^5 = x^{-8}y^9 = \frac{y^9}{x^8}
Applying negative exponents to the coefficient creates a fraction, and recognizing that x0=1x^0 = 1 simplifies the second term before multiplication.
3
Simplify the third expression by resolving the fraction inside first, then applying the external negative exponent.
(15x3y−45x−2y3)−1=(3x5y−7)−1=3−1x−5y7=y73x5 \left( \frac{15x^3y^{-4}}{5x^{-2}y^3} \right)^{-1} = (3x^5y^{-7})^{-1} = 3^{-1}x^{-5}y^7 = \frac{y^7}{3x^5}
Simplifying the internal expression first reduces algebraic errors before applying the reciprocal operation for the -1 exponent.
4
Simplify the fourth expression by multiplying the numerator terms, then dividing by the denominator.
6x−1y4⋅2x5y−23x8y−3=12x4y23x8y−3=4x4−8y2−(−3)=4x−4y5=4y5x4 \frac{6x^{-1}y^4 \cdot 2x^5y^{-2}}{3x^8y^{-3}} = \frac{12x^4y^2}{3x^8y^{-3}} = 4x^{4-8}y^{2-(-3)} = 4x^{-4}y^5 = \frac{4y^5}{x^4}
Combining like terms in the numerator through multiplication organizes the expression for a straightforward application of the quotient rule.

Key Concept

Operations with Variable Exponents
Question 4Question →

Which of the following represents the fully simplified form of the algebraic expression below, assuming x≠0x \neq 0 and y≠0y \neq 0?

(2x−3y44x2y−1)−2 \left( \frac{2x^{-3}y^4}{4x^2y^{-1}} \right)^{-2}
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Answer: 4x10y10 \frac{4x^{10}}{y^{10}}

Answer

The expression simplifies to 4x10y10 \frac{4x^{10}}{y^{10}} .
The correct answer is derived by first applying the quotient rule to simplify the expression inside the parentheses, being careful to subtract negative exponents correctly (e.g., y4−(−1)=y5y^{4 - (-1)} = y^5). Then, the negative exponent on the outside of the parentheses means taking the reciprocal of the fraction and squaring all terms, which applies to the coefficient (2 becomes 4) and the variables (x5x^5 becomes x10x^{10}).

Step-by-Step Solution

1
Simplify the numerical coefficients inside the parentheses.
24=12 \frac{2}{4} = \frac{1}{2}
Simplifying the fraction before handling variables reduces computational complexity.
2
Apply the quotient rule for exponents to the variables xx and yy.
For xx: −3−2=−5-3 - 2 = -5. For yy: 4−(−1)=4+1=54 - (-1) = 4 + 1 = 5. The expression inside the parentheses becomes 12x−5y5 \frac{1}{2} x^{-5}y^5 or y52x5 \frac{y^5}{2x^5} .
When dividing terms with the same base, subtract the exponent of the denominator from the exponent of the numerator.
3
Rewrite the entire expression with the simplified inner term.
(y52x5)−2 \left( \frac{y^5}{2x^5} \right)^{-2}
This establishes the simplified base expression before applying the final outer exponent.
4
Apply the negative exponent property to the fraction.
(2x5y5)2 \left( \frac{2x^5}{y^5} \right)^2
A negative exponent on a fraction indicates the reciprocal of that fraction raised to the positive exponent.
5
Apply the power of a quotient and product rules.
22(x5)2(y5)2=4x10y10 \frac{2^2(x^5)^2}{(y^5)^2} = \frac{4x^{10}}{y^{10}}
When raising a power to a power, multiply the exponents, and ensure the outer exponent is applied to the numerical coefficient.

Key Concept

Operations with Variable Exponents
Estimated Time:2m 0s
Question 5Question →

Simplify the following algebraic expression completely. Assume all variables represent non-zero real numbers.

(−3x−2y5z−3)−2×(6x4z−1y−3)3 \left( \frac{-3 x^{-2} y^5}{z^{-3}} \right)^{-2} \times \left( \frac{6 x^4 z^{-1}}{y^{-3}} \right)^3

Which of the following represents the fully simplified expression?

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Answer: 24x16yz9 \frac{24 x^{16}}{y z^9}

Answer

The fully simplified expression is 24x16yz9 \frac{24 x^{16}}{y z^9} .
The correct answer is found by first distributing the outer exponents to all terms inside the parentheses (including coefficients), which yields (1/9) * x^4 * y^(-10) * z^(-6) and 216 * x^12 * z^(-3) * y^9. Multiplying these expressions by combining the coefficients and adding the exponents of like variables gives 24 * x^16 * y^(-1) * z^(-9). Rewriting with positive exponents moves y and z to the denominator, resulting in the correct simplified fraction.

Step-by-Step Solution

1
Resolve negative exponents in the denominators of both fractions to bring the variables to the numerators.
(−3x−2y5z3)−2×(6x4z−1y3)3 \left( -3 x^{-2} y^5 z^3 \right)^{-2} \times \left( 6 x^4 z^{-1} y^3 \right)^3
Moving negative exponents from the denominator to the numerator simplifies the terms before applying the outer powers.
2
Apply the power of a product rule to the first expression, multiplying each inner exponent by the outer exponent of -2.
(−3)−2x(−2)(−2)y(5)(−2)z(3)(−2)=19x4y−10z−6 (-3)^{-2} x^{(-2)(-2)} y^{(5)(-2)} z^{(3)(-2)} = \frac{1}{9} x^4 y^{-10} z^{-6}
The power of a product rule states that (abc)^x = a^x b^x c^x, and the power of a power rule states that (a^x)^y = a^(xy).
3
Apply the power of a product rule to the second expression, multiplying each inner exponent by the outer exponent of 3.
63x(4)(3)z(−1)(3)y(3)(3)=216x12z−3y9 6^3 x^{(4)(3)} z^{(-1)(3)} y^{(3)(3)} = 216 x^{12} z^{-3} y^9
Distributing the cube to the coefficient and each variable's exponent prepares the expression for multiplication.
4
Multiply the resulting expressions together by grouping coefficients and like bases.
(19×216)(x4⋅x12)(y−10⋅y9)(z−6⋅z−3) \left( \frac{1}{9} \times 216 \right) \left( x^4 \cdot x^{12} \right) \left( y^{-10} \cdot y^9 \right) \left( z^{-6} \cdot z^{-3} \right)
The commutative property of multiplication allows us to group similar terms to apply the product of powers rule.
5
Simplify the groups by multiplying coefficients and adding the exponents of like bases.
24x16y−1z−9 24 x^{16} y^{-1} z^{-9}
The product of powers rule states that a^x * a^y = a^(x+y).
6
Rewrite the expression with only positive exponents.
24x16yz9 \frac{24 x^{16}}{y z^9}
Standard simplified form requires all exponents to be positive, so terms with negative exponents are moved to the denominator.

Key Concept

Applying the power of a product, power of a power, quotient of powers, and negative exponent rules to simplify complex algebraic expressions.
Question 6Question →

Simplify the following expression completely. Assume xx and yy are non-zero real numbers.

(3x2y−3)−2(−3x−2y2)3(x−4y3)−2 \frac{(3x^2y^{-3})^{-2} (-3x^{-2}y^2)^3}{(x^{-4}y^3)^{-2}}

If the simplified expression is written in the form AxByCAx^B y^C, where AA, BB, and CC are integers, what is the value of A+B+CA + B + C?

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Answer: -3

Answer

-3
The correct answer is found by systematically applying the laws of exponents. First, distribute the outer exponents to all factors inside the parentheses, being careful with signs and coefficients. The numerator simplifies to (19x−4y6)⋅(−27x−6y6)=−3x−10y12(\frac{1}{9}x^{-4}y^6) \cdot (-27x^{-6}y^6) = -3x^{-10}y^{12}. The denominator simplifies to x8y−6x^8y^{-6}. Dividing the numerator by the denominator using the quotient rule yields −3x−18y18-3x^{-18}y^{18}. Setting this equal to the target form AxByCAx^By^C yields A=−3A = -3, B=−18B = -18, and C=18C = 18. The sum A+B+CA + B + C is −3−18+18=−3-3 - 18 + 18 = -3.

Step-by-Step Solution

1
Apply the power of a product rule (ab)n=anbn(ab)^n = a^n b^n and the power of a power rule (am)n=am⋅n(a^m)^n = a^{m \cdot n} to the first factor in the numerator.
(3x2y−3)−2=3−2⋅x2(−2)⋅y−3(−2)=19x−4y6(3x^2y^{-3})^{-2} = 3^{-2} \cdot x^{2(-2)} \cdot y^{-3(-2)} = \frac{1}{9}x^{-4}y^6
To eliminate the outer negative exponent on the first term.
2
Apply the same exponent rules to the second factor in the numerator.
(−3x−2y2)3=(−3)3⋅x−2(3)⋅y2(3)=−27x−6y6(-3x^{-2}y^2)^3 = (-3)^3 \cdot x^{-2(3)} \cdot y^{2(3)} = -27x^{-6}y^6
To eliminate the outer exponent on the second term.
3
Multiply the simplified factors in the numerator using the product rule am⋅an=am+na^m \cdot a^n = a^{m+n}.
(19x−4y6)⋅(−27x−6y6)=(19⋅−27)⋅x−4−6⋅y6+6=−3x−10y12(\frac{1}{9}x^{-4}y^6) \cdot (-27x^{-6}y^6) = (\frac{1}{9} \cdot -27) \cdot x^{-4-6} \cdot y^{6+6} = -3x^{-10}y^{12}
To combine the expressions in the numerator into a single term.
4
Apply the power rules to simplify the denominator.
(x−4y3)−2=x−4(−2)⋅y3(−2)=x8y−6(x^{-4}y^3)^{-2} = x^{-4(-2)} \cdot y^{3(-2)} = x^8y^{-6}
To eliminate the outer negative exponent in the denominator.
5
Divide the numerator by the denominator using the quotient rule aman=am−n\frac{a^m}{a^n} = a^{m-n}.
−3x−10y12x8y−6=−3⋅x−10−8⋅y12−(−6)=−3x−18y18\frac{-3x^{-10}y^{12}}{x^8y^{-6}} = -3 \cdot x^{-10-8} \cdot y^{12-(-6)} = -3x^{-18}y^{18}
To reach the fully simplified algebraic form.
6
Identify the values of AA, BB, and CC to calculate the required sum.
A=−3A = -3, B=−18B = -18, C=18C = 18. The sum is −3+(−18)+18=−3-3 + (-18) + 18 = -3.
To answer the specific question asked in the stem.

Key Concept

Applying multiple rules of exponents (power of a power, negative exponents, product and quotient rules) to simplify complex algebraic expressions.
Question 7Question →
During the mathematical modeling of an electromagnetic field, an engineer derives the formula (4a3b−2)−22a−4b5\frac{(4a^3b^{-2})^{-2}}{2a^{-4}b^5} to represent the field intensity. Which of the following expressions is mathematically equivalent to this derived formula?
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Answer: 132a2b\frac{1}{32a^2b}

Answer

The mathematically equivalent expression is 132a2b\frac{1}{32a^2b}.
To find the equivalent expression, first apply the power of a power rule to the numerator. Distribute the exponent -2 to each factor: 4^(-2) * (a^3)^(-2) * (b^-2)^(-2), which simplifies to (1/16) * a^-6 * b^4. Next, divide this result by the denominator, 2a^-4 * b^5. Divide the coefficients: (1/16) / 2 = 1/32. Divide the variables by subtracting the exponents of like bases: for a, it is -6 - (-4) = -2, and for b, it is 4 - 5 = -1. This yields (1/32) * a^-2 * b^-1. Finally, move the variables with negative exponents to the denominator to attain the fully simplified form.

Step-by-Step Solution

1
Apply the power of a power rule to the expression in the numerator.
(4)−2(a3)−2(b−2)−2=116a−6b4(4)^{-2}(a^3)^{-2}(b^{-2})^{-2} = \frac{1}{16}a^{-6}b^4
When an expression in parentheses is raised to a power, the outer exponent must be applied to the numerical coefficient and multiplied by the exponent of each variable inside.
2
Divide the coefficients of the numerator and the denominator.
1/162=116×12=132\frac{1/16}{2} = \frac{1}{16} \times \frac{1}{2} = \frac{1}{32}
The constant factors can be evaluated and divided separately from the variable terms.
3
Apply the quotient of powers rule for the variable bases.
a−6−(−4)=a−2a^{-6 - (-4)} = a^{-2} and b4−5=b−1b^{4 - 5} = b^{-1}
When dividing exponential terms with the same base, subtract the exponent of the denominator from the exponent of the numerator.
4
Rewrite the expression using only positive exponents.
132a−2b−1=132a2b\frac{1}{32}a^{-2}b^{-1} = \frac{1}{32a^2b}
Final simplified algebraic expressions are standardly written by moving variables with negative exponents to the denominator.

Key Concept

Applying multiple properties of exponents, including the power of a product rule, power of a power rule, and quotient of powers rule with negative integers.
Question 8Question →

Match each unsimplified algebraic expression with its fully simplified equivalent form. Assume all variables represent non-zero real numbers.

Click a left item, then click its matching right item

Items

(3x−2y4)(5x6y−3) (3x^{-2}y^4)(5x^6y^{-3})
30x5y−22x9y3 \frac{30x^5y^{-2}}{2x^9y^3}
(−2x3y−1)4 (-2x^3y^{-1})^4
(3x−2y4)−2 \left(\frac{3x^{-2}}{y^4}\right)^{-2}

Matches

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Answer

Each unsimplified expression matches exactly one fully simplified form by applying the product, quotient, and power rules of exponents.
The correct matches are found by systematically applying the fundamental properties of exponents: adding exponents for multiplication, subtracting for division, multiplying for powers of powers, and converting negative exponents to reciprocals.

Step-by-Step Solution

1
Simplify the product of powers expression: (3x^-2y^4)(5x^6y^-3)
15x^4y
Multiply numerical coefficients together. Add exponents for variables with the same base according to the product rule.
2
Simplify the quotient of powers expression: (30x^5y^-2) / (2x^9y^3)
15 / (x^4y^5)
Divide numerical coefficients. Subtract the denominator's exponent from the numerator's exponent. Convert negative exponents to positive by moving them to the denominator.
3
Simplify the power of a product expression: (-2x^3y^-1)^4
(16x^12) / y^4
Raise every factor inside the parentheses to the 4th power. Note that an even power of a negative coefficient yields a positive result.
4
Simplify the expression with an outer negative exponent: (3x^-2 / y^4)^-2
(x^4y^8) / 9
Multiply all inner exponents by the outer exponent -2. Then take the reciprocal of terms with negative exponents to rewrite them with positive exponents.

Key Concept

Operations with Variable Exponents
Question 9Question →

Connect each unsimplified exponential operation to its mathematically equivalent simplified form.

Click a left item, then click its matching right item

Items

(3a2)(2a4)(3a^2)(2a^4)
12a62a2\frac{12a^6}{2a^2}
(2a3)2(2a^3)^2
6a−26a^{-2}

Matches

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Answer

The correct pairings are: (3a^2)(2a^4) pairs with 6a^6, (12a^6)/(2a^2) pairs with 6a^4, (2a^3)^2 pairs with 4a^6, and 6a^-2 pairs with 6/a^2.
Each pair represents a fundamental rule of exponents applied to a one-step algebraic expression involving both coefficients and variables.

Step-by-Step Solution

1
Evaluate the product (3a2)(2a4)(3a^2)(2a^4).
6a66a^6
By the product rule, multiply coefficients (3×2=63 \times 2 = 6) and add the exponents of the same base (2+4=62 + 4 = 6).
2
Evaluate the quotient 12a62a2\frac{12a^6}{2a^2}.
6a46a^4
By the quotient rule, divide coefficients (12÷2=612 \div 2 = 6) and subtract the bottom exponent from the top exponent (6−2=46 - 2 = 4).
3
Evaluate the power (2a3)2(2a^3)^2.
4a64a^6
By the power of a product rule, apply the exponent to both the coefficient (22=42^2 = 4) and the variable (3×2=63 \times 2 = 6).
4
Evaluate the negative exponent expression 6a−26a^{-2}.
6a2\frac{6}{a^2}
By the negative exponent rule, a−2a^{-2} becomes 1a2\frac{1}{a^2}. The coefficient 6 is not raised to the negative power, so it stays in the numerator.

Key Concept

Operations with Variable Exponents
Question 10Question →

Which of the following represents the simplified form of the expression 4b5⋅2b−24b^5 \cdot 2b^{-2}?

Show answer & explanation

Answer: 8b38b^3

Answer

The simplified expression is 8b38b^3.
The correct expression is found by multiplying the numerical coefficients (4×2=84 \times 2 = 8) and adding the exponents for the common base bb (5+(−2)=35 + (-2) = 3), resulting in 8b38b^3.

Step-by-Step Solution

1
Multiply the numerical coefficients.
4×2=84 \times 2 = 8
When multiplying algebraic terms, their numerical coefficients are multiplied together.
2
Apply the product rule of exponents to the variable terms.
b5⋅b−2=b5+(−2)=b3b^5 \cdot b^{-2} = b^{5 + (-2)} = b^3
When multiplying exponential expressions with the same base, add their exponents.
3
Combine the coefficient and the variable term.
8b38b^3
This is the final simplified form of the expression.

Key Concept

When multiplying two exponential terms with the same base, multiply their coefficients and add their exponents.
Question 11Question →

Match each algebraic expression with its simplified equivalent form.

Click a left item, then click its matching right item

Items

x3⋅x2x^3 \cdot x^2
x8x2\frac{x^8}{x^2}
(x4)2(x^4)^2
x−3x^{-3}

Matches

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Answer

The matches are: x3⋅x2x^3 \cdot x^2 simplifies to x5x^5, x8x2\frac{x^8}{x^2} simplifies to x6x^6, (x4)2(x^4)^2 simplifies to x8x^8, and x−3x^{-3} simplifies to 1x3\frac{1}{x^3}.
Each expression correctly pairs with its simplified form by applying the four fundamental rules of exponents: the product rule, the quotient rule, the power of a power rule, and the negative exponent rule.

Step-by-Step Solution

1
Apply the product rule for exponents to x3⋅x2x^3 \cdot x^2.
x3+2=x5x^{3+2} = x^5
Multiplying identical bases requires adding the exponents.
2
Apply the quotient rule for exponents to x8x2\frac{x^8}{x^2}.
x8−2=x6x^{8-2} = x^6
Dividing identical bases requires subtracting the denominator's exponent.
3
Apply the power rule for exponents to (x4)2(x^4)^2.
x4⋅2=x8x^{4 \cdot 2} = x^8
Raising an exponent to another power requires multiplying the exponents.
4
Apply the negative exponent rule to x−3x^{-3}.
1x3\frac{1}{x^3}
A negative exponent signifies the reciprocal of the corresponding positive exponent.

Key Concept

Operations with Variable Exponents
Estimated Time:1m 0s
Question 12Question →

Simplify the following algebraic expression:

(3y4)(−5y2)(3y^4)(-5y^2)

Which of the following is the correctly simplified expression?

Show answer & explanation

Answer: −15y6-15y^6

Answer

−15y6-15y^6
The expression is simplified by first multiplying the numerical coefficients (3×−5=−153 \times -5 = -15). Then, the product rule for exponents is applied to the variables by adding their exponents (y4×y2=y4+2=y6y^4 \times y^2 = y^{4+2} = y^6). Combining these two parts yields the correct simplified expression, −15y6-15y^6.

Step-by-Step Solution

1
Group and multiply the numerical coefficients.
3×(−5)=−153 \times (-5) = -15
The commutative property allows us to group the constant numbers together for multiplication.
2
Multiply the variable terms using the product rule for exponents.
y4×y2=y4+2=y6y^4 \times y^2 = y^{4+2} = y^6
When multiplying exponential terms with the same base, their exponents must be added.
3
Combine the resulting coefficient and variable term.
−15y6-15y^6
This represents the final simplified form of the expression.

Key Concept

Product Rule for Exponents and Integer Multiplication
Question 13Question →

Find the value of nn that makes the following equation true:

x9x3=xn\frac{x^9}{x^3} = x^n
Show answer & explanation

Answer: 6; six

Answer

6
According to the quotient rule for exponents, when you divide two exponential terms that have the same base, you subtract the exponent of the denominator from the exponent of the numerator. In this equation, both terms have the base xx. Subtracting the denominator's exponent (3) from the numerator's exponent (9) gives 9−3=69 - 3 = 6. Therefore, the left side simplifies to x6x^6. Equating this to xnx^n means that nn must equal 6.

Step-by-Step Solution

1
Apply the quotient rule for exponents, which states that when dividing terms with identical bases, you subtract the exponent in the denominator from the exponent in the numerator.
x9−3=xnx^{9-3} = x^n
To simplify the left side of the equation into a single term with base xx.
2
Perform the subtraction in the exponent.
x6=xnx^6 = x^n
Calculating 9−39 - 3 gives the simplified exponent.
3
Set the resulting exponent equal to nn.
n=6n = 6
Since the bases are identical (xx), their exponents must be equal for the equation to hold true.

Key Concept

Quotient Rule for Exponents
Question 14Question →

A square metal plate has a side length of 3x43x^4 centimeters. Which of the following expressions represents the area of the plate in square centimeters?

Show answer & explanation

Answer: 9x89x^8

Answer

9x89x^8
The area of a square is calculated by squaring its side length. When squaring a term like 3x43x^4, the power of a product rule states that both the coefficient and the variable must be raised to the second power. Squaring the coefficient 3 yields 9, and raising x4x^4 to the second power requires multiplying the exponents (4×2=84 \times 2 = 8), resulting in a final area of 9x89x^8.

Step-by-Step Solution

1
Recall the formula for the area of a square.
A=s2A = s^2
To find the area, the side length must be squared.
2
Substitute the given side length into the area formula.
A=(3x4)2A = (3x^4)^2
This sets up the mathematical expression that needs to be simplified.
3
Apply the power of a product rule and power of a power rule.
32⋅(x4)2=9x83^2 \cdot (x^4)^2 = 9x^8
The outer exponent (2) must be applied to both the numerical coefficient (32=93^2=9) and the variable exponent (4×2=84 \times 2 = 8).

Key Concept

Power of a Product and Power of a Power rules
Question 15Question →

Consider the exponential expression (y7)3(y^7)^3. If this expression is rewritten in the equivalent form yny^n, what is the numerical value of nn?

Show answer & explanation

Answer: 21; twenty-one; n=21; n = 21

Answer

The numerical value of the exponent is 21.
When raising a power to another power, the laws of exponents dictate that the two exponents must be multiplied. Multiplying the inside exponent 7 by the outside exponent 3 yields 21.

Step-by-Step Solution

1
Identify the appropriate exponent rule to simplify the expression (y7)3(y^7)^3.
The expression requires the Power of a Power property.
An exponential term with a base and an exponent is being raised entirely to another power.
2
Apply the Power of a Power rule, which states that (xa)b=xa×b(x^a)^b = x^{a \times b}.
Multiply the inner exponent by the outer exponent: 7×37 \times 3.
To raise a power to another power, you must multiply the exponents together.
3
Calculate the final numerical value.
7×3=217 \times 3 = 21, meaning the expression simplifies to y21y^{21}.
By matching y21y^{21} to the given form yny^n, we determine that n=21n = 21.

Key Concept

Power of a Power Rule for Exponents
Estimated Time:30s
Question 16Question →

Evaluate the product of the monomial expressions 4z54z^5 and 2z32z^3. Which of the following represents the simplified result?

Show answer & explanation

Answer: 8z88z^8

Answer

The expression simplifies to 8z88z^8.
When multiplying two monomial expressions, the numerical coefficients are multiplied together (4×2=84 \times 2 = 8), and the exponents of the common variable base are added together according to the product rule for exponents (5+3=85 + 3 = 8). This results in 8z88z^8.

Step-by-Step Solution

1
Identify the operation to be performed between the two terms.
The terms 4z54z^5 and 2z32z^3 are being multiplied.
Understanding the operation is required to apply the correct mathematical rules for coefficients and exponents.
2
Multiply the numerical coefficients.
4×2=84 \times 2 = 8
Coefficients in a product of monomials are multiplied together normally.
3
Multiply the variable parts by applying the product rule for exponents.
z5×z3=z5+3=z8z^5 \times z^3 = z^{5+3} = z^8
When multiplying exponential expressions with the same base, their exponents must be added.
4
Combine the calculated coefficient and the variable part.
8z88z^8
This brings the simplified pieces together into the final term.

Key Concept

Operations with Variable Exponents
Question 17Question →

Pair each basic algebraic expression with its correctly simplified result by applying the fundamental laws of exponents.

Click a left item, then click its matching right item

Items

m6⋅m2m^6 \cdot m^2
n8n5\frac{n^8}{n^5}
(p4)3(p^4)^3
q−5q^{-5}

Matches

Show answer & explanation

Answer

The correct pairings are: m6⋅m2m^6 \cdot m^2 with m8m^8; n8n5\frac{n^8}{n^5} with n3n^3; (p4)3(p^4)^3 with p12p^{12}; and q−5q^{-5} with 1q5\frac{1}{q^5}.
Each expression can be simplified by directly applying one specific fundamental property of exponents. m6⋅m2m^6 \cdot m^2 uses the product rule. n8n5\frac{n^8}{n^5} uses the quotient rule. (p4)3(p^4)^3 uses the power rule. q−5q^{-5} uses the negative exponent definition.

Step-by-Step Solution

1
Simplify the product expression m6⋅m2m^6 \cdot m^2.
m8m^8
When bases are identical and multiplied, their exponents are added together.
2
Simplify the quotient expression n8n5\frac{n^8}{n^5}.
n3n^3
When dividing identical bases, the exponent of the denominator is subtracted from the exponent of the numerator.
3
Simplify the nested power expression (p4)3(p^4)^3.
p12p^{12}
When raising an exponent to another power, the two exponents are multiplied.
4
Rewrite the negative exponent expression q−5q^{-5}.
1q5\frac{1}{q^5}
A negative exponent indicates taking the reciprocal of the base raised to the positive exponent.

Key Concept

Applying fundamental exponent rules (product, quotient, power, and negative exponents) to simplify single-variable expressions.
Question 18Question →

If the equation y8⋅yk=y15y^8 \cdot y^k = y^{15} is true for all non-zero values of yy, what is the value of kk?

Show answer & explanation

Answer: 7; seven

Answer

7
The correct value of kk is 77 because applying the product rule of exponents gives y8⋅y7=y8+7=y15y^8 \cdot y^7 = y^{8+7} = y^{15}.

Step-by-Step Solution

1
Apply the product rule for exponents to the left side of the equation.
y8⋅yk=y8+ky^8 \cdot y^k = y^{8+k}
When multiplying exponential expressions with the same base, the exponents are added.
2
Set the exponent of the simplified left side equal to the exponent of the right side.
8+k=158 + k = 15
Since the bases are equal and non-zero, their exponents must be equal.
3
Solve the linear equation for kk.
k=7k = 7
Subtracting 8 from both sides of the equation isolates the variable kk.

Key Concept

Product Rule for Exponents
Question 19Question →

Simplify the following algebraic expression completely:

(3a5)2(3a^5)^2

What is the simplified form of this expression?

Show answer & explanation

Answer: 9a109a^{10}

Answer

The simplified expression is 9a109a^{10}.
The correct result is found by distributing the exponent of 2 to both factors inside the parentheses. The coefficient 3 is squared to become 9, and the exponent 5 is multiplied by 2 to become 10, resulting in 9a109a^{10}.

Step-by-Step Solution

1
Apply the power of a product rule by distributing the outer exponent to both the numerical coefficient and the variable inside the parentheses.
32⋅(a5)23^2 \cdot (a^5)^2
When a product is raised to a power, each independent factor inside the parentheses must be raised to that power.
2
Simplify the numerical coefficient by calculating its square.
9⋅(a5)29 \cdot (a^5)^2
The exponent 2 indicates multiplying the base by itself: 3×3=93 \times 3 = 9.
3
Apply the power of a power rule to the variable by multiplying the inner exponent by the outer exponent.
9a109a^{10}
According to exponent laws, (xm)n=xm⋅n(x^m)^n = x^{m \cdot n}, so the new exponent is 5×2=105 \times 2 = 10.

Key Concept

Power of a Product and Power of a Power Rules
Question 20Question →

Consider the following algebraic equation, where x≠0x \neq 0:

(3x4)29xk=x5 \frac{(3x^4)^2}{9x^k} = x^5

What is the value of the exponent kk?

Show answer & explanation

Answer: 3; k=3; k = 3; three

Answer

3
First, the numerator (3x4)2(3x^4)^2 simplifies to 9x89x^8. Substituting this back yields 9x89xk=x5\frac{9x^8}{9x^k} = x^5. The coefficients cancel out, and applying the quotient rule to the variables gives x8−k=x5x^{8-k} = x^5. By equating the exponents, we get 8−k=58 - k = 5, which means k=3k = 3.

Step-by-Step Solution

1
Apply the power of a product rule to the numerator.
(3x4)2=32⋅(x4)2=9x8(3x^4)^2 = 3^2 \cdot (x^4)^2 = 9x^8
When raising a product to a power, you must raise every factor inside the parentheses to that power.
2
Substitute the simplified numerator back into the equation and simplify the numerical coefficients.
9x89xk=x8xk\frac{9x^8}{9x^k} = \frac{x^8}{x^k}
The coefficients 99\frac{9}{9} cancel out to 1, making it easier to isolate the variable base.
3
Apply the quotient rule for exponents on the left side of the equation.
x8xk=x8−k\frac{x^8}{x^k} = x^{8-k}
When dividing exponential expressions with the same base, you subtract the exponent in the denominator from the exponent in the numerator.
4
Set the resulting exponent equal to the exponent on the right side of the original equation and solve for kk.
8−k=5⇒k=38 - k = 5 \Rightarrow k = 3
Since the bases on both sides of the equation are the same (xx), their exponents must be equal.

Key Concept

Operations with Variable Exponents
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