GSEB Class 8 Maths Solutions Chapter 12 Exponents and Powers Exercise 12.1

Question 1. Evaluate:
(i) \( 3^{-2} \)
(ii) \( (-4)^{-2} \)
(iii) \( (\frac{1}{2})^{-5} \)
Answer:
(i) \( 3^{-2} = \frac{1}{3^2} = \frac{1}{3 \times 3} = \frac{1}{9} \)
(ii) \( (-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{(-4) \times (-4)} = \frac{1}{16} \)
(iii) \( (\frac{1}{2})^{-5} = \frac{1}{(\frac{1}{2})^5} = \frac{1}{\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}} = \frac{1}{\frac{1}{32}} = 32 \)
In simple words: To evaluate these, you need to use the rules of exponents. A negative exponent means you take the reciprocal (flip the fraction) and then make the exponent positive. For example, \( a^{-n} = \frac{1}{a^n} \).

Exam Tip: Remember that a negative base raised to an even exponent will result in a positive answer, while a negative base raised to an odd exponent will result in a negative answer.

Question 2. Simplify and express the result in power notation with positive exponent?
(i) \( (-4)^5 \div (-4)^8 \)
(ii) \( (\frac{1}{2^3})^2 \)
(iii) \( (-3)^4 \times (\frac{5}{3})^4 \)
(iv) \( (3^{-7} \div 3^{-10}) \times 3^{-5} \)
(v) \( 2^{-3} \times (-7)^{-3} \)
Answer:
(i) \( (-4)^5 \div (-4)^8 \)
Using the rule \( a^m \div a^n = a^{m-n} \)
So, \( (-4)^5 \div (-4)^8 = (-4)^{5-8} = (-4)^{-3} \)
To express with a positive exponent, we use \( a^{-n} = \frac{1}{a^n} \)
Thus, \( (-4)^{-3} = \frac{1}{(-4)^3} \)
(ii) \( (\frac{1}{2^3})^2 \)
We know that \( 1 = 1^3 \), so \( \frac{1}{2^3} = (\frac{1}{2})^3 \)
Now, \( (\frac{1}{2^3})^2 = ((\frac{1}{2})^3)^2 \)
Using the rule \( (a^m)^n = a^{mn} \)
\( ((\frac{1}{2})^3)^2 = (\frac{1}{2})^{3 \times 2} = (\frac{1}{2})^6 \)
(iii) \( (-3)^4 \times (\frac{5}{3})^4 \)
Using the rule \( a^m \times b^m = (ab)^m \)
So, \( (-3)^4 \times (\frac{5}{3})^4 = ((-3) \times \frac{5}{3})^4 \)
\( = (-1 \times 5)^4 = ((-1)^4 \times 5^4) \)
\( = (1 \times 5^4) = 5^4 \)
(iv) \( (3^{-7} \div 3^{-10}) \times 3^{-5} \)
Using \( a^m \div a^n = a^{m-n} \) and \( a^m \times a^n = a^{m+n} \)
\( (3^{-7} \div 3^{-10}) \times 3^{-5} = (3^{-7 - (-10)}) \times 3^{-5} \)
\( = (3^{-7+10}) \times 3^{-5} = 3^3 \times 3^{-5} \)
\( = 3^{3+(-5)} = 3^{3-5} = 3^{-2} \)
To express with a positive exponent, \( 3^{-2} = \frac{1}{3^2} \)
(v) \( 2^{-3} \times (-7)^{-3} \)
Using the rule \( a^m \times b^m = (ab)^m \)
\( 2^{-3} \times (-7)^{-3} = (2 \times (-7))^{-3} \)
\( = (-14)^{-3} \)
To express with a positive exponent, \( (-14)^{-3} = \frac{1}{(-14)^3} \)
In simple words: When simplifying, remember to use the exponent rules for multiplying and dividing powers with the same base or different bases but the same exponent. If your answer has a negative exponent, flip the base to make the exponent positive.

Exam Tip: Always double-check your exponent rules. A common error is mixing up \( (a^m)^n \) with \( a^m \times a^n \). Also, pay attention to negative signs with the base.

Question 3. Find the value of:
(i) \( (3^0 + 4^{-1}) \times 2^2 \)
(ii) \( (2^{-1} \times 4^{-1}) \div 2^{-2} \)
(iii) \( (\frac{1}{2})^{-2} + (\frac{1}{3})^{-2} + (\frac{1}{4})^{-2} \)
(iv) \( (3^{-1} + 4^{-1} + 5^{-1})^0 \)
(v) \( \{(\frac{-2}{3})^{-2}\}^2 \)
Answer:
(i) \( (3^0 + 4^{-1}) \times 2^2 \)
We know \( a^0 = 1 \) and \( a^{-1} = \frac{1}{a} \)
So, \( (3^0 + 4^{-1}) \times 2^2 = (1 + \frac{1}{4}) \times 4 \)
\( = (\frac{4+1}{4}) \times 4 = \frac{5}{4} \times 4 = 5 \)
(ii) \( (2^{-1} \times 4^{-1}) \div 2^{-2} \)
Using \( a^m \times b^m = (ab)^m \)
\( (2^{-1} \times 4^{-1}) = (2 \times 4)^{-1} = 8^{-1} \)
So, \( 8^{-1} \div 2^{-2} = \frac{1}{8} \div \frac{1}{2^2} \)
\( = \frac{1}{8} \div \frac{1}{4} = \frac{1}{8} \times \frac{4}{1} = \frac{4}{8} = \frac{1}{2} \)
(iii) \( (\frac{1}{2})^{-2} + (\frac{1}{3})^{-2} + (\frac{1}{4})^{-2} \)
Using \( (\frac{a}{b})^{-n} = (\frac{b}{a})^n \)
\( (\frac{1}{2})^{-2} = (2)^2 = 4 \)
\( (\frac{1}{3})^{-2} = (3)^2 = 9 \)
\( (\frac{1}{4})^{-2} = (4)^2 = 16 \)
So, \( 4 + 9 + 16 = 29 \)
(iv) \( (3^{-1} + 4^{-1} + 5^{-1})^0 \)
We know that any non-zero number raised to the power of 0 is 1. Here, \( (3^{-1} + 4^{-1} + 5^{-1}) \) is not zero.
Thus, \( (3^{-1} + 4^{-1} + 5^{-1})^0 = 1 \)
(v) \( \{(\frac{-2}{3})^{-2}\}^2 \)
Using \( (a^m)^n = a^{mn} \)
\( \{(\frac{-2}{3})^{-2}\}^2 = (\frac{-2}{3})^{-2 \times 2} = (\frac{-2}{3})^{-4} \)
To make the exponent positive, we flip the base: \( (\frac{-3}{2})^4 \)
\( = \frac{(-3)^4}{2^4} = \frac{(-3) \times (-3) \times (-3) \times (-3)}{2 \times 2 \times 2 \times 2} = \frac{81}{16} \)
In simple words: Carefully apply exponent rules step-by-step. Remember that anything to the power of zero is one, a negative exponent means taking the reciprocal, and when you have a power raised to another power, you multiply the exponents.

Exam Tip: For expressions with multiple operations, follow the order of operations (PEMDAS/BODMAS). Handle parentheses/brackets first, then exponents, and so on. Also, remember that \( (x^{-n}) = 1/x^n \).

Question 4. Evaluate
1. \( \frac{8^{-1} \times 5^3}{2^{-4}} \)
2. \( (5^{-1} \times 2^{-1}) \times 6^{-1} \)
Answer:
1. \( \frac{8^{-1} \times 5^3}{2^{-4}} \)
\( = 8^{-1} \times 5^3 \times 2^4 \)
\( = \frac{1}{8} \times (5 \times 5 \times 5) \times (2 \times 2 \times 2 \times 2) \)
\( = \frac{1}{8} \times 125 \times 16 \)
\( = 2 \times 125 = 250 \)
2. \( (5^{-1} \times 2^{-1}) \times 6^{-1} \)
Using \( a^m \times b^m = (ab)^m \)
\( (5^{-1} \times 2^{-1}) = (5 \times 2)^{-1} = 10^{-1} \)
So, \( 10^{-1} \times 6^{-1} = (10 \times 6)^{-1} = 60^{-1} \)
\( = \frac{1}{60} \)
In simple words: For the first part, move negative exponents to the other side of the fraction bar to make them positive. Then, calculate the powers and multiply. For the second part, combine terms with the same exponent, then apply the negative exponent rule.

Exam Tip: When evaluating complex expressions, it is often helpful to convert all terms with negative exponents to positive exponents first by taking their reciprocals. This can simplify the calculation process.

Question 5. Find the value of m for which \( 5^m \div 5^{-3} = 5^5 \)
Answer:
We are given the equation: \( 5^m \div 5^{-3} = 5^5 \)
Using the exponent rule \( a^p \div a^q = a^{p-q} \)
The left-hand side (LHS) becomes: \( 5^{m - (-3)} = 5^{m+3} \)
So, the equation is: \( 5^{m+3} = 5^5 \)
Since the bases are equal (both are 5), the exponents must also be equal.
Therefore, \( m+3 = 5 \)
Subtract 3 from both sides to solve for m:
\( m = 5 - 3 \)
\( m = 2 \)
Thus, the required value of m is 2.
In simple words: When you divide powers with the same base, you subtract their exponents. After doing that, if the bases on both sides of an equation are the same, then the powers themselves must also be equal.

Exam Tip: When solving equations with exponents, remember that if \( a^x = a^y \), then \( x = y \) (provided \( a \neq 0, 1, -1 \)). This property helps you equate the exponents once the bases are the same.

Question 6. Evaluate:
(i) \( \{(\frac{1}{3})^{-1} - (\frac{1}{4})^{-1}\}^{-1} \)
(ii) \( (\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4} \)
Answer:
(i) \( \{(\frac{1}{3})^{-1} - (\frac{1}{4})^{-1}\}^{-1} \)
First, evaluate the terms inside the curly brackets using \( (\frac{a}{b})^{-1} = \frac{b}{a} \)
\( (\frac{1}{3})^{-1} = 3 \)
\( (\frac{1}{4})^{-1} = 4 \)
So, the expression inside the brackets is \( (3 - 4) = -1 \)
Now, we have \( (-1)^{-1} \)
Using \( a^{-1} = \frac{1}{a} \)
\( (-1)^{-1} = \frac{1}{-1} = -1 \)
(ii) \( (\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4} \)
To make the bases the same, convert \( (\frac{8}{5})^{-4} \) to a base of \( \frac{5}{8} \).
Using \( (\frac{b}{a})^{-n} = (\frac{a}{b})^n \)
\( (\frac{8}{5})^{-4} = (\frac{5}{8})^4 \)
Now the expression is \( (\frac{5}{8})^{-7} \times (\frac{5}{8})^4 \)
Using the rule \( a^m \times a^n = a^{m+n} \)
\( = (\frac{5}{8})^{-7+4} \)
\( = (\frac{5}{8})^{-3} \)
To express with a positive exponent, we flip the base:
\( = (\frac{8}{5})^3 \)
\( = \frac{8^3}{5^3} = \frac{8 \times 8 \times 8}{5 \times 5 \times 5} = \frac{512}{125} \)
In simple words: For the first part, resolve the inner negative exponents first by flipping the fractions. Then complete the subtraction and find the reciprocal of the result. For the second part, ensure all bases are the same. Then, when multiplying, add the exponents together. Finally, change any negative exponent to a positive one by inverting the fraction.

Exam Tip: When dealing with fractional bases and negative exponents, it's often easiest to flip the fraction first to make the exponent positive. This simplifies calculations and helps avoid errors. For example, \( (\frac{a}{b})^{-n} = (\frac{b}{a})^n \).

Question 7. Simplify:
(i) \( \frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} \) (t \( \neq \) 0)
(ii) \( \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}} \)
Answer:
(i) \( \frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} \)
First, express all numbers as powers of prime factors and move negative exponents to the numerator/denominator to make them positive.
\( 25 = 5^2 \)
\( 10 = 2 \times 5 \)
\( = \frac{5^2 \times t^{-4} \times 5^3 \times t^8}{10} \)
\( = \frac{5^2 \times 5^3 \times t^{-4} \times t^8}{2 \times 5} \)
Using \( a^m \times a^n = a^{m+n} \) and \( \frac{a^m}{a^n} = a^{m-n} \)
\( = \frac{5^{2+3} \times t^{-4+8}}{2 \times 5^1} \)
\( = \frac{5^5 \times t^4}{2 \times 5^1} \)
\( = \frac{5^{5-1} \times t^4}{2} \)
\( = \frac{5^4 \times t^4}{2} \)
\( = \frac{625 t^4}{2} \)
(ii) \( \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}} \)
Express all numbers as powers of prime factors:
\( 10^{-5} = (2 \times 5)^{-5} = 2^{-5} \times 5^{-5} \)
\( 125 = 5^3 \)
\( 6^{-5} = (2 \times 3)^{-5} = 2^{-5} \times 3^{-5} \)
Substitute these into the expression:
\( = \frac{3^{-5} \times (2^{-5} \times 5^{-5}) \times 5^3}{5^{-7} \times (2^{-5} \times 3^{-5})} \)
Group terms with the same base:
\( = \frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^3}{3^{-5} \times 2^{-5} \times 5^{-7}} \)
Cancel out common terms from numerator and denominator, and use \( \frac{a^m}{a^n} = a^{m-n} \)
\( = 3^{-5-(-5)} \times 2^{-5-(-5)} \times 5^{-5-3-(-7)} \)
\( = 3^{-5+5} \times 2^{-5+5} \times 5^{-5+3+7} \)
\( = 3^0 \times 2^0 \times 5^5 \)
Since \( a^0 = 1 \)
\( = 1 \times 1 \times 5^5 = 5^5 \)
In simple words: When simplifying expressions with exponents, always try to break down numbers into their prime factors first. Then, apply the exponent rules for multiplication and division. Remember that a term can be moved from the denominator to the numerator (or vice versa) by changing the sign of its exponent.

Exam Tip: For complex expressions, always simplify the base numbers into their prime factors before applying exponent rules. This makes it easier to combine terms and track exponents correctly. For example, replace 10 with \( 2 \times 5 \) and 6 with \( 2 \times 3 \).