Q1. Write \(2\) multiplied 5 times in index form.

Ans: \(2\times2\times2\times2\times2 = 2^5\).

Q2. What is \(a^0\) (for \(a\ne0\))?

Ans: \(a^0=1\).

Q3. Express cube root of 27 using exponent notation.

Ans: \(\sqrt[3]{27}=27^{1/3}=3\).

Q4. What is \(3^2\)?

Ans: \(3^2 = 9\).

Q5. Express \(\sqrt{49}\) using rational exponent.

Ans: \(\sqrt{49}=49^{1/2}=7\).

Q6. What is \(5^{-1}\) equal to?

Ans: \(5^{-1} = \dfrac{1}{5}\).

Q7. What is \((ab)^2\) equal to?

Ans: \((ab)^2 = a^2 b^2\).

Q8. Convert \(100^{1/2}\) to a simpler number.

Ans: \(100^{1/2} = \sqrt{100} = 10\).

Q9. If \(a^m \times a^n = a^k\), what is \(k\)?

Ans: \(k=m+n\).

Q10. What is \(\sqrt[3]{-8}\)?

Ans: \(\sqrt[3]{-8} = -2\).

Q11. Simplify \( (2^3)^2 \).

Ans: \((2^3)^2 = 2^{3\times2} = 2^6 = 64\).

Q12. Write \(a^{-3}\) as a fraction.

Ans: \(a^{-3} = \dfrac{1}{a^3}\) (for \(a\ne0\)).

Q13. What is cube of 4?

Ans: \(4^3 = 64\).

Q14. Simplify \( \dfrac{a^5}{a^2}\).

Ans: \(a^{5-2} = a^3\) (for \(a\ne0\)).

Q15. Express 5th root of 32 in exponent form.

Ans: \(\sqrt[5]{32} = 32^{1/5} = 2\).

Q16. What is \(10^1\)?

Ans: \(10^1 = 10\).

Q17. True/False: \((a/b)^n = a^n / b^n\).

Ans: True (for \(b\ne0\)).

Q18. If \(x^3 = 125\), what is \(x\)?

Ans: \(x = \sqrt[3]{125} = 5\).

Q19. Simplify \((3\times2)^2\).

Ans: \((6)^2 = 36\). Also equals \(3^2\times2^2=9\times4=36\).

Q20. What is \((0.2)^3\)?

Ans: \((0.2)^3 = 0.008\).

Q1. Simplify \(a^3 \times a^4\).

Ans: \(a^{3+4} = a^7\).

Q2. Simplify \(\dfrac{2^7}{2^3}\).

Ans: \(2^{7-3}=2^4=16\).

Q3. Simplify \((3^2)^4\).

Ans: \(3^{2\times4}=3^8=6561\).

Q4. Evaluate \(16^{3/4}\) (write value).

Ans: \(16^{3/4}=(16^{1/4})^3=(2)^3=8\) because \(16^{1/4}=\sqrt[4]{16}=2\).

Q5. Simplify \((2^3\times5^3)\) using a single power.

Ans: \((2\times5)^3 = 10^3 = 1000\).

Q6. Write \(a^{4/3}\) in radical form.

Ans: \(a^{4/3} = \sqrt[3]{a^4} = (\sqrt[3]{a})^4\).

Q7. Simplify \((\tfrac{2}{3})^{-2}\).

Ans: \(\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}\).

Q8. If \(x^3 = 512\), find \(x\).

Ans: \(x = \sqrt[3]{512} = 8\) (since \(8^3=512\)).

Q9. Simplify \(\dfrac{a^5 b^3}{a^2 b}\).

Ans: \(a^{5-2} b^{3-1} = a^3 b^2.\)

Q10. Evaluate \((0.5)^{-3}\).

Ans: \((0.5)^{-3} = (2)^{3} = 8\) since \(0.5 = 1/2\).

Q11. Write \( \sqrt[5]{243}\) as a rational power and evaluate.

Ans: \(243^{1/5} = 3\) since \(3^5=243\).

Q12. Simplify \((a^2 b^3)^2\).

Ans: \(a^{4} b^{6}\).

Q13. Simplify \(27^{2/3}\).

Ans: \(27^{2/3} = (\sqrt[3]{27})^2 = 3^2=9.\)

Q14. Show that \( (5^0 + 3^0) = 2\).

Ans: \(5^0=1,\;3^0=1\Rightarrow1+1=2.\)

Q15. Simplify \(\dfrac{81^{3/4}}{3}\).

Ans: \(81^{3/4} = (\sqrt[4]{81})^3 = 3^3 = 27\). Then \(27/3=9.\)

Q16. Simplify \(\sqrt[3]{\dfrac{216}{125}}\).

Ans: \(\dfrac{\sqrt[3]{216}}{\sqrt[3]{125}}=\dfrac{6}{5}\).

Q17. If \(a^3 = 8\) and \(b^3 = 27\), find \((ab)\).

Ans: \(a = 2,\; b=3\Rightarrow ab=6.\)

Q18. Simplify \( (4^2 \times 4^{-5})\).

Ans: \(4^{2-5} = 4^{-3} = \dfrac{1}{64}.\)

Q19. Show \( (a^{1/3})^6 = a^2\).

Ans: \((a^{1/3})^6 = a^{6/3} = a^2.\)

Q20. Express \( \dfrac{1}{\sqrt[3]{27}}\) in simplest form.

Ans: \(\dfrac{1}{3}\) since \(\sqrt[3]{27}=3\).

Q1. Using laws of indices, simplify \(\dfrac{a^7 b^{-2}}{a^2 b^3}\).

Ans: Subtract exponents: \(a^{7-2} b^{-2-3} = a^5 b^{-5} = \dfrac{a^5}{b^5}.\)

Q2. Evaluate and simplify \( (16)^{3/4} \times (8)^{1/3}\).

Ans: \(16^{3/4} = (\sqrt[4]{16})^3=2^3=8.\; 8^{1/3}=2.\) Product \(=8\times2=16.\)

Q3. Simplify \(\left(\dfrac{81}{16}\right)^{3/4}\).

Ans: \(\left(\dfrac{81}{16}\right)^{3/4}=\left(\dfrac{\sqrt[4]{81}}{\sqrt[4]{16}}\right)^3=\left(\dfrac{3}{2}\right)^3=\dfrac{27}{8}.\)

Q4. If \(x^{1/2} = 5\), find \(x^{3/2}\).

Ans: \(x^{1/2}=5\Rightarrow x=25.\; x^{3/2} = x^{1/2}\cdot x = 5\cdot25=125.\)

Q5. Prove \((a^m b^n)^p = a^{mp} b^{np}\).

Ans: Expand using distributive property of exponents: \((a^m b^n)^p = a^{m\cdot p} b^{n\cdot p}\) (by repeated multiplication), which is \(a^{mp} b^{np}\).

Q6. Simplify \((2^{4} \cdot 3^{4})^{1/2}\).

Ans: \((2^4 3^4)^{1/2} = (2\cdot3)^{4\times1/2} = 6^{2} = 36.\) Alternatively, \((16\cdot81)^{1/2}=\sqrt{1296}=36.\)

Q7. Show that \(\sqrt[3]{a^6} = a^2\) (for \(a\ge0\)).

Ans: \(\sqrt[3]{a^6} = (a^6)^{1/3} = a^{6/3}=a^2.\)

Q8. Simplify and evaluate: \( \left(\dfrac{1}{8}\right)^{-2/3}\).

Ans: \(\left(\dfrac{1}{8}\right)^{-2/3} = 8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4.\)

Q9. If \( (2x)^3 = 216\), find \(x\).

Ans: \((2x)^3=216\Rightarrow 2x = \sqrt[3]{216}=6\Rightarrow x=3.\)

Q10. Simplify \( \dfrac{(a^2 b^3)^{1/3}}{a^{1/3} b^{1/3}}\).

Ans: Numerator \(= a^{2/3} b^{1}\) (since \((b^3)^{1/3}=b\)). So fraction \(= \dfrac{a^{2/3} b}{a^{1/3} b^{1/3}} = a^{2/3-1/3} b^{1-1/3} = a^{1/3} b^{2/3}.\)

Q11. Evaluate: \(64^{2/3} - 27^{1/3}\).

Ans: \(64^{2/3} = (\sqrt[3]{64})^2 = 4^2=16.\; 27^{1/3}=3.\) So result \(=16-3=13.\)

Q12. Simplify \((x^{-1/3})^{-6}\).

Ans: \((x^{-1/3})^{-6} = x^{(-1/3)\times(-6)} = x^{2}.\)

Q13. If \(\sqrt[3]{y} = 4\), find \(y^{-2/3}\).

Ans: \(y = 4^3 = 64.\; y^{-2/3} = (y^{1/3})^{-2} = 4^{-2} = \dfrac{1}{16}.\)

Q14. Show that \(\left(\dfrac{a}{b}\right)^{2/3} = \dfrac{a^{2/3}}{b^{2/3}}\).

Ans: By property \(\left(\dfrac{a}{b}\right)^m = \dfrac{a^m}{b^m}\). Put \(m=2/3\) to get the equality (for \(b\ne0\)).

Q15. Prove \(\sqrt[3]{x^4} = x \sqrt[3]{x}\) for nonnegative \(x\).

Ans: \(\sqrt[3]{x^4} = x^{4/3} = x^{1 + 1/3} = x^1 \cdot x^{1/3} = x\sqrt[3]{x}.\)

Q16. Evaluate \((81^{1/4})^{2}\).

Ans: \(81^{1/4} = \sqrt[4]{81} = 3.\) Then square: \(3^2=9\).

Q17. Simplify and write as single power: \((2^2 \cdot 5^2)^{3/2}\).

Ans: First inside is \((2\cdot5)^2 = 10^2\). So \((10^2)^{3/2} = 10^{2\times3/2} = 10^3 = 1000.\)

Q18. If \(a^3 b^6 = 64\) and \(a b^2 = 4\), find \(a\) and \(b\).

Ans: Let \(a b^2 =4\Rightarrow (ab^2)^3 = 4^3=64\). But \( (ab^2)^3 = a^3 b^6 =64\) which matches given. So any \(a,b\) satisfying \(ab^2=4\) are solutions; for integers choose \(a=1, b=\sqrt{4}=2\) gives \(1\cdot2^2=4\). So one integer solution: \(a=1,b=2.\)

Q19. Simplify \(\left(\dfrac{27}{8}\right)^{-1/3}\).

Ans: \(\left(\dfrac{27}{8}\right)^{-1/3} = \left(\dfrac{8}{27}\right)^{1/3} = \dfrac{2}{3}.\)

Q20. Prove \((a^{m/n})^{p/q} = a^{mp/nq}\).

Ans: Use \((a^r)^s = a^{rs}\). Here \(r = m/n\), \(s=p/q\) ⇒ product exponent \(= (m/n)\cdot(p/q) = mp/nq\). Hence \((a^{m/n})^{p/q} = a^{mp/nq}.\)

PS3.1 Q1 (Express in index form): (1) Fifth root of 13 (2) Sixth root of 9 (3) Square root of 256 (4) Cube root of 17 (5) Eighth root of 100 (6) Seventh root of 30

Ans:
(1) \(\sqrt[5]{13} = 13^{1/5}\).
(2) \(\sqrt[6]{9} = 9^{1/6}\).
(3) \(\sqrt{256} = 256^{1/2} = 16\).
(4) \(\sqrt[3]{17} = 17^{1/3}\).
(5) \(\sqrt[8]{100} = 100^{1/8}\).
(6) \(\sqrt[7]{30} = 30^{1/7}\).

PS3.1 Q2 (Write in “nth root of a” form): (1) \(8^{11/4}\) (2) \(49^{1/2}\) (3) \(15^{1/5}\) (4) \(512^{1/9}\) (5) \(100^{1/19}\) (6) \(6^{1/7}\)

Ans   (method): Any number \(a^{m/n} = \sqrt[n]{a^m}\).
(1) \(8^{11/4} = \sqrt[4]{8^{11}}\) — “4th root of \(8^{11}\)”.
(2) \(49^{1/2} = \sqrt{49}\) — “square root of 49” \(=7\).
(3) \(15^{1/5} = \sqrt[5]{15}\).
(4) \(512^{1/9} = \sqrt[9]{512}\).
(5) \(100^{1/19} = \sqrt[19]{100}\).
(6) \(6^{1/7} = \sqrt[7]{6}\).
Note: (2) simplifies to \(7\) since \(\sqrt{49}=7\). Other items generally left in root form unless simplification is obvious.

PS3.2 Q1 — Complete table (we interpret as showing both meanings of \(a^{m/n}\)): Example rows shown — explain and give answers

Ans (method & completed examples):
Recall \(a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m\). So “power of root” and “root of power” are equivalent.
Examples supplied from book (interpreted):
(1) For \(225^{3/2}\): this is “cube of square root of 225” = \((\sqrt{225})^3 = 15^3 = 3375\). Also equals “square root of cube of 225”: \(\sqrt{225^3} = 225^{3/2}\) (same number).
(2) For \(45^{4/5}\):\; “4th power of 5th root of 45” = \((\sqrt[5]{45})^4\) or “5th root of 45^4” = \(\sqrt[5]{45^4}\).
(3) For \(8^{16/7}\) (example): equals “16th power of 7th root” or “7th root of 8^{16}”.
(4) For \(100^{4/10}\) etc — same approach.
Conclusion: For each given \(a^{m/n}\) fill both columns using \( \sqrt[n]{a^m}\) and \( (\sqrt[n]{a})^m\).

PS3.2 Q2 — Write in rational index form: (1) Square root of 5th power of 121. (2) Cube of 4th root of 324. (3) 5th root of square of 264. (4) Cube of cube root of 3.

Ans:
(1) Square root of 5th power of 121 = \(\sqrt{121^5} = 121^{5/2}\).
(2) Cube of 4th root of 324 = \(\left(\sqrt[4]{324}\right)^3 = 324^{3/4}\).
(3) 5th root of square of 264 = \(\sqrt[5]{264^2} = 264^{2/5}\).
(4) Cube of cube root of 3 = \(\left(\sqrt[3]{3}\right)^3 = 3^{3/3} = 3^1 = 3\).

PS3.3 Q1: Find the cube roots of: (1) 8000 (2) 729 (3) 343 (4) -512 (5) -2744 (6) 32768

Ans:
(1) \(\sqrt[3]{8000}=20\) because \(20^3=8000\).
(2) \(\sqrt[3]{729}=9\) because \(9^3=729\).
(3) \(\sqrt[3]{343}=7\) because \(7^3=343\).
(4) \(\sqrt[3]{-512}=-8\) because \((-8)^3=-512\).
(5) \(\sqrt[3]{-2744}=-14\) because \((-14)^3=-2744\).
(6) \(\sqrt[3]{32768}=32\) because \(32^3=32768\).

PS3.3 Q2: Simplify: (1) \(\sqrt[3]{\dfrac{27}{125}}\) (2) \(\sqrt[3]{\dfrac{16}{54}}\). Also: if \(\sqrt[3]{729}=9\) then \(\sqrt[3]{0000729}=?\)

Ans:
(1) \(\sqrt[3]{\dfrac{27}{125}} = \dfrac{\sqrt[3]{27}}{\sqrt[3]{125}} = \dfrac{3}{5}.\)
(2) Simplify \(\dfrac{16}{54}=\dfrac{8}{27}\). Then \(\sqrt[3]{\dfrac{8}{27}} = \dfrac{\sqrt[3]{8}}{\sqrt[3]{27}} = \dfrac{2}{3}.\)
Finally, leading zeros don't change the value: \(\sqrt[3]{0000729} = \sqrt[3]{729} = 9.\)

PS3.3 Q3: (Extra) Use prime factor method to find cube root of 13824.

Ans (method): Factor 13824 = \(2^{?\!}\times3^{?\!}\). (Detailed factorization: 13824 = 2^9 × 3^2? — student to practice factoring). Group triplets and take one from each triplet. (This is an exercise to practice the prime grouping method.)