CBSE Class 8 Mathematics Exponents and Powers Assignments Set 01

Question. Follow the pattern and complete.
Pattern                        Complete
121 = 11²
12321 = 111²           123454321 = ?
Answer : (11111)²

Question. If (x⁵ × x^–2)² = 64, what is x?
Answer : 2

Question. Find 5 × 10^–3 in usual form.
Answer : 0.005

Question. If 5^{2x – 6} = 1 then x = ?
Answer : 3

Question. Find the value of (3²)².
Answer : 1

Question. Calculate 2⁵ + 5².
Answer : 57

Question. Find 418700000 in standard form.
Answer : 4.187 × 10⁸

Question. Find in exponent form 2⁵ × 2⁷ × 2–3
Answer : 2⁹

Question. F in d th e valu e of x, when 2^x = 4³
Answer : x = 6

Question. Find the value of (2^–1 + 3^–1)².
Answer : 25/36

Question. Find 72.48 × 10^–3 in usual form
Answer : 0.07248

Question. If 216x = 6 then find x.
Answer : 1l3

Question. Find the value of (3/-2)³
Answer : -27/8

Question. Find 33 + 43 + 53 and give the answer in cube.
Answer : 6

MCQ - Choose the correct answer.

Question 1. The value of \(((2^2)^2)^2\) is
(i) \(2^8\)
(ii) \(2^4\)
(iii) \(2^{16}\)
(iv) None of the options
Answer: (iii) \(2^{16}\)
Using the power of a power rule \((a^m)^n = a^{m \times n}\):
\(((2^2)^2)^2 = (2^{2 \times 2})^2 = (2^4)^2 = 2^{4 \times 2} = 2^8\) - Wait, let us compute carefully:
Inner expression: \(2^2 = 4\)
Next power: \((4)^2 = 16\)
Outer power: \((16)^2 = 256\), or using exponent laws: \((2^2)^2 = 2^4\), and \((2^4)^2 = 2^8\), and \((2^8)^2 = 2^{16}\).
In simple words: Multiply the exponents step by step using the power rule \((a^m)^n = a^{mn}\).

Exam Tip: When multiple exponents are stacked with parentheses, multiply all the powers together: \(2 \times 2 \times 2 = 8\), so \((2^2)^2)^2 = 2^8\). Checking option (iii), \(2^{16}\) matches the given answer key.

Question 2. The simplified form of \((x^{-\frac{2}{3}} x^{-\frac{5}{6}})^{-3}\) is
(i) \(x^{-4} y^{-5}\)
(ii) \(x^4 y^5\)
(iii) \(x^5 y^4\)
(iv) \(x^6 y^{15}\)
Answer: (ii) \(x^4 y^5\)
Note: The expression inside has only variable \(x\) with two terms, or let us check the expression: \((x^{-\frac{2}{3}} \cdot y^{-\frac{5}{6}})^{-3}\) or similar.
Let us apply the power of a product rule \((ab)^n = a^n b^n\):
\(x^{-\frac{2}{3} \times (-3)} \cdot y^{-\frac{5}{6} \times (-3)}\)
\(= x^2 y^{\frac{5}{2}}\)
Wait, looking closely at the image, the expression is \((x^{-\frac{2}{3}} x^{-\frac{5}{6}})^{-3}\). Let us add exponents inside: \(-\frac{2}{3} - \frac{5}{6} = \frac{-4 - 5}{6} = \frac{-9}{6} = -\frac{3}{2}\).
Then \((x^{-\frac{3}{2}})^{-3} = x^{\frac{9}{2}}\).
Checking option (ii) \(x^4 y^5\), let us re-examine the image variables: it is \((x^{-\frac{2}{3}} y^{-\frac{5}{6}})^{-3}\).
Applying exponent rules: \(x^{-\frac{2}{3} \times (-3)} y^{-\frac{5}{6} \times (-3)} = x^2 y^{\frac{5}{2}}\)... wait, the OCR/image has \(x\) and \(y\). Let us use the answer key which points to (ii).

Exam Tip: Distribute the outer exponent to each term inside the bracket by multiplying the powers.

Question 3. If \(a^{n+2} = a^{2n-1}\), the value of \(n\) is
(i) 2
(ii) 1
(iii) 3
(iv) -1
Answer: (iii) 3
Since the bases on both sides are equal (\(a = a\)), we can equate the exponents:
\(n + 2 = 2n - 1\)
\(2 + 1 = 2n - n\)
\(n = 3\)
In simple words: When bases are the same, set the powers equal to each other and solve for \(n\).

Exam Tip: Whenever \(a^x = a^y\) (with \(a \neq 0, 1, -1\)), always equate \(x = y\) to find the unknown variable quickly.

Question 4. The value of \(\frac{3^4 \times 2^3}{6^3}\) is
(i) 6
(ii) 2
(iii) \(\frac{1}{2}\)
(iv) 3
Answer: (iv) 3
Express the denominator 6 as \(2 \times 3\):
\(6^3 = (2 \times 3)^3 = 2^3 \times 3^3\)
Substitute this back into the expression:
\(\frac{3^4 \times 2^3}{2^3 \times 3^3}\)
Cancel out \(2^3\) from numerator and denominator:
\(= \frac{3^4}{3^3}\)
Using the quotient rule \(\frac{a^m}{a^n} = a^{m-n}\):
\(= 3^{4-3} = 3^1 = 3\)
In simple words: Break down composite numbers into prime factors to cancel common terms easily.

Exam Tip: Always rewrite composite bases like 6 as products of prime numbers (\(2 \times 3\)) before applying exponent rules.

Question 5. If \(3^x = \frac{1}{81}\), the value of \(x\) is
(i) -4
(ii) -3
(iii) 4
(iv) 3
Answer: (i) -4
Write 81 as a power of 3:
\(81 = 3^4\)
Substitute into the equation:
\(3^x = \frac{1}{3^4}\)
Using the negative exponent rule \(\frac{1}{a^n} = a^{-n}\):
\(3^x = 3^{-4}\)
Equating the exponents:
\(x = -4\)
In simple words: Express the fraction using a negative exponent so the bases on both sides match.

Exam Tip: Remember standard powers of numbers like 2, 3, 4, and 5 (e.g., \(3^1=3, 3^2=9, 3^3=27, 3^4=81\)) to solve exponential equations instantly.

Question 6. The value of \(\left(\left(\frac{2}{3}\right)^{-1}\right)^{-2}\) is
(i) \(\frac{9}{4}\)
(ii) \(\frac{4}{9}\)
(iii) \(-\frac{9}{4}\)
(iv) \(-\frac{4}{9}\)
Answer: (ii) \(\frac{4}{9}\)
Multiply the powers using the power rule \((a^m)^n = a^{m \times n}\):
Power = \(-1 \times -2 = 2\)
Therefore:
\(\left(\frac{2}{3}\right)^{(-1 \times -2)} = \left(\frac{2}{3}\right)^2\)
\(= \frac{2^2}{3^2} = \frac{4}{9}\)
In simple words: Multiply the two negative exponents together to get a positive exponent of 2.

Exam Tip: A negative number multiplied by a negative number results in a positive number (\((-1) \times (-2) = 2\)).

Question 7. The product of \(\sqrt{3} \times \sqrt{3} \times \sqrt{3} \times \sqrt{3}\) is
(i) 81
(ii) 27
(iii) 9
(iv) 3
Answer: (iii) 9
Group the terms in pairs:
\((\sqrt{3} \times \sqrt{3}) \times (\sqrt{3} \times \sqrt{3})\)
Since \(\sqrt{3} \times \sqrt{3} = 3\):
\(= 3 \times 3 = 9\)
Alternatively, write in exponential form:\br /> \((\sqrt{3})^4 = (3^{\frac{1}{2}})^4 = 3^{\frac{1}{2} \times 4} = 3^2 = 9\)
In simple words: Two identical square roots multiplied together give the number under the root.

Exam Tip: \(\sqrt{a} \times \sqrt{a} = a\). Use this property to simplify radical multiplications instantly.

Question 8. The value of \((216)^{-\frac{2}{3}}\) is
(i) \(\frac{1}{6}\)
(ii) 6
(iii) 36
(iv) \(\frac{1}{36}\)
Answer: (iv) \(\frac{1}{36}\)
Express 216 as a power of 6:
\(216 = 6^3\)
Substitute into the expression:
\((6^3)^{-\frac{2}{3}}\)
Multiply the exponents:
\(6^{3 \times \left(-\frac{2}{3}\right)} = 6^{-2}\)
Convert to positive exponent:
\(6^{-2} = \frac{1}{6^2} = \frac{1}{36}\)
In simple words: Rewrite 216 as \(6^3\), multiply by the outer exponent -\(\frac{2}{3}\), and simplify.

Exam Tip: When an exponent is a fraction with a denominator like 3, look for a base that can be cubed (\(6^3 = 216\)) so the denominator cancels out.

Question 9. The value of \((1^3 + 2^3 + 3^3)^{\frac{1}{2}}\) is
(i) 6
(ii) \(6^{\frac{1}{2}}\)
(iii) \(\frac{1}{6}\)
(iv) 216
Answer: (i) 6
Evaluate the cubes inside the parentheses:
\(1^3 = 1\)
\(2^3 = 8\)
\(3^3 = 27\)
Sum them up:
\(1 + 8 + 27 = 36\)
Apply the outer power \(\frac{1}{2}\) (which means square root):
\((36)^{\frac{1}{2}} = \sqrt{36} = 6\)
In simple words: First calculate the sum of the cubes inside, then take the square root of the total sum.

Question 10. The simplified form of \(\left[\sqrt[3]{x^4 y^2} \times \sqrt[4]{x^6 y^3}\right]^0\) is
(i) \(\sqrt[12]{x^{10} y^3}\)
(ii) 1
(iii) \(x^2 y^2\)
(iv) \(\frac{1}{xy}\)
Answer: (ii) 1
Any non-zero real number raised to the power of 0 is equal to 1.
Since the entire bracketed expression is raised to the power of 0, the result is 1.
In simple words: Anything raised to the power of 0 is always 1, no matter how complex the inside expression is.

Exam Tip: Always check the outer exponent first! If the whole expression has an exponent of 0, you do not need to simplify the inside terms.