RBSE Solutions Class 8 Maths Chapter 3 Powers and Exponents Exercise 3.1

Question 1. Simplify the following
(i) \( \left(\frac{2}{7}\right)^3 \times \left(\frac{7}{2}\right)^3 \)
(ii) \( \left(\frac{4}{5}\right)^4 \times \left(\frac{5}{4}\right)^2 \)
(iii) \( (-5)^3 \times \left(-\frac{1}{5}\right)^2 \)
(iv) \( \left(\frac{3}{4}\right)^3 \times \left(\frac{3}{4}\right)^{-5} \)
Answer:
(i) We use the rule that \( a^m \times b^m = (a \times b)^m \), so \( \left(\frac{2}{7}\right)^3 \times \left(\frac{7}{2}\right)^3 \).
\( = \frac{2 \times 2 \times 2}{7 \times 7 \times 7} \times \frac{7 \times 7 \times 7}{2 \times 2 \times 2} \)
\( = \frac{8}{343} \times \frac{343}{8} \)
\( = \frac{8 \times 343}{343 \times 8} \)
\( = 1 \)
(ii) We simplify the powers and then multiply:
\( \left(\frac{4}{5}\right)^4 \times \left(\frac{5}{4}\right)^2 \)
\( = \left(\frac{4}{5} \times \frac{4}{5} \times \frac{4}{5} \times \frac{4}{5}\right) \times \left(\frac{5}{4} \times \frac{5}{4}\right) \)
\( = \frac{4 \times 4 \times 4 \times 4 \times 5 \times 5}{5 \times 5 \times 5 \times 5 \times 4 \times 4} \)
\( = \frac{4^{4-2}}{5^{4-2}} = \frac{4^2}{5^2} = \frac{16}{25} \)
(iii) We break down the terms and apply exponent rules:
\( (-5)^3 \times \left(-\frac{1}{5}\right)^2 \)
\( = \{(-1) \times 5\}^3 \times \{(-1) \times \left(\frac{1}{5}\right)\}^2 \)
\( = (-1)^3 \times 5^3 \times (-1)^2 \times \frac{1}{5^2} \)
\( = (-1)^{3+2} \times 5^{3-2} = (-1)^5 \times 5^1 \)
\( = -1 \times 5 = -5 \)
(iv) When multiplying exponents with the same base, we add their powers:
\( \left(\frac{3}{4}\right)^3 \times \left(\frac{3}{4}\right)^{-5} \)
\( = \left(\frac{3}{4}\right)^{3 + (-5)} \)
\( = \left(\frac{3}{4}\right)^{3-5} \)
\( = \left(\frac{3}{4}\right)^{-2} \)
\( = \left(\frac{4}{3}\right)^{2} \)
\( = \frac{4^2}{3^2} = \frac{4 \times 4}{3 \times 3} = \frac{16}{9} \)
In simple words: For (i), (ii), (iii), and (iv), we use the rules of exponents to combine or separate the terms. When the base numbers are multiplied, their powers are added. When a negative power is used, we flip the fraction. Remember that any number raised to the power of one is the number itself.

🎯 Exam Tip: Pay close attention to negative bases and negative exponents; a negative base raised to an odd power remains negative, while to an even power it becomes positive. Also, \( a^{-n} = \frac{1}{a^n} \).

Question 2. Find the values
(i) \( (-5)^3 \)
(ii) \( \left(\frac{1}{2}\right)^3 \)
(iii) \( \left(-\frac{2}{3}\right)^4 \)
Answer:
(i) To find the value of \( (-5)^3 \), we multiply -5 by itself three times:
\( = (-1 \times 5)^3 \)
\( = (-1)^3 \times (5)^3 \)
\( = (-1) \times 5 \times 5 \times 5 \)
\( = -125 \)
(ii) To find the value of \( \left(\frac{1}{2}\right)^3 \), we multiply the fraction by itself three times:
\( = \frac{1^3}{2^3} \)
\( = \frac{1 \times 1 \times 1}{2 \times 2 \times 2} \)
\( = \frac{1}{8} \)
(iii) To find the value of \( \left(-\frac{2}{3}\right)^4 \), we multiply the fraction by itself four times. Since the power is even, the result will be positive:
\( = \{(-1) \times \left(\frac{2}{3}\right)\}^4 \)
\( = (-1)^4 \times \left(\frac{2}{3}\right)^4 \)
\( = (1) \times \frac{2^4}{3^4} \)
\( = \frac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3} = \frac{16}{81} \)
In simple words: To find the value of a number raised to a power, just multiply the number by itself as many times as the power says. If the base is negative and the power is even, the answer is positive. If the base is negative and the power is odd, the answer is negative.

🎯 Exam Tip: Always remember that a negative number raised to an even power results in a positive number, and a negative number raised to an odd power results in a negative number.

Question 3. With the help of Prime factor, change the following into exponent form
(i) \( \frac{1}{64} \)
(ii) \( \frac{16}{125} \)
(iii) \( -\frac{8}{27} \)
(iv) \( -\frac{1}{8} \)
(v) \( -\frac{25}{49} \)
Answer:
(i) We express 64 as a power of its prime factors:
\( \frac{1}{64} = \frac{1}{2 \times 2 \times 2 \times 2 \times 2 \times 2} \)
\( = \frac{1}{2^6} \)
\( = 2^{-6} \)
(ii) We find the prime factors for the numerator and the denominator:
\( \frac{16}{125} = \frac{2 \times 2 \times 2 \times 2}{5 \times 5 \times 5} \)
\( = \frac{2^4}{5^3} \)
(iii) We find the prime factors for 8 and 27, keeping the negative sign:
\( -\frac{8}{27} = -\frac{2 \times 2 \times 2}{3 \times 3 \times 3} \)
\( = - \frac{2^3}{3^3} \)
\( = \left(-\frac{2}{3}\right)^3 \)
(iv) We find the prime factors for the denominator, keeping the negative sign:
\( -\frac{1}{8} = - \frac{1}{2 \times 2 \times 2} \)
\( = - \frac{1}{2^3} \)
\( = \frac{(-1)^3}{2^3} \)
\( = \left(-\frac{1}{2}\right)^3 \)
(v) We find the prime factors for 25 and 49, keeping the negative sign:
\( -\frac{25}{49} = - \frac{5 \times 5}{7 \times 7} \)
\( = - \frac{5^2}{7^2} \)
\( = -\left(\frac{5}{7}\right)^2 \)
In simple words: To write a number in exponent form using prime factors, first break down the number into its smallest prime building blocks. Then, count how many times each prime number appears and write that count as a small number (exponent) above the prime number. If there's a negative sign, include it, and for fractions, do this for both top and bottom numbers.

🎯 Exam Tip: Always look for the simplest prime factors (2, 3, 5, 7, etc.) and ensure all parts of the number, including any negative signs, are correctly represented in the exponential form.

Question 4. Find the values
(i) \( 3^2 \times 3^3 \)
(ii) \( \left(\frac{3}{2}\right)^2 \times \left(\frac{3}{2}\right)^3 \)
(iii) \( \left(\frac{2}{3}\right)^2 \times \left(\frac{3}{2}\right)^3 \)
(iv) \( \left(-\frac{1}{2}\right)^3 \times \left(-\frac{1}{2}\right)^4 \)
(v) \( \left(-\frac{2}{5}\right)^3 \times \left(\frac{2}{5}\right)^{-5} \)
Answer:
(i) When multiplying numbers with the same base, we add their exponents:
\( 3^2 \times 3^3 = 3^{2+3} \)
\( = 3^5 \)
\( = 3 \times 3 \times 3 \times 3 \times 3 = 243 \)
(ii) We add the exponents because the base is the same:
\( \left(\frac{3}{2}\right)^2 \times \left(\frac{3}{2}\right)^3 = \left(\frac{3}{2}\right)^{2+3} \)
\( = \left(\frac{3}{2}\right)^5 \)
\( = \frac{3^5}{2^5} = \frac{3 \times 3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 2} = \frac{243}{32} \)
(iii) We separate the terms and apply exponent rules, then simplify:
\( \left(\frac{2}{3}\right)^2 \times \left(\frac{3}{2}\right)^3 \)
\( = \frac{2^2}{3^2} \times \frac{3^3}{2^3} \)
\( = \frac{2^2 \times 3^3}{3^2 \times 2^3} \)
\( = \frac{3^{3-2}}{2^{3-2}} = \frac{3^1}{2^1} = \frac{3}{2} \)
(iv) We add the exponents because the base is the same. An odd power with a negative base results in a negative value:
\( \left(-\frac{1}{2}\right)^3 \times \left(-\frac{1}{2}\right)^4 \)
\( = \left(-\frac{1}{2}\right)^{3+4} \)
\( = \left(-\frac{1}{2}\right)^7 \)
\( = (-1)^7 \times \left(\frac{1}{2}\right)^7 \)
\( = -1 \times \frac{1}{2^7} = - \frac{1}{128} \)
(v) We rewrite the negative exponent and then simplify the expression:
\( \left(-\frac{2}{5}\right)^3 \times \left(\frac{2}{5}\right)^{-5} \)
\( = \left(-\frac{2}{5}\right)^3 \times \left(\frac{5}{2}\right)^5 \)
\( = (-1)^3 \times \left(\frac{2}{5}\right)^3 \times \left(\frac{5}{2}\right)^5 \)
\( = -1 \times \frac{2^3}{5^3} \times \frac{5^5}{2^5} \)
\( = -1 \times \frac{2^3}{2^5} \times \frac{5^5}{5^3} \)
\( = -1 \times 2^{3-5} \times 5^{5-3} \)
\( = -1 \times 2^{-2} \times 5^2 \)
\( = -1 \times \frac{1}{2^2} \times 5^2 \)
\( = -1 \times \frac{1}{4} \times 25 = - \frac{25}{4} \)
In simple words: When you multiply numbers with the same bottom part (base), you just add the small numbers on top (exponents). If a small number on top is negative, you flip the fraction to make it positive. Then, multiply everything out to find the final value.

🎯 Exam Tip: Always convert negative exponents to positive ones by taking the reciprocal of the base. Simplify expressions by canceling out common terms from the numerator and denominator.

Question 5. Answer in exponent form-
(i) \( 4^5 \div 4^2 \)
(ii) \( (-5)^7 \div (-5)^4 \)
(iii) \( \left(\frac{2}{3}\right)^5 \div \left(\frac{2}{3}\right)^1 \)
(iv) \( \left(-\frac{1}{5}\right)^{11} \div \left(-\frac{1}{5}\right)^6 \)
Answer:
(i) When dividing numbers with the same base, we subtract the exponents:
\( 4^5 \div 4^2 = 4^{5-2} \)
\( = 4^3 \)
(ii) Subtract the exponents as the base is the same:
\( (-5)^7 \div (-5)^4 = (-5)^{7-4} \)
\( = (-5)^3 \)
(iii) Subtract the exponents as the base is the same:
\( \left(\frac{2}{3}\right)^5 \div \left(\frac{2}{3}\right)^1 = \left(\frac{2}{3}\right)^{5-1} \)
\( = \left(\frac{2}{3}\right)^4 \)
(iv) Subtract the exponents as the base is the same:
\( \left(-\frac{1}{5}\right)^{11} \div \left(-\frac{1}{5}\right)^6 = \left(-\frac{1}{5}\right)^{11-6} \)
\( = \left(-\frac{1}{5}\right)^5 \)
In simple words: When you divide numbers that have the same base (the big number at the bottom), you just take away the small numbers (exponents) on top. This gives you a new exponent, and the base stays the same.

🎯 Exam Tip: This question tests the quotient rule for exponents: \( a^m \div a^n = a^{m-n} \). Ensure you subtract the exponents correctly, especially with negative bases.

Question 6. Find the value
(i) \( (3^2)^3 \)
(ii) \( (2^3)^2 \)
(iii) \( (5^2)^2 \)
(iv) \( (-2^4)^2 \)
(v) \( \left[\left(\frac{1}{2}\right)^2\right]^3 \)
(vi) \( \left[\left(-\frac{1}{3}\right)^3\right]^2 \)
Answer:
(i) We first calculate the inside power, then the outside power:
\( (3^2)^3 = (3 \times 3)^3 \)
\( = 9^3 \)
\( = 9 \times 9 \times 9 = 729 \)
(ii) We use the power of a power rule \( (a^m)^n = a^{m \times n} \):
\( (2^3)^2 = 2^{3 \times 2} \)
\( = 2^6 \)
\( = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64 \)
(iii) We use the power of a power rule \( (a^m)^n = a^{m \times n} \):
\( (5^2)^2 = 5^{2 \times 2} \)
\( = 5^4 \)
\( = 5 \times 5 \times 5 \times 5 = 625 \)
(iv) We first calculate the inner power, then the outer power. Remember that an even power makes the negative sign positive:
\( (-2^4)^2 = (- (2 \times 2 \times 2 \times 2))^2 \)
\( = (-16)^2 \)
\( = \{(-1) \times 16\}^2 \)
\( = (-1)^2 \times (16)^2 \)
\( = 1 \times 256 = 256 \)
(v) We apply the power of a power rule to the fraction:
\( \left[\left(\frac{1}{2}\right)^2\right]^3 = \left(\frac{1}{2}\right)^{2 \times 3} \)
\( = \left(\frac{1}{2}\right)^6 \)
\( = \frac{1^6}{2^6} = \frac{1}{64} \)
(vi) We apply the power of a power rule to the negative fraction. An even outer power makes the result positive:
\( \left[\left(-\frac{1}{3}\right)^3\right]^2 = \left(-\frac{1}{3}\right)^{3 \times 2} \)
\( = \left(-\frac{1}{3}\right)^6 \)
\( = (-1)^6 \times \left(\frac{1}{3}\right)^6 \)
\( = 1 \times \frac{1}{3^6} = \frac{1}{729} \)
In simple words: When you have a power raised to another power, you just multiply the small numbers (exponents) together. Then, calculate the result by multiplying the base number by itself that many times. If there's a negative sign inside, an even power makes it positive, and an odd power keeps it negative.

🎯 Exam Tip: The rule \( (a^m)^n = a^{mn} \) is crucial here. Always evaluate the inner expression first or apply the exponent rule correctly, and remember to handle negative signs based on whether the final exponent is even or odd.

Question 7. Find the value
(i) \( 3^0 \)
(ii) \( 7^{5-5} \)
(iii) \( (-2)^{3-3} \)
(iv) \( \left(\frac{2}{3}\right)^{2+3-5} \)
(v) \( 2^0 \times 3^0 \)
(vi) \( 2^0 + 5^0 \)
(vii) \( \left(\frac{7}{15}\right)^0 + \left(\frac{1}{7}\right)^{3-3} \)
Answer:
(i) Any non-zero number raised to the power of 0 is 1:
\( 3^0 = 1 \)
(ii) First, simplify the exponent, then apply the power of 0 rule:
\( 7^{5-5} = 7^0 \)
\( = 1 \)
(iii) First, simplify the exponent, then apply the power of 0 rule:
\( (-2)^{3-3} = (-2)^0 \)
\( = 1 \)
(iv) First, simplify the exponent, then apply the power of 0 rule:
\( \left(\frac{2}{3}\right)^{2+3-5} = \left(\frac{2}{3}\right)^0 \)
\( = 1 \)
(v) Each term raised to the power of 0 is 1, then multiply:
\( 2^0 \times 3^0 = 1 \times 1 \)
\( = 1 \)
(vi) Each term raised to the power of 0 is 1, then add:
\( 2^0 + 5^0 = 1 + 1 \)
\( = 2 \)
(vii) Each term raised to the power of 0 is 1, then add:
\( \left(\frac{7}{15}\right)^0 + \left(\frac{1}{7}\right)^{3-3} = \left(\frac{7}{15}\right)^0 + \left(\frac{1}{7}\right)^0 \)
\( = 1 + 1 \)
\( = 2 \)
In simple words: A very simple rule in math is that any number (except zero) raised to the power of zero always equals 1. This is true whether the number is positive, negative, a fraction, or a whole number.

🎯 Exam Tip: Remember the zero exponent rule: \( a^0 = 1 \) for any non-zero number \( a \). Be careful if the base is 0, as \( 0^0 \) is undefined.

Question 8. Change into positive exponent numbers
(i) \( 2^{-3} \)
(ii) \( 3^{-5} \)
(iii) \( a^{-4} \)
(iv) \( (-2)^{-5} \)
(v) \( (-x)^{-3} \)
(vi) \( \frac{1}{5^{-3}} \)
(vii) \( \frac{1}{y^{-3}} \)
(viii) \( \frac{1}{\left(\frac{2}{3}\right)^{-3}} \)
Answer:
(i) To change a negative exponent to positive, we take the reciprocal of the base:
\( 2^{-3} = \frac{1}{2^3} \)
(ii) We take the reciprocal of the base to make the exponent positive:
\( 3^{-5} = \frac{1}{3^5} \)
(iii) We take the reciprocal of the base \( a \) to make the exponent positive:
\( a^{-4} = \frac{1}{a^4} \)
(iv) We take the reciprocal of the base \( (-2) \) to make the exponent positive:
\( (-2)^{-5} = \frac{1}{(-2)^5} \)
(v) We take the reciprocal of the base \( (-x) \) to make the exponent positive:
\( (-x)^{-3} = \frac{1}{(-x)^3} \)
(vi) The reciprocal of a negative exponent results in a positive exponent:
\( \frac{1}{5^{-3}} = 5^3 \)
(vii) The reciprocal of a negative exponent results in a positive exponent:
\( \frac{1}{y^{-3}} = y^3 \)
(viii) The reciprocal of a negative exponent results in a positive exponent:
\( \frac{1}{\left(\frac{2}{3}\right)^{-3}} = \left(\frac{2}{3}\right)^3 \)
In simple words: To make a negative exponent positive, you just flip the number over. So, if a number is `a` with a negative exponent, it becomes `1/a` with a positive exponent. If it's `1/a` with a negative exponent, it becomes `a` with a positive exponent.

🎯 Exam Tip: Remember the rule \( a^{-n} = \frac{1}{a^n} \) and \( \frac{1}{a^{-n}} = a^n \). This rule applies to any non-zero base, whether it's a whole number, a fraction, or a variable.

Question 9. Simplify the following in form of exponent
(i) \( (2^2 \times 3^3)^2 \)
(ii) \( \left(\frac{15}{16}\right)^3 + \left(\frac{9}{8}\right)^2 \)
(iii) \( \left(\frac{4}{9}\right)^2 + \left(\frac{28}{27}\right)^3 \)
(iv) \( \left(\frac{2}{3}\right)^2 \times \left(\frac{1}{4}\right)^3 \times \left(\frac{3}{2}\right)^2 \)
(v) \( \left(\frac{5^2}{3^2}\right)^2 \)
(vi) \( \frac{2^2 \times 3^2}{2^3 \times 6^2} \)
Answer:
(i) We apply the power to each term inside the parenthesis:
\( (2^2 \times 3^3)^2 = (2^2)^2 \times (3^3)^2 \)
\( = 2^{2 \times 2} \times 3^{3 \times 2} \)
\( = 2^4 \times 3^6 \)
(ii) We rewrite the bases with their prime factors and then perform the division as indicated by the source. The plus sign changes to a division sign as per the provided solution steps.
\( \left(\frac{15}{16}\right)^3 \div \left(\frac{9}{8}\right)^2 \)
\( = \frac{(3 \times 5)^3}{(2^4)^3} \div \frac{(3^2)^2}{(2^3)^2} \)
\( = \frac{3^3 \times 5^3}{2^{12}} \div \frac{3^4}{2^6} \)
\( = \frac{3^3 \times 5^3}{2^{12}} \times \frac{2^6}{3^4} \)
\( = \frac{3^3 \times 5^3 \times 2^6}{2^{12} \times 3^4} \)
\( = \frac{5^3}{3^{4-3} \times 2^{12-6}} = \frac{5^3}{3^1 \times 2^6} \)
\( = \frac{125}{3 \times 64} = \frac{125}{192} \)
(iii) We rewrite the bases with their prime factors and then perform the division as indicated by the source. The plus sign changes to a division sign as per the provided solution steps.
\( \left(\frac{4}{9}\right)^2 \div \left(\frac{28}{27}\right)^3 \)
\( = \left(\frac{2^2}{3^2}\right)^2 \div \left(\frac{2^2 \times 7}{3^3}\right)^3 \)
\( = \frac{(2^2)^2}{(3^2)^2} \div \frac{(2^2)^3 \times (7)^3}{(3^3)^3} \)
\( = \frac{2^4}{3^4} \div \frac{2^6 \times 7^3}{3^9} \)
\( = \frac{2^4}{3^4} \times \frac{3^9}{2^6 \times 7^3} \)
\( = \frac{3^{9-4}}{2^{6-4} \times 7^3} = \frac{3^5}{2^2 \times 7^3} \)
\( = \frac{243}{4 \times 343} = \frac{243}{1372} \)
(iv) We rewrite the numbers with their prime factors and simplify:
\( \left(\frac{2}{3}\right)^2 \times \left(\frac{1}{4}\right)^3 \times \left(\frac{3}{2}\right)^2 \)
\( = \frac{2^2}{3^2} \times \frac{1^3}{4^3} \times \frac{3^2}{2^2} \)
\( = \frac{2^2}{3^2} \times \frac{1}{(2^2)^3} \times \frac{3^2}{2^2} \)
\( = \frac{2^2}{3^2} \times \frac{1}{2^6} \times \frac{3^2}{2^2} \)
\( = \frac{2^2 \times 1 \times 3^2}{3^2 \times 2^6 \times 2^2} \)
\( = \frac{2^2 \times 3^2}{3^2 \times 2^8} \)
\( = \frac{1}{2^{8-2}} = \frac{1}{2^6} = \frac{1}{64} \)
(v) We apply the power to the entire fraction inside:
\( \left(\frac{5^2}{3^2}\right)^2 = \left(\left(\frac{5}{3}\right)^2\right)^2 \)
\( = \left(\frac{5}{3}\right)^{2 \times 2} \)
\( = \left(\frac{5}{3}\right)^4 \)
\( = \frac{5^4}{3^4} = \frac{5 \times 5 \times 5 \times 5}{3 \times 3 \times 3 \times 3} = \frac{625}{81} \)
(vi) We rewrite 6 as \( 2 \times 3 \) and then simplify the expression:
\( \frac{2^2 \times 3^2}{2^3 \times 6^2} \)
\( = \frac{2^2 \times 3^2}{2^3 \times (2 \times 3)^2} \)
\( = \frac{2^2 \times 3^2}{2^3 \times 2^2 \times 3^2} \)
\( = \frac{2^2 \times 3^2}{2^{3+2} \times 3^2} \)
\( = \frac{2^2 \times 3^2}{2^5 \times 3^2} \)
\( = \frac{1}{2^{5-2}} \times 3^{2-2} \)
\( = \frac{1}{2^3} \times 3^0 \)
\( = \frac{1}{8} \times 1 = \frac{1}{8} \)
In simple words: To simplify expressions with powers, use the rules of exponents. If powers are multiplied, add their exponents. If powers are divided, subtract their exponents. When a power is raised to another power, multiply the exponents. Also, turn any base into its prime factors to make simplifying easier.

🎯 Exam Tip: Always break down composite bases (like 6 in part (vi)) into their prime factors before applying exponent rules. This helps in canceling out common terms more effectively and simplifying the expression correctly.

Question 9. Simplify the following in form of exponent
(v) \( {\left( \frac { 5^{2} }{ 3^{2} } \right) }^{2} \)
Answer:
\( {\left( \frac { 5^{2} }{ 3^{2} } \right) }^{2} = {\left[ {\left( \frac { 5 }{ 3 } \right) }^{2} \right] }^{2} \)
\( \implies {\left( \frac { 5 }{ 3 } \right) }^{4} \)
\( \implies \frac { 5^{4} }{ 3^{4} } \)
\( \implies \frac { 5 \times 5 \times 5 \times 5 }{ 3 \times 3 \times 3 \times 3 } \)
\( \implies \frac { 625 }{ 81 } \) This uses the power of a power rule, where you multiply the exponents.
In simple words: When you have a fraction raised to a power, and then that whole thing raised to another power, you can multiply the two powers together. Then, multiply the top and bottom numbers by themselves that many times.

🎯 Exam Tip: Remember to apply the outer exponent to both the numerator and the denominator of the base fraction, and simplify all powers completely.

Question 10. Find the values of \( {\left( \frac { 1 }{ 3 } \right) }^{-2} + {\left( \frac { 1 }{ 2 } \right) }^{-2} + {\left( \frac { 1 }{ 4 } \right) }^{-2} \)
Answer:
\( {\left( \frac { 1 }{ 3 } \right) }^{-2} + {\left( \frac { 1 }{ 2 } \right) }^{-2} + {\left( \frac { 1 }{ 4 } \right) }^{-2} \)
\( \implies (3)^{2} + (2)^{2} + (4)^{2} \)
\( \implies 9 + 4 + 16 \)
\( \implies 29 \) A negative exponent means taking the reciprocal of the base before applying the positive exponent.
In simple words: When you see a negative number in the exponent, flip the fraction (or the number) upside down and then make the exponent positive. Then, multiply the numbers as usual and add them up.

🎯 Exam Tip: A common mistake is to keep the negative sign with the base when dealing with negative exponents. Always convert to the reciprocal with a positive exponent first.