CBSE Class 7 Mathematics Exponents And Powers Worksheet Set 09

Exponents

Question 1. Value of -5 raised to the power of 4 is :
(a) -625
(b) 25
(c) -125
(d) 625
Answer: (d) 625
\((-5)^4 = (-5) \times (-5) \times (-5) \times (-5) = 625\)
In simple words: Multiplying a negative number an even number of times results in a positive product.

Exam Tip: Remember that even powers of negative numbers are always positive, and odd powers are negative.

Question 2. (-1)^31 equals to :
(a) 31
(b) -1
(c) 1
(d) -31
Answer: (b) -1
\((-1)^{31} = -1\)
In simple words: Since 31 is an odd number, -1 multiplied by itself 31 times remains -1.

Exam Tip: For any base of -1, an odd exponent gives -1, and an even exponent gives 1.

Question 3. What power of -3 is 729?
(a) 3
(b) 4
(c) 6
(d) 5
Answer: (c) 6
Prime factorization or successive multiplication:
\((-3)^1 = -3\)
\((-3)^2 = 9\)
\((-3)^3 = -27\)
\((-3)^4 = 81\)
\((-3)^5 = -243\)
\((-3)^6 = 729\)
In simple words: Multiply -3 continuously until you get 729, which takes 6 multiplications.

Exam Tip: Check the sign of the target number; a positive target with a negative base always requires an even exponent.

Question 4. Value of (-1)^15 x (-1)^16 equals to;
(a) 1
(b) -1
(c) -240
(d) 240
Answer: (b) -1
Using law of exponents \(a^m \times a^n = a^{m+n}\):
\((-1)^{15} \times (-1)^{16} = (-1)^{15+16} = (-1)^{31} = -1\)
In simple words: Add the exponents when bases are the same, then evaluate the resulting power.

Exam Tip: Always apply exponent laws first to simplify expressions before calculating final values.

Question 5. If 2^4 + 3^2 = 5^x then x equals :
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2
\(2^4 = 16\)
\(3^2 = 9\)
\(16 + 9 = 25\)
\(5^x = 25 = 5^2\)
Therefore, \(x = 2\)
In simple words: Evaluate the left side numbers first, find their sum, and write it as a power of 5.

Exam Tip: Do not add exponents when adding bases; calculate each term separately.

Question 4. Use power notation to express each of the following:
Answer:
(a) \((-2/3) \times (-2/3) \times (-2/3) \times (-2/3) = (-2/3)^4\)
(b) \(m \times m \times m \times m \times m \times m = m^6\)
(c) \(\frac{5 \times 5 \times 5 \times 5 \times 5 \times 5}{5 \times 5 \times 5 \times 5} = \frac{5^6}{5^4} = 5^{6-4} = 5^2\)
In simple words: Count how many times each number or variable is multiplied and write it as the exponent.

Exam Tip: For fraction expressions, simplify numerator and denominator powers using laws of exponents.

Question 5. Express each of the following in exponential form:
Answer:
(a) \(1024 = 2^{10}\)
(b) \(-100000 = -10^5\)
(c) \(\frac{1331}{343} = \frac{11^3}{7^3} = \left(\frac{11}{7}\right)^3\)
(d) \(-\frac{1}{125} = -\frac{1}{5^3} = \left(-\frac{1}{5}\right)^3\)
In simple words: Break down each number into its prime factors to find the correct base and power.

Exam Tip: Memorize common powers of numbers like 2, 3, 5, and 11 to solve these quickly.

Question 6. Evaluate ; (a) (3)^6 - (2)^5 (b) 4^3 + 5^3 (c) 8^3 x 2^4 (d) 3^7 \div 9^2 (e) \frac{3^5 \times 4^7}{6^3 \times 8} (f) (5^3 \times 2^3) \div (5^2 \times 2) (f) [Note: second (f)] \left(\frac{3}{2}\right)^5 \times 16 \times \left(\frac{2}{3}\right)^4 (g) \left(\frac{4}{7}\right)^2 \times \left(\frac{1}{5}\right)^2 \times \frac{35}{64} (h) \left(\frac{1}{3}\right)^4 \times \left(-\frac{27}{4}\right)^2 \times \left(-\frac{8}{3}\right)^2 (i) \left[ (1/4)^3 \right]^2 \div \left[ (1/4)^2 \right]^3 (j) \frac{4^2 \times 4^3}{2^6} (k) (-1/3)^2 \times (-1/3)^3 \times (-1/3)^0
Answer:
(a) \(3^6 - 2^5 = 729 - 32 = 697\)
(b) \(4^3 + 5^3 = 64 + 125 = 189\)
(c) \(8^3 \times 2^4 = (2^3)^3 \times 2^4 = 2^9 \times 2^4 = 2^{13} = 8192\)
(d) \(3^7 \div 9^2 = 3^7 \div (3^2)^2 = 3^7 \div 3^4 = 3^{7-4} = 3^3 = 27\)
(e) \(\frac{3^5 \times (2^2)^7}{(2 \times 3)^3 \times 2^3} = \frac{3^5 \times 2^{14}}{2^3 \times 3^3 \times 2^3} = \frac{3^5 \times 2^{14}}{3^3 \times 2^6} = 3^{5-3} \times 2^{14-6} = 3^2 \times 2^8 = 9 \times 256 = 2304\)
(f) \((5^3 \times 2^3) \div (5^2 \times 2) = \frac{5^3 \times 2^3}{5^2 \times 2^1} = 5^{3-2} \times 2^{3-1} = 5^1 \times 2^2 = 5 \times 4 = 20\)
(f - duplicate label) \(\left(\frac{3}{2}\right)^5 \times 16 \times \left(\frac{2}{3}\right)^4 = \left(\frac{3}{2}\right)^5 \times \left(\frac{2}{3}\right)^4 \times 2^4 = \left(\frac{3}{2}\right)^{5-4} \times \frac{3}{2} \times 2^4 = \frac{3}{2} \times 16 = 24\)
(g) \(\left(\frac{4}{7}\right)^2 \times \left(\frac{1}{5}\right)^2 \times \frac{35}{64} = \frac{16}{49} \times \frac{1}{25} \times \frac{35}{64} = \frac{16 \times 1 \times 35}{49 \times 25 \times 64} = \frac{5}{35 \times 25} = \frac{1}{7 \times 25} = \frac{1}{175}\)
(h) \(\left(\frac{1}{3}\right)^4 \times \left(-\frac{27}{4}\right)^2 \times \left(-\frac{8}{3}\right)^2 = \frac{1}{81} \times \frac{729}{16} \times \frac{64}{9} = \frac{729 \times 64}{81 \times 16 \times 9} = \frac{729 \times 64}{729 \times 16} = \frac{64}{16} = 4\)
(i) \(\left[ (1/4)^3 \right]^2 \div \left[ (1/4)^2 \right]^3 = (1/4)^6 \div (1/4)^6 = 1\)
(j) \(\frac{4^2 \times 4^3}{2^6} = \frac{4^5}{(2^2)^3} = \frac{(2^2)^5}{2^6} = \frac{2^{10}}{2^6} = 2^{10-6} = 2^4 = 16\)
(k) \((-1/3)^2 \times (-1/3)^3 \times (-1/3)^0 = (-1/3)^{2+3+0} = (-1/3)^5 = -\frac{1}{243}\)
In simple words: Convert all bases to prime numbers and apply exponent rules to simplify before calculating.

Exam Tip: Any non-zero number raised to the power of 0 is always equal to 1.

Question 7. Which is greater : (a) 3^6 or 6^3? (b) -(5^2) or (-5)^2?
Answer:
(a) \(3^6 = 729\) and \(6^3 = 216\). Since \(729 > 216\), \(3^6\) is greater.
(b) \(-(5^2) = -25\) and \((-5)^2 = 25\). Since \(25 > -25\), \((-5)^2\) is greater.
In simple words: Calculate the exact numerical value of both expressions to compare them accurately.

Exam Tip: Pay close attention to parentheses; \(-(5^2)\) applies the negative sign after squaring 5, whereas \((-5)^2\) squares the negative sign as well.

Question 8. Express each of the following as product of powers of prime factors: (a) 81 x 128 (b) 256 x 121
Answer:
(a) \(81 = 3^4\) and \(128 = 2^7\). Therefore, \(81 \times 128 = 3^4 \times 2^7\)
(b) \(256 = 2^8\) and \(121 = 11^2\). Therefore, \(256 \times 121 = 2^8 \times 11^2\)
In simple words: Find the prime factorization of each number separately and combine them using exponents.

Exam Tutor Tip: Always express composite numbers as powers of their prime factors (2, 3, 5, 7, 11, etc.).

Question 9. Find the value of x: a) \{(3)^3\}^7 = (3)^{7x} b) (-2)^{11} \div (-2)^9 = (-2)^{2x}
Answer:
a) \(\{(3)^3\}^7 = (3)^{21} = (3)^{7x}\)
Equating exponents: \(7x = 21 \implies x = 3\)
b) \((-2)^{11-9} = (-2)^2 = (-2)^{2x}\)
Equating exponents: \(2x = 2 \implies x = 1\)
In simple words: Simplify both sides so the bases are equal, then equate the exponents to solve for x.

Exam Tip: If the bases on both sides of an equation are equal, their exponents must also be equal.

Question 10. Using laws of exponents simplify: (a^5 \times b^3 \times b^6) \div (a^2 \times b^2)
Answer:
\(= \frac{a^5 \times b^{3+6}}{a^2 \times b^2} = \frac{a^5 \times b^9}{a^2 \times b^2}\)
\(= a^{5-2} \times b^{9-2} = a^3b^7\)
In simple words: Combine like bases in the numerator first, then subtract the denominator exponents.

Exam Tip: Group identical variables together before applying division rules.

Question 11. Express the following in standard form: (i) 782000000 (ii) 609.99 (iii) 0.234 \times 10^5 (iv) 5 million
Answer:
(i) \(782000000 = 7.82 \times 10^8\)
(ii) \(609.99 = 6.0999 \times 10^2\)
(iii) \(0.234 \times 10^5 = 2.34 \times 10^{-1} \times 10^5 = 2.34 \times 10^4\)
(iv) \(5 \text{ million} = 5,000,000 = 5 \times 10^6\)
In simple words: Move the decimal point so that only one non-zero digit remains to its left, and adjust the power of 10 accordingly.

Exam Tip: Standard form is always written as \(k \times 10^n\) where \(1 \le k < 10\) and n is an integer.

Question 12. Write in usual form: (i) 9.876 \times 10^6 (ii) 1.3 \times 10^9
Answer:
(i) \(9.876 \times 10^6 = 9,876,000\)
(ii) \(1.3 \times 10^9 = 1,300,000,000\)
In simple words: Shift the decimal point to the right by the number of places indicated by the positive exponent.

Exam Tip: Fill empty decimal places with zeros when shifting the decimal point.

Question 13. Convert into positive power and then solve: a) (5/6)^{-2} \times (5/6)^{-4} b) In the above question we have converted negative into positive. How can you convert your negative approach or attitude into positive approach. Give examples.
Answer:
a) Converting negative exponents to positive using \((a/b)^{-m} = (b/a)^m\):
\(= \left(\frac{6}{5}\right)^2 \times \left(\frac{6}{5}\right)^4\)
\(= \left(\frac{6}{5}\right)^{2+4} = \left(\frac{6}{5}\right)^6 = \frac{46656}{15625}\)
b) Just as negative exponents are flipped to become positive and useful, negative human attitudes can be flipped through optimism, learning from mistakes, and surrounding oneself with positive influences. For example, failing a test (a negative outcome) can be turned into a positive learning experience by analyzing errors and studying harder.
In simple words: Reciprocate the fraction to change the negative exponent to positive, then solve normally.

Exam Tip: Reciprocal rule states that \(a^{-m} = \left(\frac{1}{a}\right)^m\), which instantly turns negative powers positive.