GSEB Class 7 Maths Solutions Chapter 13 Exponents and Powers Exercise 13.2

Question 1. Using laws of exponents, simplify and write the answer in exponential form:
(i) \( 3^2 \times 3^4 \times 3^8 \)
(ii) \( 6^{15} \div 6^{10} \)
(iii) \( a^3 \times a^2 \)
(iv) \( 7^x \times 7^2 \)
(v) \( (5^2)^3 \div 5^3 \)
(vi) \( 2^5 \times 5^5 \)
(vii) \( a^4 \times b^4 \)
(viii) \( (3^4)^3 \)
(ix) \( (2^{20} \div 2^{15}) \times 2^3 \)
(x) \( 8^t \div 8^2 \)
Answer:
(i) We have: \( 3^2 \times 3^4 \times 3^8 = 3^{2+4+8} \) (Because \( a^m \times a^n \times a^r = a^{m+n+r} \)). Thus, \( 3^2 \times 3^4 \times 3^8 = 3^{14} \).
(ii) We have: \( 6^{15} \div 6^{10} = 6^{15-10} = 6^5 \) (Because \( a^m \div a^n = a^{m-n} \)). Thus, \( 6^{15} \div 6^{10} = 6^5 \).
(iii) We have: \( a^3 \times a^2 = a^{3+2} = a^5 \) (Because \( a^m \times a^n = a^{m+n} \)). Thus, \( a^3 \times a^2 = a^5 \).
(iv) We have: \( 7^x \times 7^2 = 7^{x+2} \) (Because \( a^m \times a^n = a^{m+n} \)). Thus, \( 7^x \times 7^2 = 7^{x+2} \).
(v) We have: \( (5^2)^3 \div 5^3 = 5^{2 \times 3} \div 5^3 \) (Because \( (a^m)^n = a^{mn} \)).
\( \implies 5^6 \div 5^3 \)
\( \implies 5^{6-3} = 5^3 \) (Because \( a^m \div a^n = a^{m-n} \)). Thus, \( (5^2)^3 \div 5^3 = 5^3 \).
(vi) We have: \( 2^5 \times 5^5 = (2 \times 5)^5 = (10)^5 \) (Because \( a^m \times b^m = (ab)^m \)). Thus, \( 2^5 \times 5^5 = 10^5 \).
(vii) We have: \( a^4 \times b^4 = (ab)^4 \) (Because \( a^m \times b^m = (ab)^m \)). Thus, \( a^4 \times b^4 = (ab)^4 \).
(viii) We have: \( (3^4)^3 = 3^{4 \times 3} = 3^{12} \) (Because \( (a^m)^n = a^{mn} \)). Thus, \( (3^4)^3 = 3^{12} \).
(ix) We have: \( (2^{20} \div 2^{15}) \times 2^3 \) (Because \( a^m \div a^n = a^{m-n} \)).
\( \implies (2^{20-15}) \times 2^3 \)
\( \implies 2^5 \times 2^3 \)
\( \implies 2^{5+3} \) (Because \( a^m \times a^n = a^{m+n} \)).
\( \implies 2^8 \). Thus, \( (2^{20} \div 2^{15}) \times 2^3 = 2^8 \).
(x) We have: \( 8^t \div 8^2 = 8^{t-2} \) (Because \( a^m \div a^n = a^{m-n} \)). Thus, \( 8^t \div 8^2 = 8^{t-2} \).
In simple words: When multiplying numbers with the same base, you just add their exponents. When dividing, you subtract them. If you raise a power to another power, you multiply the exponents. Also, if different bases have the same exponent, you can multiply the bases first, then apply the exponent.

Exam Tip: Always remember the basic laws of exponents. They simplify complex calculations significantly. Pay close attention to whether the bases are the same for addition/subtraction of exponents, or if exponents are the same for combining bases.

Question 2. Simplify and express each of the following in exponential form:
(i) \( \frac{2^{3} \times 3^{4} \times 4}{3 \times 32} \)
(ii) \( [(5^2)^3 \times 5^4] \div 5^7 \)
(iii) \( 25^4 \div 5^3 \)
(iv) \( \frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}} \)
(v) \( \frac{3^{7}}{3^{4} \times 3^{3}} \)
(vi) \( 2^0 + 3^0 + 4^0 \)
(vii) \( 2^0 \times 3^0 \times 4^0 \)
(viii) \( (3^0 + 2^0) \times 5^0 \)
(ix) \( \frac{2^{8} \times a^{5}}{4^{3} \times a^{3}} \)
(x) \( \left[\frac{a^{8}}{a^{3}}\right] \times a^{8} \)
(xi) \( \frac{4^{5} \times a^{8} b^{3}}{4^{5} \times a^{5} b^{2}} \)
(xii) \( (2^3 \times 2)^2 \)
Answer:
(i) First, rewrite numbers as powers: \( 32 = 2^5 \) and \( 4 = 2^2 \).
Now, \( \frac{2^{3} \times 3^{4} \times 4}{3 \times 32} = \frac{2^{3} \times 3^{4} \times 2^{2}}{3^1 \times 2^{5}} \)
\( \implies 2^{3+2-5} \times 3^{4-1} \) (Applying the laws \( a^m \times a^n = a^{m+n} \) and \( a^m \div a^n = a^{m-n} \)).
\( \implies 2^0 \times 3^3 \)
\( \implies 1 \times 3^3 = 3^3 \) (Since any non-zero number raised to the power of 0 is 1).
(ii) First, use the power rule: \( (5^2)^3 = 5^{2 \times 3} = 5^6 \) (Because \( (a^m)^n = a^{mn} \)).
Then, \( [(5^2)^3 \times 5^4] \div 5^7 = [5^6 \times 5^4] \div 5^7 \)
\( \implies 5^{6+4} \div 5^7 \) (Because \( a^m \times a^n = a^{m+n} \)).
\( \implies 5^{10} \div 5^7 \)
\( \implies 5^{10-7} \) (Because \( a^m \div a^n = a^{m-n} \)).
\( \implies 5^3 \).
(iii) First, express the base 25 as a power of 5: \( 25 = 5^2 \).
Then, \( 25^4 \div 5^3 = (5^2)^4 \div 5^3 \)
\( \implies 5^{2 \times 4} \div 5^3 \) (Because \( (a^m)^n = a^{mn} \)).
\( \implies 5^8 \div 5^3 \)
\( \implies 5^{8-3} \) (Because \( a^m \div a^n = a^{m-n} \)).
\( \implies 5^5 \).
(iv) First, rewrite the number 21 as a product of prime factors: \( 21 = 3 \times 7 \).
Then, \( \frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}} = \frac{3^1 \times 7^2 \times 11^{8}}{3^1 \times 7^1 \times 11^{3}} \)
\( \implies 3^{1-1} \times 7^{2-1} \times 11^{8-3} \) (Applying the law \( a^m \div a^n = a^{m-n} \)).
\( \implies 3^0 \times 7^1 \times 11^5 \)
\( \implies 1 \times 7 \times 11^5 = 7 \times 11^5 \) (Since \( a^0 = 1 \)).
(v) First, combine the terms in the denominator: \( 3^4 \times 3^3 = 3^{4+3} = 3^7 \) (Because \( a^m \times a^n = a^{m+n} \)).
Then, \( \frac{3^{7}}{3^{4} \times 3^{3}} = \frac{3^{7}}{3^{7}} \)
\( \implies 3^{7-7} \) (Because \( a^m \div a^n = a^{m-n} \)).
\( \implies 3^0 = 1 \) (Since \( a^0 = 1 \)).
(vi) Any non-zero number raised to the power of 0 is 1. So, \( 2^0 = 1 \), \( 3^0 = 1 \), and \( 4^0 = 1 \).
Then, \( 2^0 + 3^0 + 4^0 = 1 + 1 + 1 = 3 \).
(vii) Any non-zero number raised to the power of 0 is 1. So, \( 2^0 = 1 \), \( 3^0 = 1 \), and \( 4^0 = 1 \).
Then, \( 2^0 \times 3^0 \times 4^0 = 1 \times 1 \times 1 = 1 \).
(viii) Any non-zero number raised to the power of 0 is 1. So, \( 3^0 = 1 \), \( 2^0 = 1 \), and \( 5^0 = 1 \).
Then, \( (3^0 + 2^0) \times 5^0 = (1 + 1) \times 1 \)
\( \implies 2 \times 1 = 2 \).
(ix) First, express 4 as a power of 2: \( 4 = 2^2 \). So, \( 4^3 = (2^2)^3 = 2^{2 \times 3} = 2^6 \) (Because \( (a^m)^n = a^{mn} \)).
Then, \( \frac{2^{8} \times a^{5}}{4^{3} \times a^{3}} = \frac{2^{8} \times a^{5}}{2^{6} \times a^{3}} \)
\( \implies 2^{8-6} \times a^{5-3} \) (Because \( a^m \div a^n = a^{m-n} \)).
\( \implies 2^2 \times a^2 \)
\( \implies (2a)^2 \) (Because \( a^m \times b^m = (ab)^m \)).
(x) First, simplify the division within the brackets: \( \left[\frac{a^{8}}{a^{3}}\right] = a^{8-3} = a^5 \) (Because \( a^m \div a^n = a^{m-n} \)).
Then, \( \left[\frac{a^{8}}{a^{3}}\right] \times a^{8} = a^5 \times a^8 \)
\( \implies a^{5+8} \) (Because \( a^m \times a^n = a^{m+n} \)).
\( \implies a^{13} \).
(xi) \( \frac{4^{5} \times a^{8} b^{3}}{4^{5} \times a^{5} b^{2}} \)
\( \implies 4^{5-5} \times a^{8-5} \times b^{3-2} \) (Because \( a^m \div a^n = a^{m-n} \)).
\( \implies 4^0 \times a^3 \times b^1 \)
\( \implies 1 \times a^3 \times b = a^3b \) (Since \( a^0 = 1 \)).
(xii) First, simplify the multiplication inside the parenthesis: \( 2^3 \times 2 = 2^3 \times 2^1 = 2^{3+1} = 2^4 \) (Because \( a^m \times a^n = a^{m+n} \)).
Then, \( (2^3 \times 2)^2 = (2^4)^2 \)
\( \implies 2^{4 \times 2} \) (Because \( (a^m)^n = a^{mn} \)).
\( \implies 2^8 \).
In simple words: To simplify expressions with exponents, always remember the core rules: add exponents when multiplying with the same base, subtract when dividing, and multiply when raising a power to another power. Also, any number (except zero) raised to the power of zero is always 1.

Exam Tip: Break down complex expressions into simpler steps, applying one exponent law at a time. It is often helpful to convert all numbers to their prime factorization first to make simplification easier.

Question 3. Say true or false and justify your answer:
(i) \( 10 \times 10^{11} = 100^{11} \)
(ii) \( 2^3 > 5^2 \)
(iii) \( 2^3 \times 3^2 = 6^5 \)
(iv) \( 3^0 = (1000)^0 \)
Answer:
(i) We calculate \( 10 \times 10^{11} = 10^{1+11} = 10^{12} \). The right side is \( 100^{11} = (10^2)^{11} = 10^{2 \times 11} = 10^{22} \). Since \( 10^{12} \ne 10^{22} \), the statement is **False**.
(ii) We calculate \( 2^3 = 2 \times 2 \times 2 = 8 \). We also calculate \( 5^2 = 5 \times 5 = 25 \). Since \( 8 < 25 \), it means \( 2^3 < 5^2 \). Therefore, the statement \( 2^3 > 5^2 \) is **False**.
(iii) We calculate \( 2^3 \times 3^2 = (2 \times 2 \times 2) \times (3 \times 3) = 8 \times 9 = 72 \). We also calculate \( 6^5 = 6 \times 6 \times 6 \times 6 \times 6 = 7776 \). Since \( 72 \ne 7776 \), the statement \( 2^3 \times 3^2 = 6^5 \) is **False**.
(iv) We know that any non-zero number raised to the power of 0 is 1. So, \( 3^0 = 1 \) and \( (1000)^0 = 1 \). Since \( 1 = 1 \), the statement \( 3^0 = (1000)^0 \) is **True**.
In simple words: To check if a statement is true or false, you need to work out both sides of the equation or inequality using the rules of exponents. Then, compare the results. Remember that any number (except 0) raised to the power of zero is always one.

Exam Tip: For true/false questions, always show your full calculations for both sides of the expression to justify your answer. A simple 'true' or 'false' without justification will not fetch full marks.

Question 4. Express the following as a product of prime factors only in exponential form:
(i) \( 108 \times 192 \)
(ii) \( 270 \)
(iii) \( 729 \times 64 \)
(iv) \( 768 \)
Answer:
(i) First, find the prime factorization of each number:
\( 108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3 \).
\( 192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3^1 \).
Now, multiply these exponential forms: \( 108 \times 192 = (2^2 \times 3^3) \times (2^6 \times 3^1) \).
\( \implies 2^{2+6} \times 3^{3+1} \) (Adding exponents for the same bases).
\( \implies 2^8 \times 3^4 \).
(ii) Find the prime factorization of 270:
\( 270 = 2 \times 3 \times 3 \times 3 \times 5 = 2^1 \times 3^3 \times 5^1 \).
\( \implies 2 \times 3^3 \times 5 \).
(iii) First, find the prime factorization of each number:
\( 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6 \).
\( 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6 \).
Now, multiply these exponential forms: \( 729 \times 64 = 3^6 \times 2^6 \).
(iv) Find the prime factorization of 768:
\( 768 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^8 \times 3^1 \).
\( \implies 2^8 \times 3 \).
In simple words: To write a number as a product of prime factors in exponential form, break it down into its smallest prime numbers. Count how many times each prime number appears and write that count as its power. If you have two numbers to multiply, find the prime factors for each separately, then combine and add the powers of matching prime factors.

Exam Tip: Use a division ladder or factor tree to accurately find the prime factors. Ensure all numbers are broken down to their prime components before writing them in exponential form.

Question 5. Simplify:
(i) \( \frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 7} \)
(ii) \( \frac{25 \times 5^{2} \times t^{8}}{10^{3} \times t^{4}} \)
(iii) \( \frac{3^{5} \times 10^{5} \times 25}{5^{7} \times 6^{5}} \)
Answer:
(i) First, rewrite bases as powers of prime numbers: \( (2^5)^2 = 2^{10} \) (Power of a power rule) and \( 8^3 = (2^3)^3 = 2^9 \).
Now, substitute these into the expression: \( \frac{2^{10} \times 7^{3}}{2^{9} \times 7^{1}} \)
\( \implies 2^{10-9} \times 7^{3-1} \) (Applying the division rule for exponents).
\( \implies 2^1 \times 7^2 \)
\( \implies 2 \times 49 = 98 \).
(ii) First, rewrite bases as powers of prime numbers: \( 25 = 5^2 \) and \( 10^3 = (2 \times 5)^3 = 2^3 \times 5^3 \).
Now, substitute these into the expression: \( \frac{5^2 \times 5^{2} \times t^{8}}{2^3 \times 5^3 \times t^{4}} \)
\( \implies \frac{5^{2+2} \times t^{8}}{2^3 \times 5^3 \times t^{4}} \)
\( \implies \frac{5^4 \times t^{8}}{2^3 \times 5^3 \times t^{4}} \)
\( \implies \frac{5^{4-3} \times t^{8-4}}{2^3} \) (Applying division rule for exponents).
\( \implies \frac{5^1 \times t^4}{2^3} \)
\( \implies \frac{5t^4}{8} \).
(iii) First, rewrite bases as powers of prime numbers:
\( 10^5 = (2 \times 5)^5 = 2^5 \times 5^5 \).
\( 25 = 5^2 \).
\( 6^5 = (2 \times 3)^5 = 2^5 \times 3^5 \).
Now, substitute these into the expression: \( \frac{3^5 \times (2^5 \times 5^5) \times 5^2}{5^7 \times (2^5 \times 3^5)} \)
\( \implies \frac{3^5 \times 2^5 \times 5^{5+2}}{5^7 \times 2^5 \times 3^5} \)
\( \implies \frac{3^5 \times 2^5 \times 5^7}{3^5 \times 2^5 \times 5^7} \)
\( \implies 3^{5-5} \times 2^{5-5} \times 5^{7-7} \) (Applying division rule for exponents).
\( \implies 3^0 \times 2^0 \times 5^0 \)
\( \implies 1 \times 1 \times 1 = 1 \) (Since any non-zero number to the power of 0 is 1).
In simple words: To simplify expressions with exponents, first break down any composite bases into their prime factors. Then, use the rules of exponents to combine terms with the same base: add exponents for multiplication, subtract for division, and multiply for powers of powers. Any term raised to the power of zero becomes one.

Exam Tip: When simplifying expressions, always ensure all bases are prime numbers or single variables before applying exponent rules. This makes the process much clearer and reduces calculation errors.