Number Systems — NCERT Solutions | Class 9 Mathematics | Academia Aeternum
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Class 9 Mathematics Exercise-1.5 NCERT Solutions Olympiad Board Exam
Chapter 1

Number Systems

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

3 Questions
5–10 min Ideal time
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Q1
NUMERIC2 marks

Find the following:

  1. \(\displaystyle 64^{\frac{1}{2}}\)
  2. \(\displaystyle 32^{\frac{1}{5}}\)
  3. \(\displaystyle 125^{\frac{1}{3}}\)
📘 Concept & Theory Theory Used ›

Fractional exponents are closely related to roots.

The exponent \(\frac{1}{n}\) represents the nth root of a number.

\[ a^{\frac{1}{n}}=\sqrt[n]{a} \]

We also use the important law of exponents:

\[ \left(a^m\right)^n=a^{mn} \]

To solve such questions:

  • First express the number as powers of prime numbers.
  • Rewrite the expression in exponential form.
  • Apply exponent laws carefully.
  • Simplify step by step.
📝 solution Solution (i) ›
Find \(64^{\frac{1}{2}}\)
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Express 64 as a perfect square.
  2. Apply exponent law.
  3. Simplify step by step.
✏️ Solution Complete Solution ›

We know:

\[ 64=8\times 8 \]

Therefore,

\[ \begin{aligned} 64^{\frac{1}{2}} &=\left(8\times 8\right)^{\frac{1}{2}} \\ &=\left(8^2\right)^{\frac{1}{2}} \end{aligned} \]

Using the law:

\[ \left(a^m\right)^n=a^{mn} \]

Applying the law:

\[ \begin{aligned} \left(8^2\right)^{\frac{1}{2}} &=8^{2\times \frac{1}{2}} \\ &=8^{\frac{2}{2}} \\ &=8^1 \\ &=8 \end{aligned} \]

Final Answer: \[ 64^{\frac{1}{2}}=8 \]
📝 solution Solution (ii) ›
Find \(32^{\frac{1}{5}}\)
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Express 32 as a power of 2.
  2. Apply exponent law.
  3. Simplify carefully.
✏️ Solution Complete Solution ›

Prime factorisation of \(32\):

\[ 32=2\times 2\times 2\times 2\times 2 \]

Therefore,

\[ \begin{aligned} 32^{\frac{1}{5}} &=\left(2\times2\times2\times2\times2\right)^{\frac{1}{5}} \\ &=\left(2^5\right)^{\frac{1}{5}} \end{aligned} \]

Using:

\[ \left(a^m\right)^n=a^{mn} \]

Applying the formula:

\[ \begin{aligned} \left(2^5\right)^{\frac{1}{5}} &=2^{5\times \frac{1}{5}} \\ &=2^{\frac{5}{5}} \\ &=2^1 \\ &=2 \end{aligned} \]

Final Answer: \[ 32^{\frac{1}{5}}=2 \]
📝 solution Solution (iii) ›
Find \(125^{\frac{1}{3}}\)
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Express 125 as powers of 5.
  2. Rewrite in exponential form.
  3. Apply exponent law step by step.
✏️ Solution Complete Solution ›

Prime factorisation of \(125\):

\[ 125=5\times5\times5 \]

Therefore,

\[ \begin{aligned} 125^{\frac{1}{3}} &=\left(5\times5\times5\right)^{\frac{1}{3}} \\ &=\left(5^3\right)^{\frac{1}{3}} \end{aligned} \]

Using:

\[ \left(a^m\right)^n=a^{mn} \]

Applying the formula:

\[ \begin{aligned} \left(5^3\right)^{\frac{1}{3}} &=5^{3\times\frac{1}{3}} \\ &=5^{\frac{3}{3}} \\ &=5^1 \\ &=5 \end{aligned} \]

Final Answer: \[ 125^{\frac{1}{3}}=5 \]
🎯 Exam Significance Exam Significance ›
  • Fractional exponents are foundational concepts for higher algebra.
  • These concepts are frequently asked in school examinations and board exams.
  • Competitive examinations like Olympiads, NTSE, JEE Foundation, and other aptitude tests often ask simplification problems based on exponent laws.
  • Mastery of exponent rules improves speed and accuracy in mathematics.
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Q2 →
Q2
NUMERIC3 marks
Question text...
📘 Concept & Theory Theory Revision ›

Fractional exponents represent roots and powers together.

Important formulas used in these questions:

\[ a^{\frac{m}{n}}=\left(\sqrt[n]{a}\right)^m \]

\[ \left(a^m\right)^n=a^{mn} \]

\[ a^{-m}=\frac{1}{a^m} \]

Strategy for solving:

  • Express the number as powers of prime factors.
  • Rewrite in exponential form.
  • Apply exponent laws carefully.
  • Simplify step by step without skipping steps.
📝 solution Solution (i) ›
Find \(9^{\frac{3}{2}}\)
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Express 9 as a power of 3.
  2. Apply the exponent law.
  3. Simplify step by step.
✏️ Solution Complete Solution ›

We know:

\[ 9=3\times3 \]

Therefore,

\[ \begin{aligned} 9^{\frac{3}{2}} &=\left(3\times3\right)^{\frac{3}{2}} \\ &=\left(3^2\right)^{\frac{3}{2}} \end{aligned} \]

Using:

\[ \left(a^m\right)^n=a^{mn} \]

Applying the formula:

\[ \begin{aligned} \left(3^2\right)^{\frac{3}{2}} &=3^{2\times\frac{3}{2}} \\ &=3^{\frac{2\times3}{2}} \\ &=3^{\frac{\cancel{2}\times3}{\cancel{2}}} \\ &=3^3 \\ &=3\times3\times3 \\ &=27 \end{aligned} \]

Final Answer: \[ 9^{\frac{3}{2}}=27 \]
📝 solution Solution (ii) ›
Find \(32^{\frac{2}{5}}\)
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Express 32 as powers of 2.
  2. Apply exponent law carefully.
  3. Simplify completely.
✏️ Solution Complete Solution ›

Prime factorisation of \(32\):

\[ 32=2\times2\times2\times2\times2 \]

Thus,

\[ \begin{aligned} 32^{\frac{2}{5}} &=\left(2\times2\times2\times2\times2\right)^{\frac{2}{5}} \\ &=\left(2^5\right)^{\frac{2}{5}} \end{aligned} \]

Using:

\[ \left(a^m\right)^n=a^{mn} \]

Applying the formula:

\[ \begin{aligned} \left(2^5\right)^{\frac{2}{5}} &=2^{5\times\frac{2}{5}} \\ &=2^{\frac{5\times2}{5}} \\ &=2^{\frac{\cancel{5}\times2}{\cancel{5}}} \\ &=2^2 \\ &=2\times2 \\ &=4 \end{aligned} \]

Final Answer: \[ 32^{\frac{2}{5}}=4 \]
📝 solution Solution (iii) ›
Find \(16^{\frac{3}{4}}\)
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Express 16 as powers of 2.
  2. Apply exponent law.
  3. Simplify step by step.
✏️ Solution Complete Solution ›

Prime factorisation of \(16\):

\[ 16=2\times2\times2\times2 \]

Therefore,

\[ \begin{aligned} 16^{\frac{3}{4}} &=\left(2\times2\times2\times2\right)^{\frac{3}{4}} \\ &=\left(2^4\right)^{\frac{3}{4}} \end{aligned} \]

Using:

\[ \left(a^m\right)^n=a^{mn} \]

Applying the formula:

\[ \begin{aligned} \left(2^4\right)^{\frac{3}{4}} &=2^{4\times\frac{3}{4}} \\ &=2^{\frac{4\times3}{4}} \\ &=2^{\frac{\cancel{4}\times3}{\cancel{4}}} \\ &=2^3 \\ &=2\times2\times2 \\ &=8 \end{aligned} \]

Final Answer: \[ 16^{\frac{3}{4}}=8 \]
📝 solution Solution (iv) ›
Find \(125^{\frac{-1}{3}}\)
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Convert negative exponent into reciprocal form.
  2. Express 125 as powers of 5.
  3. Apply exponent law carefully.
✏️ Solution Complete Solution ›

Using:

\[ a^{-m}=\frac{1}{a^m} \]

Therefore,

\[ \begin{aligned} 125^{\frac{-1}{3}} &=\frac{1}{125^{\frac{1}{3}}} \end{aligned} \]

Prime factorisation of \(125\):

\[ 125=5\times5\times5 \]

Hence,

\[ \begin{aligned} \frac{1}{125^{\frac{1}{3}}} &=\frac{1}{\left(5\times5\times5\right)^{\frac{1}{3}}} \\ &=\frac{1}{\left(5^3\right)^{\frac{1}{3}}} \end{aligned} \]

Using:

\[ \left(a^m\right)^n=a^{mn} \]

Applying the formula:

\[ \begin{aligned} \frac{1}{\left(5^3\right)^{\frac{1}{3}}} &=\frac{1}{5^{3\times\frac{1}{3}}} \\ &=\frac{1}{5^{\frac{3}{3}}} \\ &=\frac{1}{5^{\frac{\cancel{3}}{\cancel{3}}}} \\ &=\frac{1}{5^1} \\ &=\frac{1}{5} \end{aligned} \]

Final Answer: \[ 125^{\frac{-1}{3}}=\frac{1}{5} \]
🎯 Exam Significance Exam Significance ›
  • Exponent laws are very important for CBSE board examinations.
  • Questions based on rational exponents are frequently asked in Olympiads and scholarship examinations.
  • These concepts are also used in algebra, logarithms, surds, and higher mathematics.
  • Strong understanding improves speed in competitive aptitude examinations.
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Q3 →
Q3
NUMERIC3 marks
Simplify the Following
📘 Concept & Theory Important Laws of Exponents ›

The following exponent laws are used frequently in simplification:

\[ a^m\cdot a^n=a^{m+n} \]

\[ \frac{a^m}{a^n}=a^{m-n} \]

\[ \left(a^m\right)^n=a^{mn} \]

\[ \left(\frac{a}{b}\right)^m=\frac{a^m}{b^m} \]

\[ a^m\cdot b^m=(ab)^m \]

📝 solution Solution (i) ›
Simplify \[ 2^{\frac{2}{3}}\cdot2^{\frac{1}{5}} \]
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Observe that both terms have the same base.
  2. Use the product law of exponents.
  3. Add the powers carefully.
✏️ Solution Complete Solution ›

Since the bases are same, we use:

\[ a^m\cdot a^n=a^{m+n} \]

Therefore,

\[ \begin{aligned} 2^{\frac{2}{3}}\cdot2^{\frac{1}{5}} &=2^{\frac{2}{3}+\frac{1}{5}} \end{aligned} \]

Taking LCM of \(3\) and \(5\):

\[ \text{LCM}(3,5)=15 \]

Converting fractions:

\[ \begin{aligned} \frac{2}{3} &=\frac{2\times5}{3\times5} \\ &=\frac{10}{15} \end{aligned} \]

\[ \begin{aligned} \frac{1}{5} &=\frac{1\times3}{5\times3} \\ &=\frac{3}{15} \end{aligned} \]

Adding the exponents:

\[ \begin{aligned} 2^{\frac{10}{15}+\frac{3}{15}} &=2^{\frac{13}{15}} \end{aligned} \]

Final Answer: \[ 2^{\frac{2}{3}}\cdot2^{\frac{1}{5}} =2^{\frac{13}{15}} \]
📝 solution Solution (ii) ›
Simplify \[ \left(\frac{1}{3^3}\right)^7 \]
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Apply power to numerator and denominator separately.
  2. Use exponent multiplication law.
  3. Rewrite in simplified form.
✏️ Solution Complete Solution ›

Using:

\[ \left(\frac{a}{b}\right)^m=\frac{a^m}{b^m} \]

Therefore,

\[ \begin{aligned} \left(\frac{1}{3^3}\right)^7 &=\frac{1^7}{\left(3^3\right)^7} \end{aligned} \]

Since:

\[ 1^7=1 \]

We get:

\[ \begin{aligned} \frac{1}{\left(3^3\right)^7} \end{aligned} \]

Using:

\[ \left(a^m\right)^n=a^{mn} \]

Therefore,

\[ \begin{aligned} \frac{1}{\left(3^3\right)^7} &=\frac{1}{3^{3\times7}} \\ &=\frac{1}{3^{21}} \end{aligned} \]

Using negative exponent form:

\[ \frac{1}{a^m}=a^{-m} \]

Hence,

\[ \begin{aligned} \frac{1}{3^{21}} &=3^{-21} \end{aligned} \]

Final Answer: \[ \left(\frac{1}{3^3}\right)^7=3^{-21} \]
📝 solution Solution (iii) ›
Simplify \[ \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} \]
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Observe same bases.
  2. Use division law of exponents.
  3. Subtract the powers carefully.
✏️ Solution Complete Solution ›

Using:

\[ \frac{a^m}{a^n}=a^{m-n} \]

Therefore,

\[ \begin{aligned} \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} &=11^{\frac{1}{2}-\frac{1}{4}} \end{aligned} \]

Taking LCM of \(2\) and \(4\):

\[ \text{LCM}(2,4)=4 \]

Converting fractions:

\[ \frac{1}{2}=\frac{2}{4} \]

Thus,

\[ \begin{aligned} 11^{\frac{2}{4}-\frac{1}{4}} &=11^{\frac{1}{4}} \end{aligned} \]

Final Answer: \[ \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} =11^{\frac{1}{4}} \]
📝 solution Solution (iv) ›
Simplify \[7^{\frac{1}{2}}\cdot8^{\frac{1}{2}}\]
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Use same exponent property.
  2. Combine the numbers inside one exponent.
  3. Simplify the product.
✏️ Solution Complete Solution ›

Using:

\[ a^m\cdot b^m=(ab)^m \]

Therefore,

\[ \begin{aligned} 7^{\frac{1}{2}}\cdot8^{\frac{1}{2}} &=(7\times8)^{\frac{1}{2}} \\ &=56^{\frac{1}{2}} \end{aligned} \]

Prime factorisation of \(56\):

\[ 56=2\times2\times2\times7 \]

Therefore,

\[ \begin{aligned} 56^{\frac{1}{2}} &=(2\times2\times2\times7)^{\frac{1}{2}} \\ &=(2^2\times14)^{\frac{1}{2}} \\ &=\left(2^2\right)^{\frac{1}{2}}\times14^{\frac{1}{2}} \end{aligned} \]

Using:

\[ \left(a^m\right)^n=a^{mn} \]

We get:

\[ \begin{aligned} \left(2^2\right)^{\frac{1}{2}} &=2^{2\times\frac{1}{2}} \\ &=2^{\frac{\cancel{2}}{\cancel{2}}} \\ &=2^1 \\ &=2 \end{aligned} \]

Hence,

\[ \begin{aligned} 56^{\frac{1}{2}} &=2\times14^{\frac{1}{2}} \\ &=2\sqrt{14} \end{aligned} \]

Final Answer: \[ 7^{\frac{1}{2}}\cdot8^{\frac{1}{2}} =2\sqrt{14} \]
🎯 Exam Significance Exam Significance ›
  • Exponent laws are among the most frequently used algebraic identities in school mathematics.
  • These simplifications are important for board examinations because they test conceptual clarity and calculation skills.
  • Competitive examinations often include questions involving rational exponents and surds.
  • Strong command over exponent rules increases calculation speed in higher mathematics.
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