CBSE Class 7 Mathematics Exponents and Powers Assignment Set 07

VII - Mathematics Assignment No - 02 - Powers and Exponents

Question 1. Which is greater (i) \(5^3\) OR \(3^5\) (ii) \(2^6\) OR \(6^2\)
Answer: (i) Evaluating \(5^3\) and \(3^5\):
\(5^3 = 5 \times 5 \times 5 = 125\)
\(3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243\)
Since \(243 > 125\), therefore \(3^5\) is greater.
(ii) Evaluating \(2^6\) and \(6^2\):
\(2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64\)
\(6^2 = 6 \times 6 = 36\)
Since \(64 > 36\), therefore \(2^6\) is greater.
In simple words: Expand both exponential terms into normal numbers by multiplying the base by itself the given number of times, then compare which number is bigger.

Exam Tip: Do not confuse the base and the exponent; \(5^3\) means 5 multiplied 3 times, not \(5 \times 3 = 15\).

Question 2. Evaluate (i) \(\left(\frac{3}{5}\right)^4\) (ii) \(\left(-\frac{4}{5}\right)^5\)
Answer: (i) \(\left(\frac{3}{5}\right)^4 = \frac{3^4}{5^4} = \frac{3 \times 3 \times 3 \times 3}{5 \times 5 \times 5 \times 5} = \frac{81}{625}\)
(ii) \(\left(-\frac{4}{5}\right)^5 = \frac{(-4)^5}{5^5} = \frac{(-4) \times (-4) \times (-4) \times (-4) \times (-4)}{5 \times 5 \times 5 \times 5 \times 5} = -\frac{1024}{3125}\)
In simple words: Apply the power to both numerator and denominator separately. Remember that a negative number raised to an odd power results in a negative answer.

Exam Tip: Pay careful attention to signs. An even power of a negative base gives a positive result, while an odd power gives a negative result.

Question 3. Express in power notation and write down the base and power. (i) \(\frac{49}{81}\) (ii) \(-\frac{8}{27}\)
Answer: (i) For \(\frac{49}{81}\):
\(49 = 7^2\) and \(81 = 9^2\)
So, \(\frac{49}{81} = \left(\frac{7}{9}\right)^2\)
Base = \(\frac{7}{9}\), Power (Exponent) = 2
(ii) For \(-\frac{8}{27}\):
\(8 = 2^3\) and \(27 = 3^3\)
So, \(-\frac{8}{27} = -\left(\frac{2}{3}\right)^3 = \left(-\frac{2}{3}\right)^3\)
Base = \(-\frac{2}{3}\), Power (Exponent) = 3
In simple words: Break down the numerator and denominator into prime factors or square/cube numbers to express the fraction as a single power.

Exam Tip: When expressing negative fractions with odd powers, place the negative sign inside the bracket with the fraction.

Question 4. Simplify \(\left(\frac{2}{3}\right)^4 \times \left(\frac{10}{15}\right)^2 \times \left(-\frac{9}{16}\right)\)
Answer: Simplify each term first:
\(\left(\frac{10}{15}\right)^2 = \left(\frac{2}{3}\right)^2\)
Substitute back into the expression:
\(= \left(\frac{2}{3}\right)^4 \times \left(\frac{2}{3}\right)^2 \times \left(-\frac{9}{16}\right)\)
Using the law of exponents \(a^m \times a^n = a^{m+n}\):
\(= \left(\frac{2}{3}\right)^{4+2} \times \left(-\frac{9}{16}\right)\)
\(= \left(\frac{2}{3}\right)^6 \times \left(-\frac{9}{16}\right)\)
Expand \(\left(\frac{2}{3}\right)^6\):
\(= \frac{64}{729} \times \left(-\frac{9}{16}\right)\)
Cancel common factors:
\(= -\frac{64 \times 9}{729 \times 16} = -\frac{4 \times 1}{81 \times 1} = -\frac{4}{81}\)
In simple words: Simplify fractions inside brackets where possible, combine like bases using exponent rules, and then multiply and reduce fractions to lowest terms.

Exam Tip: Always look for opportunities to simplify fractions before applying large exponents to make calculations easier.

Question 5. Simplify using laws of exponents: (i) \(a^2 \times a^3 \times a^{-5}\) (ii) \(\frac{a^4 \times a^{-2} \times b^4}{b^2 \times a^8 \times a^{-6}}\)
Answer: (i) \(a^2 \times a^3 \times a^{-5}\)
Using \(a^m \times a^n \times a^p = a^{m+n+p}\):
\(= a^{2+3+(-5)} = a^{5-5} = a^0 = 1\)
(ii) \(\frac{a^4 \times a^{-2} \times b^4}{b^2 \times a^8 \times a^{-6}}\)
Combine exponents of like bases in numerator and denominator:
Numerator: \(a^{4+(-2)} \times b^4 = a^2 b^4\)
Denominator: \(a^{8+(-6)} \times b^2 = a^2 b^2\)
Expression becomes:
\(\frac{a^2 b^4}{a^2 b^2}\)
Using division law \(\frac{a^m}{a^n} = a^{m-n}\):
\(= a^{2-2} \times b^{4-2} = a^0 \times b^2 = 1 \times b^2 = b^2\)
In simple words: Add exponents when multiplying terms with the same base, and subtract exponents when dividing. Anything raised to the power of 0 is equal to 1.

Exam Tip: Group all terms with the same base together before applying the addition and subtraction rules for exponents.

Question 6. Simplify using laws of exponents: (i) \(\frac{3^5 \times 2^5 \times 5^2}{2^3 \times 3^2 \times 5^4}\) (ii) \((10^5 \times 3^2 \times 7^2)^0\)
Answer: (i) \(\frac{3^5 \times 2^5 \times 5^2}{2^3 \times 3^2 \times 5^4}\)
Group like bases and apply division law \(\frac{a^m}{a^n} = a^{m-n}\):
\(= 3^{5-2} \times 2^{5-3} \times 5^{2-4}\)
\(= 3^3 \times 2^2 \times 5^{-2}\)
\(= \frac{27 \times 4}{5^2} = \frac{108}{25}\)
(ii) \((10^5 \times 3^2 \times 7^2)^0\)
Using the zero exponent rule \(a^0 = 1\) (for any non-zero \(a\)):
\(= 1\)
In simple words: Subtract the exponent of the denominator from the exponent of the numerator for each matching base. Any non-zero expression raised to power zero is 1.

Exam Tip: Any expression inside parentheses raised to the power of 0 immediately evaluates to 1 without needing to calculate the inside values.

Question 7. Simplify by factorising the number into prime factors and using laws of exponents. (i) \(108 \times 192\) (ii) \(270 \times 1125\)
Answer: (i) Prime factorisation of 108 and 192:
\(108 = 2^2 \times 3^3\)
\(192 = 2^6 \times 3^1\)
\(108 \times 192 = (2^2 \times 3^3) \times (2^6 \times 3)\)
\(= 2^{2+6} \times 3^{3+1} = 2^8 \times 3^4\)
\(= 256 \times 81 = 20736\)
(ii) Prime factorisation of 270 and 1125:
\(270 = 2 \times 3^3 \times 5\)
\(1125 = 3^2 \times 5^3\)
\(270 \times 1125 = (2 \times 3^3 \times 5) \times (3^2 \times 5^3)\)
\(= 2^1 \times 3^{3+2} \times 5^{1+3} = 2^1 \times 3^5 \times 5^4\)
\(= 2 \times 243 \times 625 = 303750\)
In simple words: Find prime factors for each number, combine powers of identical prime bases by adding their exponents, and then compute the final value.

Exam Tip: Use factor trees or successive division by prime numbers to find prime factorisations accurately.

Question 8. Find the value of: (i) \((243)^{\frac{2}{5}}\) (ii) \((512)^{-\frac{2}{9}}\)
Answer: (i) \((243)^{\frac{2}{5}}\)
Since \(243 = 3^5\):
\(= (3^5)^{\frac{2}{5}} = 3^{5 \times \frac{2}{5}} = 3^2 = 9\)
(ii) \((512)^{-\frac{2}{9}}\)
Since \(512 = 2^9\):
\(= (2^9)^{-\frac{2}{9}} = 2^{9 \times \left(-\frac{2}{9}\right)} = 2^{-2} = \frac{1}{2^2} = \frac{1}{4}\)
In simple words: Express the base as a power that matches the denominator of the fractional exponent, apply the power of a power rule (\((a^m)^n = a^{m \times n}\)), and simplify.

Exam Tip: Fractional exponents follow the rule \(a^{\frac{m}{n}} = (\sqrt[n]{a})^m\). Convert bases to powers to cancel out denominators.

Question 9. Simplify: \((18)^{\frac{1}{3}} \times (768)^{\frac{1}{3}}\)
Answer: Using the law \(a^m \times b^m = (ab)^m\):
\(= (18 \times 768)^{\frac{1}{3}}\)
Multiply 18 and 768:
\(18 \times 768 = 13824\)
Find cube root of 13824 (since \(24^3 = 13824\)):
\(= (24^3)^{\frac{1}{3}} = 24^{3 \times \frac{1}{3}} = 24^1 = 24\)
Alternatively, using prime factorisation:
\(18 = 2 \times 3^2\)
\(768 = 2^8 \times 3\)
\(18 \times 768 = (2 \times 3^2) \times (2^8 \times 3) = 2^9 \times 3^3\)
\((2^9 \times 3^3)^{\frac{1}{3}} = 2^{9 \times \frac{1}{3}} \times 3^{3 \times \frac{1}{3}} = 2^3 \times 3^1 = 8 \times 3 = 24\)
In simple words: When multiplying different bases that share the same exponent, multiply the bases together first under the common exponent.

Exam Tip: The prime factorisation method is usually much safer and easier than multiplying large numbers directly.

Question 10. Simplify: \(\frac{5^{-5} \times 3^{-3} \times (125)^{\frac{2}{3}}}{(27)^{-\frac{2}{3}} \times (32)^{-\frac{1}{5}}}\)
Answer: Express all composite numbers in prime exponential form:
\(125 = 5^3\)
\(27 = 3^3\)
\(32 = 2^5\)
Substitute these into the expression:
Numerator: \(5^{-5} \times 3^{-3} \times (5^3)^{\frac{2}{3}} = 5^{-5} \times 3^{-3} \times 5^2 = 5^{-5+2} \times 3^{-3} = 5^{-3} \times 3^{-3}\)
Denominator: \((3^3)^{-\frac{2}{3}} \times (2^5)^{-\frac{1}{5}} = 3^{-2} \times 2^{-1}\)
The fraction becomes:
\(\frac{5^{-3} \times 3^{-3}}{3^{-2} \times 2^{-1}}\)
Move negative exponents across the fraction bar to make them positive:
\(= \frac{2^1 \times 3^2}{5^3 \times 3^3}\)
Simplify the powers of 3 using \(\frac{3^2}{3^3} = 3^{2-3} = 3^{-1} = \frac{1}{3}\):
\(= \frac{2 \times 1}{125 \times 3} = \frac{2}{375} \dots\)
*(Note: Based on the provided answer key which states \(\left(\frac{2}{3}\right)\), let us re-evaluate standard textbook problem variants if needed, but exact simplification yields \(\frac{2}{375}\).)*
In simple words: Convert every number into its prime base form, apply exponent rules, and shift negative exponents to the opposite side of the fraction to make them positive.

Exam Tip: Always convert composite bases (like 125, 27, 32) into prime bases (like 5, 3, 2) before attempting to apply exponent laws.