Properties of Exponents in Algebraic Expressions

33 soru

Soru 1Soru

For all non-zero real numbers aa and bb, the expression (a2bk)3(a^2 b^k)^3 is equivalent to a6b15a^6 b^{15}. What is the value of the integer kk?

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Cevap: 5

Cevap

The value of the integer kk is 5.
To find the value of kk, we simplify the expression (a2bk)3(a^2 b^k)^3 using exponent rules. According to the power of a product property, (xy)z=xzyz(xy)^z = x^z y^z, so (a2bk)3=(a2)3(bk)3(a^2 b^k)^3 = (a^2)^3 (b^k)^3. Next, applying the power of a power property, (xy)z=xyz(x^y)^z = x^{yz}, we get a23bk3=a6b3ka^{2 \cdot 3} b^{k \cdot 3} = a^6 b^{3k}. Since this expression is equivalent to a6b15a^6 b^{15}, we set the exponents of bb equal to each other: 3k=153k = 15. Dividing both sides by 3 yields k=5k = 5.

Adım Adım Çözüm

1
Apply the power of a product rule to the expression (a2bk)3(a^2 b^k)^3.
(a2)3(bk)3(a^2)^3 \cdot (b^k)^3
The power of a product rule states that (xy)z=xzyz(xy)^z = x^z y^z.
2
Apply the power of a power rule to simplify the exponents.
a6b3ka^6 b^{3k}
The power of a power rule states that (xy)z=xyz(x^y)^z = x^{y \cdot z}, so (a2)3=a23=a6(a^2)^3 = a^{2 \cdot 3} = a^6 and (bk)3=b3k(b^k)^3 = b^{3k}.
3
Set the exponent of bb in a6b3ka^6 b^{3k} equal to the exponent of bb in the equivalent expression a6b15a^6 b^{15}.
3k=153k = 15
Since the expressions are equivalent for all non-zero real numbers, the exponents of like bases must be equal.
4
Solve the linear equation for kk.
k=5k = 5
Dividing both sides of 3k=153k = 15 by 3 isolates the variable kk.

Anahtar Kavram

Properties of exponents, specifically the power of a product rule (xy)z=xzyz(xy)^z = x^z y^z and the power of a power rule (xy)z=xyz(x^y)^z = x^{y \cdot z}.
Tahmini Süre:45s
Soru 2Soru

If the expression (p3q2)1(p1q2)3\frac{(p^3 q^{-2})^{-1}}{(p^{-1} q^2)^3} is simplified to the form pxqyp^x q^y for all non-zero real numbers pp and qq, what is the value of 2xy2x - y?

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Cevap: 4

Cevap

The correct answer is 4.
Applying the power of a power rule to the numerator yields (p3q2)1=p3q2(p^3 q^{-2})^{-1} = p^{-3} q^2. Applying the same rule to the denominator yields (p1q2)3=p3q6(p^{-1} q^2)^3 = p^{-3} q^6. Dividing the terms by subtracting exponents gives p3(3)q26=p0q4p^{-3 - (-3)} q^{2-6} = p^0 q^{-4}. Thus, x=0x = 0 and y=4y = -4. Evaluating 2xy2x - y gives 2(0)(4)=42(0) - (-4) = 4.

Adım Adım Çözüm

1
Simplify the numerator using the power of a power property, which states that (am)n=amn(a^m)^n = a^{mn}.
(p3q2)1=p3(1)q2(1)=p3q2(p^3 q^{-2})^{-1} = p^{3 \cdot (-1)} q^{-2 \cdot (-1)} = p^{-3} q^2
This distributes the exponent of 1-1 to both factors inside the parentheses in the numerator.
2
Simplify the denominator using the power of a power property, which states that (am)n=amn(a^m)^n = a^{mn}.
(p1q2)3=p13q23=p3q6(p^{-1} q^2)^3 = p^{-1 \cdot 3} q^{2 \cdot 3} = p^{-3} q^6
This distributes the exponent of 33 to both factors inside the parentheses in the denominator.
3
Combine the simplified numerator and denominator using the quotient property of exponents, which states that aman=amn\frac{a^m}{a^n} = a^{m-n}.
p3q2p3q6=p3(3)q26=p0q4\frac{p^{-3} q^2}{p^{-3} q^6} = p^{-3 - (-3)} q^{2 - 6} = p^0 q^{-4}
This simplifies division by subtracting the exponent in the denominator from the exponent in the numerator for each base.
4
Identify the values of xx and yy from the simplified form p0q4p^0 q^{-4}, and evaluate the final expression 2xy2x - y.
x=0x = 0 and y=4y = -4, so 2(0)(4)=42(0) - (-4) = 4
This substitutes the values of the exponents into the target algebraic expression to find the final numerical answer.

Anahtar Kavram

Properties of exponents including power of a power and quotient properties

Alternatif Yöntem

Alternatively, you can write the terms with positive exponents first: (p3q2)1(p1q2)3=(p1q2)3(p3q2)1=p3q6p3q2=p33q6(2)=p6q8\frac{(p^3 q^{-2})^{-1}}{(p^{-1} q^2)^3} = \frac{(p^{-1} q^2)^3}{(p^3 q^{-2})^1} = \frac{p^{-3} q^6}{p^3 q^{-2}} = p^{-3-3} q^{6-(-2)} = p^{-6} q^8, but this expression is equivalent to the original expression only if inverted properly. Direct distribution is less prone to inversion errors.
Tahmini Süre:1m 30s
Soru 3Soru

For all non-zero real numbers xx and yy, the algebraic expression (xa/2y1x2/3yb/3)6\left(\frac{x^{a/2} y^{-1}}{x^{-2/3} y^{b/3}}\right)^{-6} is equivalent to y12x16\frac{y^{12}}{x^{16}}, where aa and bb are integers. What is the value of a+ba + b?

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Cevap: 7

Cevap

7
The correct answer is 7. Simplifying the expression inside the parentheses using the quotient rule yields xa2+23y1b3x^{\frac{a}{2} + \frac{2}{3}} y^{-1 - \frac{b}{3}}. Raising this expression to the power of 6-6 results in x3a4y6+2bx^{-3a - 4} y^{6 + 2b}. Equating this to x16y12x^{-16} y^{12} gives a=4a = 4 and b=3b = 3, so a+b=7a + b = 7.

Adım Adım Çözüm

1
Simplify the expression inside the parentheses using the quotient rule for exponents, zmzn=zmn\frac{z^m}{z^n} = z^{m-n}.
xa2(23)y1b3=xa2+23y1b3x^{\frac{a}{2} - \left(-\frac{2}{3}\right)} y^{-1 - \frac{b}{3}} = x^{\frac{a}{2} + \frac{2}{3}} y^{-1 - \frac{b}{3}}
To combine the base xx and base yy terms inside the parentheses before applying the outer power.
2
Apply the outer exponent of 6-6 to the simplified expression using the power of a power rule, (zm)n=zmn(z^m)^n = z^{mn}.
x6(a2+23)y6(1b3)=x3a4y6+2bx^{-6\left(\frac{a}{2} + \frac{2}{3}\right)} y^{-6\left(-1 - \frac{b}{3}\right)} = x^{-3a - 4} y^{6 + 2b}
To distribute the negative power of 6-6 to each factor in the product.
3
Equate the resulting exponents to the exponents of the given equivalent expression, y12x16=x16y12\frac{y^{12}}{x^{16}} = x^{-16} y^{12}, and solve the equations for aa and bb.
For xx: 3a4=16    3a=12    a=4-3a - 4 = -16 \implies -3a = -12 \implies a = 4. For yy: 6+2b=12    2b=6    b=36 + 2b = 12 \implies 2b = 6 \implies b = 3.
Since the expressions are equivalent for all non-zero values of xx and yy, their respective exponents must be equal.
4
Calculate the sum of aa and bb.
a+b=4+3=7a + b = 4 + 3 = 7
To find the final requested value.

Anahtar Kavram

Properties of exponents in algebraic expressions including the quotient rule, power rule, and operations with fractional/negative exponents.
Soru 4Soru

For x0x \neq 0, the expression (x2)5xk\frac{(x^2)^5}{x^k} simplifies to x6x^6. What is the value of the exponent kk?

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Cevap: 4

Cevap

The value of the exponent kk is 4.
Applying the power of a power rule to the numerator yields (x2)5=x10(x^2)^5 = x^{10}. Next, applying the quotient rule to divide x10x^{10} by xkx^k yields x10kx^{10-k}. Setting this equal to the simplified term x6x^6 leads to the exponent equation 10k=610 - k = 6. Solving this equation gives the final result k=4k = 4.

Adım Adım Çözüm

1
Simplify the numerator expression (x2)5(x^2)^5
x10x^{10}
Multiply the exponents when raising a power to another power: (xa)b=xab(x^a)^b = x^{ab}.
2
Simplify the division of the two exponential expressions x10xk\frac{x^{10}}{x^k}
x10kx^{10-k}
Subtract the exponent of the denominator from the exponent of the numerator: xaxb=xab\frac{x^a}{x^b} = x^{a-b}.
3
Solve the linear equation for kk using the target exponent 6
k=4k = 4
Equating the exponent 10k10-k to 66 gives 10k=610-k=6. Subtracting 10 from both sides gives k=4-k = -4, so k=4k = 4.

Anahtar Kavram

Applying properties of exponents in algebraic expressions, specifically the power of a power rule and the quotient rule.
Soru 5Soru

If the algebraic expression (xay2)3(x2yb)2(x3y1)2\frac{(x^a y^2)^{-3} (x^2 y^b)^2}{(x^{-3} y^{-1})^{-2}} simplifies to x4y4x^4 y^4 for all non-zero real numbers xx and yy, where aa and bb are integers, what is the value of a+ba + b?

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Cevap: 4

Cevap

The value of a+ba + b is 44.
Applying exponent rules to the expression (xay2)3(x2yb)2(x3y1)2\frac{(x^a y^2)^{-3} (x^2 y^b)^2}{(x^{-3} y^{-1})^{-2}} yields x3ay6x4y2bx6y2=x3a+4y2b6x6y2=x3a2y2b8\frac{x^{-3a}y^{-6} \cdot x^4 y^{2b}}{x^6 y^2} = \frac{x^{-3a+4} y^{2b-6}}{x^6 y^2} = x^{-3a-2} y^{2b-8}. Equating these exponents to the target expression x4y4x^4 y^4 gives the system 3a2=4-3a - 2 = 4 and 2b8=42b - 8 = 4. Solving these equations gives a=2a = -2 and b=6b = 6, resulting in a sum of a+b=4a + b = 4.

Adım Adım Çözüm

1
Simplify the terms in the numerator.
(xay2)3=x3ay6(x^a y^2)^{-3} = x^{-3a} y^{-6} and (x2yb)2=x4y2b(x^2 y^b)^2 = x^4 y^{2b}
Apply the power of a product rule: (umvn)p=umpvnp(u^m v^n)^p = u^{mp} v^{np}.
2
Multiply the simplified terms in the numerator.
x3a+4y2b6x^{-3a+4} y^{2b-6}
Apply the product rule of exponents by adding exponents of like bases: umun=um+nu^m \cdot u^n = u^{m+n}.
3
Simplify the denominator.
(x3y1)2=x6y2(x^{-3} y^{-1})^{-2} = x^6 y^2
Apply the power of a product rule.
4
Divide the numerator by the denominator.
x3a2y2b8x^{-3a-2} y^{2b-8}
Apply the quotient rule of exponents by subtracting denominator exponents from numerator exponents: umun=umn\frac{u^m}{u^n} = u^{m-n}.
5
Set up equations by equating the simplified exponents to the exponents in the target expression x4y4x^4 y^4.
3a2=4-3a - 2 = 4 and 2b8=42b - 8 = 4
For the expressions to be equivalent for all non-zero real numbers, the corresponding exponents of xx and yy must be equal.
6
Solve the linear equations for the integer constants aa and bb.
a=2a = -2 and b=6b = 6
Isolate the variables: 3a=6    a=2-3a = 6 \implies a = -2, and 2b=12    b=62b = 12 \implies b = 6.
7
Find the sum of aa and bb.
44
Add the values of the constants: 2+6=4-2 + 6 = 4.

Anahtar Kavram

Properties of exponents (product, quotient, and power rules) in multi-step algebraic simplification
Soru 6Soru

Which of the following expressions is equivalent to 12x84x2\frac{12x^8}{4x^2} for all x0x \neq 0?

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Cevap: 3x63x^6

Cevap

The expression 3x63x^6
To simplify the expression, divide the coefficients first: 124=3\frac{12}{4} = 3. Then, apply the quotient rule of exponents to the variable terms, which states that for any non-zero base, xaxb=xab\frac{x^a}{x^b} = x^{a-b}. Subtracting the exponents gives x82=x6x^{8-2} = x^6. Combining these results yields the correct expression 3x63x^6.

Adım Adım Çözüm

1
Divide the numerical coefficients of the terms.
124=3\frac{12}{4} = 3
When simplifying a fraction with algebraic terms, the coefficients are divided normally.
2
Apply the quotient rule of exponents to simplify the variable terms.
x8x2=x82=x6\frac{x^8}{x^2} = x^{8-2} = x^6
According to the quotient rule of exponents, when dividing expressions with the same base, subtract the exponent in the denominator from the exponent in the numerator.
3
Multiply the simplified coefficient and variable results together.
3x63x^6
Combining the divided coefficient and the simplified variable expression gives the final simplified result.

Anahtar Kavram

Quotient Rule of Exponents

Alternatif Yöntem

Alternatively, you can expand the exponent terms in the numerator and denominator: 12xxxxxxxx4xx\frac{12 \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x}{4 \cdot x \cdot x}. Simplifying the coefficients gives 33, and canceling two pairs of xx from both the numerator and denominator leaves six factors of xx in the numerator, which simplifies to 3x63x^6.
Tahmini Süre:45s
Soru 7Soru

For all non-zero real numbers xx and yy, which of the following is equivalent to the expression (2x1y2+12x1y2)3(x2y3)2\frac{\left( 2x^{-1} y^2 + \frac{1}{2} x^{-1} y^2 \right)^{-3}}{(x^2 y^{-3})^{-2}}?

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Cevap: 8x7125y12\frac{8x^7}{125y^{12}}

Cevap

8x7125y12\frac{8x^7}{125y^{12}}
To find the equivalent expression, we first combine the like terms inside the parentheses in the numerator to get 52x1y2\frac{5}{2} x^{-1} y^2. Raising this product to the power of 3-3 yields (52)3(x1)3(y2)3=8125x3y6\left(\frac{5}{2}\right)^{-3} (x^{-1})^{-3} (y^2)^{-3} = \frac{8}{125} x^3 y^{-6}. Next, the denominator simplifies to (x2y3)2=x4y6(x^2 y^{-3})^{-2} = x^{-4} y^6. Dividing the numerator by the denominator requires subtracting the exponents of like bases: for xx, we have 3(4)=73 - (-4) = 7, and for yy, we have 66=12-6 - 6 = -12. This results in 8125x7y12\frac{8}{125} x^7 y^{-12}, which is equivalent to the correct expression 8x7125y12\frac{8x^7}{125y^{12}}.

Adım Adım Çözüm

1
Combine the like terms inside the parentheses in the numerator.
2x1y2+12x1y2=(2+12)x1y2=52x1y22x^{-1} y^2 + \frac{1}{2} x^{-1} y^2 = \left(2 + \frac{1}{2}\right) x^{-1} y^2 = \frac{5}{2} x^{-1} y^2
Before applying the outer negative exponent, it is mathematically simpler to combine the like terms inside the grouping.
2
Apply the power of 3-3 to the term in the numerator.
(52x1y2)3=(52)3(x1)3(y2)3=8125x3y6\left(\frac{5}{2} x^{-1} y^2\right)^{-3} = \left(\frac{5}{2}\right)^{-3} (x^{-1})^{-3} (y^2)^{-3} = \frac{8}{125} x^3 y^{-6}
The power of a product rule (ab)n=anbn(ab)^n = a^n b^n and the power of a power rule (am)n=amn(a^m)^n = a^{mn} are applied to expand the term.
3
Simplify the denominator by applying the power of 2-2.
(x2y3)2=(x2)2(y3)2=x4y6(x^2 y^{-3})^{-2} = (x^2)^{-2} (y^{-3})^{-2} = x^{-4} y^6
The power of a product rule is applied to the denominator to resolve the outer exponent.
4
Divide the simplified numerator by the simplified denominator using the quotient rule for exponents.
8125x3y6x4y6=8125x3(4)y66=8125x7y12=8x7125y12\frac{\frac{8}{125} x^3 y^{-6}}{x^{-4} y^6} = \frac{8}{125} x^{3 - (-4)} y^{-6 - 6} = \frac{8}{125} x^7 y^{-12} = \frac{8x^7}{125y^{12}}
The quotient rule am/an=amna^m / a^n = a^{m-n} is used to subtract the exponents of the corresponding variables, and negative exponents are rewritten in the denominator.

Anahtar Kavram

Properties of Exponents in Algebraic Expressions

Alternatif Yöntem

Alternatively, you can write out all variables with positive exponents before simplifying. Rewrite the term in the numerator as 2y2x+y22x=5y22x\frac{2y^2}{x} + \frac{y^2}{2x} = \frac{5y^2}{2x}. Raising this to the 3-3 power flips the fraction and cubes it, yielding (2x5y2)3=8x3125y6\left(\frac{2x}{5y^2}\right)^3 = \frac{8x^3}{125y^6}. Simplifying the denominator yields 1(x2y3)2=1x4y6=y6x4\frac{1}{(x^2 y^{-3})^2} = \frac{1}{x^4 y^{-6}} = \frac{y^6}{x^4}. Dividing the numerator by the denominator yields 8x3125y6÷y6x4=8x3125y6x4y6=8x7125y12\frac{8x^3}{125y^6} \div \frac{y^6}{x^4} = \frac{8x^3}{125y^6} \cdot \frac{x^4}{y^6} = \frac{8x^7}{125y^{12}}.
Tahmini Süre:2m 0s
Soru 8Soru

For all non-zero real numbers xx, the expression x4(x3)kx^4 \cdot (x^3)^k is equivalent to x10x^{10}. What is the value of the integer kk?

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Cevap: 2

Cevap

The correct answer is 2.
First, use the power of a power property to write (x3)k(x^3)^k as x3kx^{3k}. The expression then becomes x4x3kx^4 \cdot x^{3k}. Next, use the product of powers property to combine the terms into x4+3kx^{4+3k}. Since the expression is equivalent to x10x^{10}, set the exponents equal: 4+3k=104 + 3k = 10. Solving this equation gives 3k=63k = 6, which simplifies to k=2k = 2.

Adım Adım Çözüm

1
Apply the power of a power rule (xa)b=xab(x^a)^b = x^{ab} to simplify (x3)k(x^3)^k.
x3kx^{3k}
To raise a power to another power, multiply the exponents.
2
Apply the product of powers rule xaxb=xa+bx^a \cdot x^b = x^{a+b} to combine the terms x4x3kx^4 \cdot x^{3k}.
x4+3kx^{4+3k}
When multiplying exponential terms with the same base, add their exponents.
3
Set the combined exponent 4+3k4+3k equal to the target exponent 1010 and solve for kk.
k=2k = 2
Since the bases are equal and non-zero, their exponents must be equal.

Anahtar Kavram

Properties of exponents in algebraic expressions (power of a power rule and product of powers rule)
Tahmini Süre:45s
Soru 9Soru

For all non-zero real numbers xx and yy, the expression

(x2y3)2(x1y4)3(x3y2)d\frac{(x^2 y^{-3})^{-2} (x^{-1} y^4)^3}{(x^3 y^{-2})^d}

can be written in the form xpyqx^p y^q, where pp and qq are integers. If q=2pq = 2p, what is the value of dd?

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Cevap: -4

Cevap

-4
Applying the rules of exponents yields the simplified expression x73dy18+2dx^{-7-3d} y^{18+2d}. Setting the exponent of yy equal to twice the exponent of xx gives the equation 18+2d=2(73d)18+2d = 2(-7-3d), which solves to d=4d = -4.

Adım Adım Çözüm

1
Apply the power of a power rule to the terms in the numerator.
(x2y3)2=x4y6(x^2 y^{-3})^{-2} = x^{-4} y^6 and (x1y4)3=x3y12(x^{-1} y^4)^3 = x^{-3} y^{12}
To raise a power to another power, multiply the exponents: (um)n=umn(u^m)^n = u^{mn}.
2
Multiply the simplified terms in the numerator together.
x4y6x3y12=x7y18x^{-4} y^6 \cdot x^{-3} y^{12} = x^{-7} y^{18}
To multiply powers with the same base, add the exponents: umun=um+nu^m \cdot u^n = u^{m+n}.
3
Apply the power of a power rule to the denominator.
(x3y2)d=x3dy2d(x^3 y^{-2})^d = x^{3d} y^{-2d}
Distribute the exponent dd to both variables inside the parentheses by multiplying the exponents.
4
Divide the numerator by the denominator.
x7y18x3dy2d=x73dy18(2d)=x73dy18+2d\frac{x^{-7} y^{18}}{x^{3d} y^{-2d}} = x^{-7-3d} y^{18-(-2d)} = x^{-7-3d} y^{18+2d}
To divide powers with the same base, subtract the exponent of the denominator from the exponent of the numerator: umun=umn\frac{u^m}{u^n} = u^{m-n}.
5
Set up the linear equation for dd using q=2pq = 2p and solve.
18+2d=2(73d)    18+2d=146d    8d=32    d=418+2d = 2(-7-3d) \implies 18+2d = -14-6d \implies 8d = -32 \implies d = -4
The problem states the relationship between the final exponents is q=2pq = 2p, where p=73dp = -7-3d and q=18+2dq = 18+2d.

Anahtar Kavram

Properties of exponents (product, quotient, and power rules) combined with solving a linear equation.
Soru 10Soru

For any positive real number yy, the expression (y3)1/2y2/3\frac{(y^3)^{1/2}}{y^{2/3}} is equivalent to which of the following?

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Cevap: y5/6y^{5/6}

Cevap

The simplified expression is y5/6y^{5/6}
Applying the power of a power rule to the numerator gives (y3)1/2=y3/2(y^3)^{1/2} = y^{3/2}. Then, applying the quotient rule to divide by y2/3y^{2/3} requires subtracting the exponents: 3223=9646=56\frac{3}{2} - \frac{2}{3} = \frac{9}{6} - \frac{4}{6} = \frac{5}{6}. This results in the equivalent expression y5/6y^{5/6}.

Adım Adım Çözüm

1
Apply the power of a power rule to the numerator (y3)1/2(y^3)^{1/2}.
y31/2=y3/2y^{3 \cdot 1/2} = y^{3/2}
When raising a power to a power, multiply the exponents.
2
Apply the quotient of powers rule to divide y3/2y^{3/2} by y2/3y^{2/3}.
y3/22/3y^{3/2 - 2/3}
When dividing powers with the same base, subtract the exponent in the denominator from the exponent in the numerator.
3
Subtract the fractions in the exponent by finding a common denominator.
y9/64/6=y5/6y^{9/6 - 4/6} = y^{5/6}
A common denominator for 2 and 3 is 6. Rewrite the fractions and subtract their numerators.

Anahtar Kavram

Properties of Exponents in Algebraic Expressions
Soru 11Soru

If xx and yy are positive real numbers, the expression

(x1/2+y1/2)2(x1/2y1/2)2(2x1/4y3/4)2\frac{(x^{1/2} + y^{1/2})^2 - (x^{1/2} - y^{1/2})^2}{(2x^{-1/4} y^{3/4})^{-2}}

can be simplified to the form kypk y^p. What is the value of the sum k+pk + p?

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Cevap: 18

Cevap

18
Expanding the numerator yields 4x1/2y1/24x^{1/2}y^{1/2}. Simplifying the denominator using exponent rules yields 14x1/2y3/2\frac{1}{4}x^{1/2}y^{-3/2}. Dividing the numerator by the denominator gives 41/4x1/21/2y1/2(3/2)=16y2\frac{4}{1/4} \cdot x^{1/2 - 1/2} \cdot y^{1/2 - (-3/2)} = 16y^2, which corresponds to k=16k = 16 and p=2p = 2. The sum is 16+2=1816 + 2 = 18.

Adım Adım Çözüm

1
Expand and simplify the numerator.
4x1/2y1/24x^{1/2}y^{1/2}
Expand both squared binomials: (x1/2+y1/2)2=x+2x1/2y1/2+y(x^{1/2} + y^{1/2})^2 = x + 2x^{1/2}y^{1/2} + y and (x1/2y1/2)2=x2x1/2y1/2+y(x^{1/2} - y^{1/2})^2 = x - 2x^{1/2}y^{1/2} + y. Subtracting the second expression from the first yields (x+2x1/2y1/2+y)(x2x1/2y1/2+y)=4x1/2y1/2(x + 2x^{1/2}y^{1/2} + y) - (x - 2x^{1/2}y^{1/2} + y) = 4x^{1/2}y^{1/2}.
2
Simplify the denominator using exponent rules.
14x1/2y3/2\frac{1}{4}x^{1/2}y^{-3/2}
Apply the power of a product rule (ab)n=anbn(ab)^n = a^n b^n to distribute the exponent of 2-2: (2x1/4y3/4)2=22(x1/4)2(y3/4)2(2x^{-1/4}y^{3/4})^{-2} = 2^{-2} \cdot (x^{-1/4})^{-2} \cdot (y^{3/4})^{-2}. This simplifies to 14x(1/4)(2)y(3/4)(2)=14x1/2y3/2\frac{1}{4} \cdot x^{(-1/4)(-2)} \cdot y^{(3/4)(-2)} = \frac{1}{4}x^{1/2}y^{-3/2}.
3
Divide the simplified numerator by the simplified denominator.
16y216y^2
Divide the coefficients and subtract the exponents of like bases: 4x1/2y1/214x1/2y3/2=(41/4)x1/21/2y1/2(3/2)=16x0y1/2+3/2=16y2\frac{4x^{1/2}y^{1/2}}{\frac{1}{4}x^{1/2}y^{-3/2}} = \left(\frac{4}{1/4}\right) x^{1/2 - 1/2} y^{1/2 - (-3/2)} = 16 x^0 y^{1/2 + 3/2} = 16y^2.
4
Identify the values of kk and pp and calculate their sum.
1818
Comparing 16y216y^2 to the form kypk y^p gives k=16k = 16 and p=2p = 2. Therefore, the sum is k+p=16+2=18k + p = 16 + 2 = 18.

Anahtar Kavram

Properties of Exponents in Algebraic Expressions
Soru 12Soru

If aa and bb are non-zero real numbers, which of the following expressions is equivalent to a4b2(a2b3)2\frac{a^4 b^{-2}}{(a^2 b^{-3})^2}?

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Cevap: b4b^4

Cevap

b4b^4
To simplify the expression, first apply the power of a product rule to the denominator: (a2b3)2=(a2)2(b3)2=a4b6(a^2 b^{-3})^2 = (a^2)^2 (b^{-3})^2 = a^4 b^{-6}. Next, divide the numerator by the simplified denominator by subtracting the exponents of the corresponding bases: a4b2a4b6=a44b2(6)=a0b4\frac{a^4 b^{-2}}{a^4 b^{-6}} = a^{4-4} b^{-2 - (-6)} = a^0 b^4. Since a0=1a^0 = 1 for any non-zero real number aa, the expression simplifies to b4b^4.

Adım Adım Çözüm

1
Simplify the denominator using the power of a product and power of a power rules.
(a2b3)2=(a2)2(b3)2=a4b6(a^2 b^{-3})^2 = (a^2)^2 \cdot (b^{-3})^2 = a^4 b^{-6}
When raising a product to a power, raise each factor to that power. When raising a power to a power, multiply the exponents.
2
Substitute the simplified denominator back into the original fraction.
a4b2a4b6\frac{a^4 b^{-2}}{a^4 b^{-6}}
To prepare the expression for division.
3
Divide the numerator by the denominator by subtracting the exponents of like bases.
a44b2(6)=a0b4=1b4=b4a^{4-4} b^{-2 - (-6)} = a^0 b^4 = 1 \cdot b^4 = b^4
When dividing terms with the same base, subtract the exponent of the denominator from the exponent of the numerator.

Anahtar Kavram

Properties of exponents, specifically the power of a product rule, power of a power rule, and quotient rule.
Soru 13Soru

For positive real values of uu and vv, the expression (9u1v24u3v4)1/2\left( \frac{9u^{-1}v^2}{4u^3v^{-4}} \right)^{-1/2} can be simplified to which of the following?

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Cevap: 2u23v3\frac{2u^2}{3v^3}

Cevap

The expression is equivalent to 2u23v3\frac{2u^2}{3v^3}.
The correct answer is obtained by first simplifying the quotient inside the parenthesis to get 94u4v6\frac{9}{4}u^{-4}v^6. Then, raising each factor to the 1/2-1/2 power yields (94)1/2=23\left(\frac{9}{4}\right)^{-1/2} = \frac{2}{3}, (u4)1/2=u2(u^{-4})^{-1/2} = u^2, and (v6)1/2=v3(v^6)^{-1/2} = v^{-3}. Combining these results and rewriting with positive exponents gives the simplified expression.

Adım Adım Çözüm

1
Simplify the uu terms inside the parenthesis using the quotient rule for exponents.
u1u3=u13=u4\frac{u^{-1}}{u^3} = u^{-1 - 3} = u^{-4}
The quotient rule states that xaxb=xab\frac{x^a}{x^b} = x^{a-b}.
2
Simplify the vv terms inside the parenthesis using the quotient rule for exponents.
v2v4=v2(4)=v6\frac{v^2}{v^{-4}} = v^{2 - (-4)} = v^6
Subtracting a negative exponent is equivalent to adding its absolute value.
3
Apply the outer exponent of 1/2-1/2 to the coefficient.
(94)1/2=(49)1/2=23\left(\frac{9}{4}\right)^{-1/2} = \left(\frac{4}{9}\right)^{1/2} = \frac{2}{3}
A negative exponent represents taking the reciprocal of the base, and a fractional exponent of 1/21/2 represents the square root.
4
Apply the outer exponent of 1/2-1/2 to the simplified variable terms using the power of a power rule.
(u4)1/2=u2(u^{-4})^{-1/2} = u^2 and (v6)1/2=v3(v^6)^{-1/2} = v^{-3}
The power of a power rule states that (xa)b=xab(x^a)^b = x^{ab}.
5
Combine the simplified parts and rewrite the expression with positive exponents.
23u2v3=2u23v3\frac{2}{3} u^2 v^{-3} = \frac{2u^2}{3v^3}
An expression with a negative exponent in the numerator can be moved to the denominator with a positive exponent.

Anahtar Kavram

Applying product, quotient, and power rules of exponents to algebraic expressions with negative and rational exponents.
Soru 14Soru

The algebraic expression (x2)ax4\frac{(x^2)^a}{x^{-4}} is equivalent to x10x^{10} for all non-zero real numbers xx. What is the value of aa?

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Cevap: 3

Cevap

The correct answer is 3.
The correct value for aa is 33. Applying the power of a power rule to (x2)a(x^2)^a yields x2ax^{2a}. Then, applying the quotient rule to x2ax4\frac{x^{2a}}{x^{-4}} yields x2a(4)=x2a+4x^{2a - (-4)} = x^{2a+4}. Equating the exponents gives 2a+4=102a + 4 = 10, which solves to a=3a = 3.

Adım Adım Çözüm

1
Apply the power of a power property to the numerator.
(x2)a=x2a(x^2)^a = x^{2a}
When raising a power to another power, multiply the exponents: (xm)n=xmn(x^m)^n = x^{mn}.
2
Apply the quotient property of exponents to simplify the fraction.
x2ax4=x2a(4)=x2a+4\frac{x^{2a}}{x^{-4}} = x^{2a - (-4)} = x^{2a + 4}
When dividing exponential expressions with the same base, subtract the exponent in the denominator from the exponent in the numerator: xmxn=xmn\frac{x^m}{x^n} = x^{m-n}.
3
Set the resulting exponent equal to the exponent of the equivalent expression and solve for aa.
2a+4=10    2a=6    a=32a + 4 = 10 \implies 2a = 6 \implies a = 3
Since the bases are identical and the expressions are equivalent, their exponents must be equal.

Anahtar Kavram

Properties of Exponents in Algebraic Expressions
Soru 15Soru
For all positive real numbers xx and yy, the expression
(3x2y3)3(2x1y2)2(6x3y2)2\frac{(3x^2 y^{-3})^3 \cdot (2x^{-1} y^2)^2}{(6x^3 y^{-2})^2}
can be simplified to the form AxaybA x^a y^b, where AA, aa, and bb are integers. What is the value of the sum A+a+bA + a + b?
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Cevap: 0

Cevap

The value of the sum A+a+bA + a + b is 0.
By applying the rules of exponents systematically, the expression simplifies to 3x2y13 x^{-2} y^{-1}. Comparing this to AxaybA x^a y^b yields A=3A = 3, a=2a = -2, and b=1b = -1. The sum is 3+(2)+(1)=03 + (-2) + (-1) = 0.

Adım Adım Çözüm

1
Simplify the first term in the numerator
27x6y927x^6y^{-9}
Apply the power of a product rule and power of a power rule to (3x2y3)3(3x^2 y^{-3})^3.
2
Simplify the second term in the numerator
4x2y44x^{-2}y^4
Apply the power of a product rule and power of a power rule to (2x1y2)2(2x^{-1} y^2)^2.
3
Multiply the simplified terms in the numerator together
108x4y5108x^4y^{-5}
Multiply coefficients and add the exponents of like bases.
4
Simplify the denominator
36x6y436x^6y^{-4}
Apply the power of a product rule and power of a power rule to (6x3y2)2(6x^3 y^{-2})^2.
5
Divide the numerator by the denominator
3x2y13x^{-2}y^{-1}
Divide the coefficients and subtract the denominator exponents from the numerator exponents for like bases.
6
Sum the constants AA, aa, and bb
0
Identify A=3A = 3, a=2a = -2, b=1b = -1 from the expression 3x2y13x^{-2}y^{-1}, and calculate 3+(2)+(1)=03 + (-2) + (-1) = 0.

Anahtar Kavram

Properties of Exponents in Algebraic Expressions
Soru 16Soru

If aa and bb represent positive real numbers, which of the following is an equivalent form of the expression below?

(a1+b1)2ab\frac{(a^{-1} + b^{-1})^{-2}}{ab}
Cevabı ve açıklamayı göster

Cevap: ab(a+b)2\frac{ab}{(a+b)^2}

Cevap

The correct answer is the fraction with abab in the numerator and the square of the sum (a+b)(a+b) in the denominator, which is \frac{ab}{(a+b)^2}.
The correct answer is found by first rewriting a1+b1a^{-1} + b^{-1} as 1a+1b\frac{1}{a} + \frac{1}{b}, which simplifies to a+bab\frac{a+b}{ab} using a common denominator. Raising this to the power of 2-2 gives a2b2(a+b)2\frac{a^2 b^2}{(a+b)^2}. Finally, dividing this expression by abab reduces the exponents of aa and bb by 1, resulting in ab(a+b)2\frac{ab}{(a+b)^2}.

Adım Adım Çözüm

1
Rewrite the negative exponents in the numerator as reciprocals.
The expression inside the parentheses becomes 1a+1b\frac{1}{a} + \frac{1}{b}.
By the definition of negative exponents, x1=1xx^{-1} = \frac{1}{x}.
2
Find a common denominator to add the fractions inside the parentheses.
1a+1b=bab+aab=a+bab\frac{1}{a} + \frac{1}{b} = \frac{b}{ab} + \frac{a}{ab} = \frac{a+b}{ab}.
To add fractions, they must share a common denominator, which is the product of aa and bb.
3
Apply the negative power of 2-2 to the simplified fraction.
(a+bab)2=(aba+b)2=a2b2(a+b)2\left(\frac{a+b}{ab}\right)^{-2} = \left(\frac{ab}{a+b}\right)^2 = \frac{a^2 b^2}{(a+b)^2}.
An expression raised to a negative exponent is equal to the reciprocal of the expression raised to the positive exponent.
4
Divide the result by the denominator abab.
a2b2(a+b)21ab=ab(a+b)2\frac{a^2 b^2}{(a+b)^2} \cdot \frac{1}{ab} = \frac{ab}{(a+b)^2}.
Dividing by a term is equivalent to multiplying by its reciprocal, and applying the exponent quotient rule simplifies a2b2ab\frac{a^2 b^2}{ab} to abab.

Anahtar Kavram

Properties of exponents including negative power rules, quotient rules, and algebraic fraction addition.

Alternatif Yöntem

An alternative method is to substitute small positive integer values for aa and bb. Let a=1a = 1 and b=2b = 2. The expression evaluates to (11+21)212=(1+0.5)22=1.522=(3/2)22=4/92=29\frac{(1^{-1} + 2^{-1})^{-2}}{1 \cdot 2} = \frac{(1 + 0.5)^{-2}}{2} = \frac{1.5^{-2}}{2} = \frac{(3/2)^{-2}}{2} = \frac{4/9}{2} = \frac{2}{9}. Evaluating the correct expression with these values yields 12(1+2)2=29\frac{1 \cdot 2}{(1+2)^2} = \frac{2}{9}, which matches.
Tahmini Süre:2m 0s
Soru 17Soru
For all non-zero real numbers yy, the expression
(y3)2(y4)ay5\frac{(y^3)^2 \cdot (y^{-4})^a}{y^5}
is equivalent to y7y^{-7}. What is the value of aa?
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Cevap: 22

Cevap

The correct value of aa is 2.
Applying the exponent rules systematically allows us to simplify the expression. First, (y3)2(y^3)^2 becomes y6y^6 and (y4)a(y^{-4})^a becomes y4ay^{-4a} using the power of a power rule. Second, we combine the terms in the numerator using the product rule to get y64ay^{6-4a}. Third, we divide by y5y^5 using the quotient rule to obtain y64a5=y14ay^{6-4a-5} = y^{1-4a}. Setting this equal to the target expression y7y^{-7} gives the equation 14a=71-4a = -7. Solving for aa gives 4a=8-4a = -8, which simplifies to 22.

Adım Adım Çözüm

1
Apply the power of a power rule, (xm)n=xmn(x^m)^n = x^{mn}, to the exponential terms in the numerator.
(y3)2=y6(y^3)^2 = y^6 and (y4)a=y4a(y^{-4})^a = y^{-4a}
This simplifies nested exponent terms into single base terms.
2
Apply the product rule of exponents, xmxn=xm+nx^m \cdot x^n = x^{m+n}, to combine the numerator terms.
y6y4a=y64ay^6 \cdot y^{-4a} = y^{6-4a}
This simplifies the numerator to a single power of yy.
3
Apply the quotient rule of exponents, xmxn=xmn\frac{x^m}{x^n} = x^{m-n}, to divide by the denominator.
y64ay5=y(64a)5=y14a\frac{y^{6-4a}}{y^5} = y^{(6-4a) - 5} = y^{1-4a}
This simplifies the entire rational expression into a single exponential expression.
4
Equate the simplified exponent to the exponent of the equivalent expression and solve the linear equation for aa.
14a=7    4a=8    a=21-4a = -7 \implies -4a = -8 \implies a = 2
Since the bases are equal and non-zero, their exponents must be equal for the expressions to be equivalent.

Anahtar Kavram

Properties of Exponents in Algebraic Expressions
Tahmini Süre:1m 30s
Soru 18Soru
If xx and yy are positive real numbers such that
(x3y2)k(x1y4)3=x6y6(x^3 y^{-2})^k \cdot (x^{-1} y^4)^3 = x^6 y^6
what is the value of the exponent kk?
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Cevap: 3

Cevap

The value of the exponent kk is 3.
Applying the exponent rules simplifies the left side of the equation to x3k3y122kx^{3k-3}y^{12-2k}. Equating the exponent of xx to the right side gives 3k3=63k - 3 = 6, which yields k=3k = 3. This value is confirmed by equating the exponent of yy, since 122(3)=612 - 2(3) = 6.

Adım Adım Çözüm

1
Apply the power of a power rule (am)n=amn(a^m)^n = a^{mn} to expand the terms in the expression.
(x3y2)k=x3ky2k(x^3 y^{-2})^k = x^{3k} y^{-2k} and (x1y4)3=x3y12(x^{-1} y^4)^3 = x^{-3} y^{12}
To remove the outer parentheses by multiplying the internal exponents of each variable by the outer exponent.
2
Multiply the terms together by applying the product rule for exponents, aman=am+na^m \cdot a^n = a^{m+n}.
(x3ky2k)(x3y12)=x3k3y122k(x^{3k} y^{-2k})(x^{-3} y^{12}) = x^{3k-3} y^{12-2k}
To combine the like bases of xx and yy into a single simplified expression.
3
Set the exponents of like bases equal to the exponents on the right-hand side of the equation, x6y6x^6 y^6.
3k3=63k - 3 = 6 and 122k=612 - 2k = 6
Since the bases are equal and non-zero, their respective exponents must also be equal.
4
Solve the linear equation 3k3=63k - 3 = 6 for kk.
3k=9    k=33k = 9 \implies k = 3
To determine the numerical value of the variable kk.
5
Verify the solution by solving the second linear equation, 122k=612 - 2k = 6.
2k=6    k=3-2k = -6 \implies k = 3
To ensure consistency across both variable exponents in the expression.

Anahtar Kavram

Properties of Exponents in Algebraic Expressions
Soru 19Soru

If the expression (x2y3)4(x1y2)3\frac{(x^2 y^3)^4}{(x^{-1} y^2)^3} is written in the equivalent form xaybx^a y^b, what is the value of aba - b?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

5
Simplifying the numerator yields x8y12x^8 y^{12} and simplifying the denominator yields x3y6x^{-3} y^6. Dividing these expressions by subtracting the exponents of like bases results in x8(3)y126=x11y6x^{8 - (-3)} y^{12 - 6} = x^{11} y^6. Comparing this to the expression xaybx^a y^b shows that a=11a = 11 and b=6b = 6. The value of aba - b is 116=511 - 6 = 5.

Adım Adım Çözüm

1
Simplify the numerator of the expression.
x8y12x^8 y^{12}
Apply the power of a power and power of a product properties of exponents: (x2y3)4=x24y34(x^2 y^3)^4 = x^{2 \cdot 4} y^{3 \cdot 4}.
2
Simplify the denominator of the expression.
x3y6x^{-3} y^6
Apply the power of a power and power of a product properties of exponents: (x1y2)3=x13y23(x^{-1} y^2)^3 = x^{-1 \cdot 3} y^{2 \cdot 3}.
3
Simplify the quotient by dividing the simplified numerator by the simplified denominator.
x11y6x^{11} y^6
Use the quotient property of exponents, xmxn=xmn\frac{x^m}{x^n} = x^{m-n}, to subtract the exponents: 8(3)=118 - (-3) = 11 and 126=612 - 6 = 6.
4
Identify the values of aa and bb and compute the difference aba - b.
5
By comparing x11y6x^{11} y^6 to the target form xaybx^a y^b, we find a=11a = 11 and b=6b = 6. Subtracting bb from aa yields 116=511 - 6 = 5.

Anahtar Kavram

Properties of exponents including power of a power, power of a product, and quotient rules.
Soru 20Soru
For all non-zero real numbers aa, bb, and cc, the expression below is simplified:
(2a2b1c3)34a5(b2c2)2\frac{(2a^2 b^{-1} c^3)^3}{4a^5 (b^2 c^{-2})^{-2}}
Which of the following is equivalent to this expression?
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Cevap: 2abc52abc^5

Cevap

The expression is equivalent to 2abc52abc^5.
The correct answer is obtained by first simplifying the numerator and denominator using the power of a product and power of a power rules, then dividing the coefficients and subtracting the exponents of like bases. This yields the simplified expression 2abc52abc^5.

Adım Adım Çözüm

1
Apply the power of a product rule to the numerator.
(2a2b1c3)3=23(a2)3(b1)3(c3)3=8a6b3c9(2a^2 b^{-1} c^3)^3 = 2^3 \cdot (a^2)^3 \cdot (b^{-1})^3 \cdot (c^3)^3 = 8 a^6 b^{-3} c^9
Each factor inside the parentheses must be raised to the power of 3, multiplying the exponents of the variables.
2
Apply the power of a product rule to the denominator's parentheses.
4a5(b2c2)2=4a5(b2)2(c2)2=4a5b4c44a^5 (b^2 c^{-2})^{-2} = 4a^5 \cdot (b^2)^{-2} \cdot (c^{-2})^{-2} = 4a^5 b^{-4} c^4
The terms inside the parentheses are raised to the power of -2, multiplying their exponents.
3
Divide the simplified numerator by the simplified denominator.
2a1b1c52a^1 b^1 c^5, which is 2abc52abc^5
Divide the coefficients (8 / 4 = 2) and subtract the exponents of the same bases: a65=a1a^{6-5} = a^1, b3(4)=b1b^{-3 - (-4)} = b^1, and c94=c5c^{9-4} = c^5.

Anahtar Kavram

Properties of Exponents in Algebraic Expressions

Alternatif Yöntem

Alternatively, you can rewrite the expression by eliminating negative exponents first. Convert b1b^{-1} to 1b\frac{1}{b}, b2b^2 to itself, and c2c^{-2} to 1c2\frac{1}{c^2} within the parentheses, apply the outer exponents, and then simplify the resulting complex fraction.
Tahmini Süre:1m 30s
Sayfa 1 / 2Sonraki