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Exponents Practice Questions

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Updated on Oct 3, 2011

To review these concepts, go to Exponents Study Guide.

Exponents Practice Questions

Problems

Practice 1

Solve:

  1. (b2)3
  2. (n6)6
  3. 3(v7)4
  4. 4(r5)10
  5. 5(3w8)2

Practice 2

  1. (pqr)9
  2. (3j6k2)3
  3. –10(2s8t5)5
  4. (u7)–6
  5. 5(e–5)12
  6. (6a–11)3
  7. (2z–2)–2
  8. (g3h–4)–4
  9. (4p3q8)0
  10. 2(k13m0n–1)–1

Solutions

Practice 1

  1. b2 is raised to the third power. Multiply the exponents: (2)(3) = 6, and (b2)3 = b6.

  2. n6 is raised to the sixth power. Multiply the exponents: (6)(6) = 36, and (n6)6 = n36.

  3. 3v7 is raised to the fourth power.

    Raise 3 to the fourth power and raise v7 to the fourth power: 34 = 81.

    To find (v7)4, multiply the exponents: (7)(4) = 28, v7)4 = v28, and (3v7)4 = 81v28.

  4. r5 is raised to the tenth power and multiplied by 4.

    Exponents come first in the order of operations, so begin by raising r5 to the tenth power.

    Multiply the exponents: (5)(10) = 50, and (r5)10 = r50.

    Finally, multiply by 4: (4)(r50) = 4r50.

  5. 3w8 is raised to the second power and multiplied by 4. Work with the exponent before multiplying.

    Raise 3 to the second power and raise w8 to the second power. 32 = 9.

    To find (w8)2, multiply the exponents: (8)(2) = 16, (w8)2 = w16, and (3w8)2 = 9w16.

    Finally, multiply 9w16 by 5: (5)(9w16) = 45w16.

Practice 2

  1. pqr is raised to the ninth power. Each variable has an exponent of 1, and each exponent must be multiplied by 9. Because (1)(9) = 9, the exponent of each variable is 9.

    (pqr)9 = p9q9r9

  2. (3j6k2) is raised to the third power. Each variable must have its exponent multiplied by 3, and the coefficient 3 must also be raised to the third power.

    33 = 27

    (6)(3) = 18, so (j6)3 = j18

    (2)(3) = 6, so (k2)3 = k6

    (3j6k2)3 = 27j18k6

  3. (2s8t5) is raised to the fifth power, and then multiplied by –10. Each variable must have its exponent multiplied by 5, and the coefficient 2 must also be raised to the fifth power.

    25 = 32

    (8)(5) = 40, so (s8)5 = s40

    (5)(5) = 25, so (t5)5 = t25

    (2s8t5)5 = 32s40t25

    Finally, multiply by –10:

    (–10)(32s40t25) = –320s40t25

  4. (u7) is raised to the negative sixth power. Multiply 7 by –6:

    (7)(–6) = –42, so (u7)–6 = u–42

  5. (e–5) is raised to the 12th power, and then multiplied by 5. Multiply –5 by 12:

    (–5)(12) = –60, so (e–5)12 = e–60

    Multiply e–60 by its coefficient, 5:

    (5)(e–60) = 5e–60

  6. (6a–11) is raised to the third power. The variable a must have its exponent multiplied by 3, and the coefficient 6 must also be raised to the third power.

    63 = 216

    (–11)(3) = –33, so (a–11)3 = a–33

    (6a–11)3 = 216a–33

  7. (2z–2) is raised to the negative second power. The variable z must have its exponent multiplied by –2, and the coefficient 2 must also be raised to the negative second power. To raise 2 to the negative second power, place the 2 with an exponent of positive 2 in the denominator of a fraction with a numerator of 1:

    (–2)(–2) = 4, so (z–2)–2 = z4

  8. (g3h–4) is raised to the negative eighth power. The variables g and h must each have their exponent multiplied by –8.

    (3)(–8) = –24, so (g3)–8 = g–24

    (–4)(–8) = 32, so (h–4)–8 = h32

    (g3h–4)–8 = g–24h32

  9. The entire term in parentheses is raised to the zero power. Any quantity raised to the zero power is equal to 1.

  10. (k13m0n–1) is raised to the negative 1 power, and then multiplied by 2. The variables k and n must each have their exponent multiplied by –1. The variable m is raised to the zero power, so it is equal to 1. Because k13 and n–1 are multiplied by m0=1, the m term drops out of the expression.

    (13)(–1) = –13, so (k13)–1 = k–13

    (–1)(–1) = 1, so (n–1)–1 = n1, or n

    The expression is now 2(k–13n). Multiply by 2:

    (2)(k–13n) = 2k–13n

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