RBSE Solutions Class 9 Maths Chapter 2 Number System Exercise 2.3

Question 1. Find the value of
(i) \( 81^{\frac{1}{2}} \)
(ii) \( 64^{\frac{1}{6}} \)
(iii) \( 125^{\frac{1}{3}} \)
Answer:
(i) We need to find the value of \( 81^{\frac{1}{2}} \). We know that \( 81 \) is \( 9^2 \).
So, \( 81^{\frac{1}{2}} = (9^2)^{\frac{1}{2}} \)
Using the exponent rule \( (a^m)^n = a^{mn} \), we multiply the powers:
\( = 9^{2 \times \frac{1}{2}} \)
\( = 9^1 \)
\( = 9 \)
(ii) Next, we find the value of \( 64^{\frac{1}{6}} \). We know that \( 64 \) is \( 2^6 \).
So, \( 64^{\frac{1}{6}} = (2^6)^{\frac{1}{6}} \)
Again, using the exponent rule, we multiply the powers:
\( = 2^{6 \times \frac{1}{6}} \)
\( = 2^1 \)
\( = 2 \)
(iii) Finally, we find the value of \( 125^{\frac{1}{3}} \). We know that \( 125 \) is \( 5^3 \).
So, \( 125^{\frac{1}{3}} = (5^3)^{\frac{1}{3}} \)
Multiplying the powers:
\( = 5^{3 \times \frac{1}{3}} \)
\( = 5^1 \)
\( = 5 \)
In simple words: To find the value, rewrite the base number as a power (like 81 as \( 9^2 \)). Then, use the rule that when you have a power raised to another power, you multiply the exponents. This simplifies the number quickly.

🎯 Exam Tip: Always look for the smallest prime base for the given number to simplify calculations effectively.

Question 2. Find the value of
(i) \( 4^{\frac{3}{2}} \)
(ii) \( 32^{\frac{2}{5}} \)
(iii) \( 16^{\frac{3}{4}} \)
Answer:
(i) We need to find the value of \( 4^{\frac{3}{2}} \). We can write \( 4 \) as \( 2^2 \).
So, \( 4^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} \)
Using the exponent rule \( (a^m)^n = a^{mn} \), we multiply the powers:
\( = 2^{2 \times \frac{3}{2}} \)
\( = 2^3 \)
\( = 2 \times 2 \times 2 \)
\( = 8 \)
(ii) Next, we find the value of \( 32^{\frac{2}{5}} \). We know that \( 32 \) is \( 2^5 \).
So, \( 32^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} \)
Again, using the exponent rule, we multiply the powers:
\( = 2^{5 \times \frac{2}{5}} \)
\( = 2^2 \)
\( = 2 \times 2 \)
\( = 4 \)
(iii) Finally, we find the value of \( 16^{\frac{3}{4}} \). We know that \( 16 \) is \( 2^4 \).
So, \( 16^{\frac{3}{4}} = (2^4)^{\frac{3}{4}} \)
Multiplying the powers:
\( = 2^{4 \times \frac{3}{4}} \)
\( = 2^3 \)
\( = 2 \times 2 \times 2 \)
\( = 8 \)
In simple words: First, change the base number into its prime factor form (like 4 to \( 2^2 \)). Then, multiply the exponents together using the power of a power rule. This will simplify the calculation to find the final value easily.

🎯 Exam Tip: Always simplify the base to its smallest possible integer base first (e.g., \( 4 = 2^2 \), \( 32 = 2^5 \), \( 16 = 2^4 \)) before applying fractional exponents, as this makes the multiplication of powers much easier.

Question 3. Simplify each of the following
(i) \( 2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}} \)
(ii) \( \left(\frac{1}{3^3}\right)^7 \)
Answer:
(i) We have the expression \( 2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}} \).
Using the exponent rule \( a^m \cdot a^n = a^{m+n} \), we add the powers since the base is the same:
\( = 2^{\frac{2}{3} + \frac{1}{3}} \)
\( = 2^{\frac{2+1}{3}} \)
\( = 2^{\frac{3}{3}} \)
\( = 2^1 \)
\( = 2 \)
(ii) We have the expression \( \left(\frac{1}{3^3}\right)^7 \).
We can first write \( \frac{1}{3^3} \) as \( 3^{-3} \).
So, \( \left(\frac{1}{3^3}\right)^7 = (3^{-3})^7 \)
Using the exponent rule \( (a^m)^n = a^{mn} \), we multiply the powers:
\( = 3^{-3 \times 7} \)
\( = 3^{-21} \)
Alternatively, we can apply the power to both the numerator and denominator:
\( = \frac{1^7}{(3^3)^7} \)
\( = \frac{1}{3^{3 \times 7}} \)
\( = \frac{1}{3^{21}} \)
Both \( 3^{-21} \) and \( \frac{1}{3^{21}} \) are correct ways to express the answer. The key is to apply the rules of exponents properly.
In simple words: For the first part, when you multiply numbers with the same base, you just add their powers. For the second part, turn the fraction into a negative power, then multiply the powers together. This helps make the expression much simpler.

🎯 Exam Tip: Remember two key exponent rules: \( a^m \cdot a^n = a^{m+n} \) for multiplying powers with the same base, and \( (a^m)^n = a^{mn} \) for a power raised to another power. Also, \( \frac{1}{a^n} = a^{-n} \).

Question 4. Find the value of x in the following
\( \left(\frac{3}{5}\right)^x \left(\frac{5}{3}\right)^{2x} = \frac{125}{27} \)
Answer:
We are given the equation: \( \left(\frac{3}{5}\right)^x \left(\frac{5}{3}\right)^{2x} = \frac{125}{27} \)
Let's convert all terms to have a common base, either \( \frac{3}{5} \) or \( \frac{5}{3} \). We will use \( \frac{5}{3} \).
We know that \( \frac{3}{5} = \left(\frac{5}{3}\right)^{-1} \).
So, \( \left(\frac{3}{5}\right)^x = \left(\left(\frac{5}{3}\right)^{-1}\right)^x = \left(\frac{5}{3}\right)^{-x} \).
For the right side of the equation, we can write \( \frac{125}{27} \) as \( \left(\frac{5}{3}\right)^3 \), because \( 5^3 = 125 \) and \( 3^3 = 27 \).
Now, substitute these into the original equation:
\( \left(\frac{5}{3}\right)^{-x} \left(\frac{5}{3}\right)^{2x} = \left(\frac{5}{3}\right)^3 \)
Using the exponent rule \( a^m \cdot a^n = a^{m+n} \), we add the powers on the left side:
\( \left(\frac{5}{3}\right)^{-x + 2x} = \left(\frac{5}{3}\right)^3 \)
\( \left(\frac{5}{3}\right)^x = \left(\frac{5}{3}\right)^3 \)
Since the bases are equal, their exponents must also be equal:
\( \implies x = 3 \)
The value of x is 3.
In simple words: To solve this, make all the fraction bases the same. Remember that turning a fraction upside down makes its power negative. Once all bases are the same, you can just set the powers equal to each other to find 'x'.

🎯 Exam Tip: When solving equations with exponents and fractions, always try to express all numbers with a common base. Remember that \( \left(\frac{a}{b}\right)^n = \left(\frac{b}{a}\right)^{-n} \).