# LAWS OF EXPONENTS WORKSHEET

Problems 1-10 : Simplify each expression using the laws of exponents.

Problem 1 :

2m2  2m3

Problem 2 :

m4  2m-3

Problem 3 :

(4a3)2

Problem 4 :

(x3)0

Problem 5 :

Problem 6 :

(x2y-1)4

Problem 7 :

(3x)(2x1/2)

Problem 8 :

Problem 9 :

(6ab2c3)(4b-2c-3d)

Problem 10 :

(x-ayb)3(x-3y-2)-a

Problem 11 :

Find the value of y :

(y2)2.5 = -32

Problem 12 :

h(x) = 2x

The function is h is defined above. What is h(5) - h(3)?

Problem 13 :

If 3x  = 10, what is the value of 3x-3?

Problem 14 :

If x-1/3 = 5/2, then find the value of x.

Problem 15 :

If 5x/25= 125, then solve for x in terms of y.

Problem 16 :

If a and b are positive even integers, which of the following is greatest ?

A) (-2a)b

B) (-2a)2b

C) (2a)b

D) 2a2b

Problem 17 :

If √(x√x) = xa, then find the value of a.

Problem 18 :

If x2 = y3 and x3z = y9, then find the value of z.

Problem 19 :

If n3 = x, n4 = 20x and n > 0, then find the value of n.

Problem 20 :

If x2y3 = 10 and x3y= 8, what is the value of x5y5?

Problem 21 :

If x8y7 = 333 and x7y= 3, what is the value of xy?

Problem 22 :

If 2x + 3 - 2x = k(2x), what is the value of k?

Problem 23 :

If xac ⋅ xbc = x30, x > 1 and a + b = 5, what is the value of c?

Problem 24 :

If (√9)-7 ⋅ (√3)-4 = 3k, then find the value of k.

Problem 25 :

In the equation 2√(x - 2) = 3√2, if x ≥ 2, then find the value of x.

Problem 26 :

x - 2 = √x

In the equation above x ≥ 0, find all the values of x which satisfy the equation.

= 2m2  2m3

= (2 ⋅ 2)(mm3)

= 4m2 + 3

= 4m5

= m4  2m-3

2(mm-3)

= 2m4 - 3

= 2m1

= (4a3)

42(a3)2

= 16a⋅ 2

= 16a6

= (x3)0

Anything to the zeroth power is equal to 1.

= 1

= (x2y-1)4

(x2)4(y-1)4

x8y-4

= (3x)(2x1/2)

= 6x1 + 1/2

= 6x3/2

= (6ab2c3)(4b-2c-3d)

= 24ab2 - 2c3 - 3d

= 24ab0c0d

= 24a(1)(1)d

= (x-ayb)3(x-3y-2)-a

= (x-a)3(yb)3(x-3)-a(y-2)-a

= x-3ay3bx3ay2a

= x-3a + 3ay3b + 2a

= x30y3b + 2a

= (1)y2a + 3b

= y2a + 3b

(y2)2.5 = -32

y2 x 2.5= (-2)5

y= (-2)5

Since there is same exponent on both sides, bases can be equated.

y = -2

h(x) = 2x

h(5) - h(3) :

= 223

= (2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2) - (2 ⋅ 2 ⋅ 2)

= 32 - 8

= 24

3x-3 = 3x/33

= 3x/(3 ⋅ 3 ⋅ 3)

= 3x/27

Substitute 3= 10.

= 10/27

x-1/3 = 5/2

x = (5/2)-3/1

x = (5/2)-3

x = (2/5)3

Distribute the exponent to numerator and denominator.

x = 23 / 53

x = 8/125

5x/25= 125

5x/(52)= 53

5x/52= 53

5x - 2y = 53

Since there is same base on both sides, exponents can be equated.

x - 2y = 3

x = 2y + 3

Because a and b are positive even integers, better we can assume some values for a and b and go through each choice.

Let a = 2 and b = 2.

Substitute a = 2 and b = 2 in each option.

A : [-2(2)]2 = (-4)2 = 16

B : [-2(2)]2(2) = (-4)4 = 256

C : [2(2)]2 = (4)2 = 16

D : 2(2)2(2) = 2(2)4 = 2(16) = 32

So, option B is the greatest.

√(x√x)  = xa

√(x ⋅ x1/2)  = xa

√(x1 + 1/2)  = xa

√(x3/2) = xa

(x3/2)1/2 = xa

x3/4 = xa

3/4 = a

x3z = y9

x3z = y3(3)

x3z = (y3)3

Substitute xfor y3.

x3z = (x2)3

x3z = x6

3z = 6

Divide each side by 3.

z = 2

n4 = 20x

n⋅ n = 20x

Substitute x for n3.

x ⋅ n = 20x

nx = 20x

Divide each side by x.

n = 20

x2y3 = 10 ----(1)

x3y= 8 ----(2)

Multiply (1) and (2).

x2y x3y2 = 10 ⋅ 8

x2+ 3y3 + 2 = 80

x5y5 = 80

x8y7 = 333 ----(1)

x7y= 3 ----(2)

Divide (1) and (2).

(x8y7)/(x7y6) = 333/3

x8 - 7y7- 6 = 111

xy = 111

2x + 3 - 2x = k(2x)

2x ⋅ 2- 2x = k(2x)

Factor 2on the left side.

2x(2- 1) = k(2x)

Divide both sides by 2x.

2- 1 = k

8 - 1 = k

7 =  k

xac ⋅ xbc = x30

xac + bc = x30

Since there is same base (x) on both sides, exponents can be equated.

ac + bc = 30

c(a + b) = 30

Substitute a + b = 5.

c(5) = 30

Divide both sides by 5.

c = 6

(91/2)-7 ⋅ (31/2)-4 = 3k

(9)-7/2 ⋅ (3)-4/2 = 3k

(32)-7/2 ⋅ 3-2 = 3k

3⋅ (-7/2) ⋅ 3-2 = 3k

3-7 ⋅ 3-2 = 3k

3-7 - 2 = 3k

3-9 = 3k

k = -9

2√(x - 2) = 3√2

Square both sides to get rid of the radicals.

[2√(x - 2)]2 = (3√2)2

2⋅ [√(x - 2)]2 = 3⋅ (√2)2

4 ⋅ (x - 2) = 9 ⋅ 2

4x - 8 = 18

4x = 26

Divide each side by 4.

x = 6.5

x - 2 = √x

Square both sides to get rid of the square root on the right side.

(x - 2)2 = (√x)2

(x - 2)(x - 2) = x

x2 - 2x - 2x + 4 = x

x2 - 4x + 4 = x

Subtrtact x from both sides.

x2 - 5x + 4 = 0

Factor and solve.

x2 - 4x - x + 4 = 0

x(x - 4) - 1(x - 4) = 0

(x - 4)(x - 1) = 0

x - 4 = 0  or  x - 1 = 0

x = 4  or  x = 1

Verify the solutions with the original equation.

 x = 4 :x - 2 = √x4 - 2 = √42 = 2 ✔ x = 1 :x - 2 = √x1 - 2 = √1-1 = 1 ✘

x = 2 staisfies the original squation and x = 1 doesn't.

Therefore, the solution is x = 2.

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