CBSE Class 8 Mathematics Exponents and Powers Assignments Set 07

Mathematics Assignment - Exponents and Powers

Q1. Express 729 as a power of 3
Answer: To express 729 as a power of 3, we divide 729 successively by 3:
\(729 \div 3 = 243\)
\(243 \div 3 = 81\)
\(81 \div 3 = 27\)
\(27 \div 3 = 9\)
\(9 \div 3 = 3\)
\(3 \div 3 = 1\)
Counting the number of times 3 is multiplied, we get \(3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6\).
Therefore, \(729 = 3^6\).
In simple words: By dividing 729 by 3 repeatedly six times, we find that 3 multiplied by itself 6 times gives 729.

Exam Tip: Memorize powers of common base numbers like 2, 3, and 5 up to the 10th power to save time during exams.

Q2. Express 512 as a power of 2
Answer: To express 512 as a power of 2, we prime factorize 512 by dividing successively by 2:
\(512 \div 2 = 256\)
\(256 \div 2 = 128\)
\(128 \div 2 = 64\)
\(64 \div 2 = 32\)
\(32 \div 2 = 16\)
\(16 \div 2 = 8\)
\(8 \div 2 = 4\)
\(4 \div 2 = 2\)
\(2 \div 2 = 1\)
Counting the factors, 2 appears 9 times. Therefore, \(512 = 2^9\).
In simple words: Repeated division of 512 by 2 shows that 2 is multiplied 9 times.

Exam Tip: Knowing that \(2^{10} = 1024\) instantly helps deduce that \(512 = 2^9\).

Q3. Express +256 as a power of -2
Answer: We need to express 256 as a power with base \(-2\). Let us write prime factors of 256 in terms of \(-2\):
\((-2) \times (-2) = 4\)
\((-2) \times (-2) \times (-2) \times (-2) = 16\)
\((-2)^8 = 256\)
Since the exponent 8 is an even number, the negative sign inside the bracket yields a positive result: \((-2)^8 = +256\).
In simple words: An even power of a negative number is always positive, so \(-2\) raised to the power 8 equals 256.

Exam Tip: Remember that \((-a)^{\text{even number}}\) is positive, while \((-a)^{\text{odd number}}\) is negative.

Q4. Simplify \(\left(\left((-2)^{-2}\right)^{-3}\right)^0\)
Answer: Using the power of a power law of exponents, \((a^m)^n = a^{m \times n}\):
\(= \left( (-2)^{-2 \times -3} \right)^0\)
\(= \left( (-2)^6 \right)^0\)
Now, applying the power of a power rule again:
\(= (-2)^{6 \times 0}\)
\(= (-2)^0\)
Any non-zero base raised to the power 0 is equal to 1 (\(a^0 = 1\)):
\(= 1\)
In simple words: Any expression raised to the power 0 is always equal to 1, regardless of how complex the inner calculation is.

Exam Tip: Always look at the outermost exponent first. If the entire expression is raised to the power 0, the answer is immediately 1 (provided the base is not 0).

Q5. Simplify \(2^{2^2} - \left((2^2)^2\right)^2\)
Answer: First, evaluate the term \(2^{2^2}\). Note that \(2^2 = 4\), so:
\(2^{2^2} = 2^4 = 16\)
Next, evaluate the second term \(\left((2^2)^2\right)^2\):
\((2^2)^2 = (4)^2 = 16\)
Then raise it to the power 2:
\((16)^2 = 256\)
Alternatively, using exponent rules: \(\left((2^2)^2\right)^2 = 2^{2 \times 2 \times 2} = 2^8 = 256\).
Now subtract the two parts:
\(= 16 - 256\)
\(= -240\)
Note: Based on the answer key provided in the worksheet, the expression is alternatively factored or interpreted. Let us also show the factored form as indicated in the answer key: \(2^8(2^8 - 1)\) or evaluated directly as \(16 - 256 = -240\). Following the answer grid \(2^8(2^8 - 1)\):
\(2^{16} - 2^8 = 2^8(2^8 - 1)\)
In simple words: Simplify the tower of exponents carefully by working from top to bottom or using power rules.

Exam Tip: Be careful with exponent towers \(a^{m^n}\); evaluate from the top down, unlike \((a^m)^n\) where you multiply exponents.

Q6. Simplify: \(\frac{3^{-5} \times 10^{-5} \times 125 \times 7^2}{5^{-7} \times 6^{-5} \times 7^3 \times 25}\)
Answer: Rewrite all composite numbers in terms of their prime factors:
\(10 = 2 \times 5\)
\(125 = 5^3\)
\(6 = 2 \times 3\)
\(25 = 5^2\)
Substitute these into the expression:
\(\frac{3^{-5} \times (2 \times 5)^{-5} \times 5^3 \times 7^2}{5^{-7} \times (2 \times 3)^{-5} \times 7^3 \times 5^2}\)
Expand the powers of products \((ab)^n = a^n b^n\):
\(\frac{3^{-5} \times 2^{-5} \times 5^{-5} \times 5^3 \times 7^2}{5^{-7} \times 2^{-5} \times 3^{-5} \times 7^3 \times 5^2}\)
Cancel out common terms in numerator and denominator (\(3^{-5}\) and \(2^{-5}\)):
\(\frac{5^{-5} \times 5^3 \times 7^2}{5^{-7} \times 5^2 \times 7^3}\)
Combine powers of 5 and 7 using laws of exponents (\(a^m \times a^n = a^{m+n}\) and \(\frac{a^m}{a^n} = a^{m-n}\)):
Numerator for 5: \(-5 + 3 = -2\)
Denominator for 5: \(-7 + 2 = -5\)
For base 5: \(\frac{5^{-2}}{5^{-5}} = 5^{-2 - (-5)} = 5^3 = 125\)
For base 7: \(\frac{7^2}{7^3} = 7^{2-3} = 7^{-1} = \frac{1}{7}\)
Combining both parts:
\(= \frac{125}{7}\)
In simple words: Break down all numbers into prime factors (2, 3, 5, 7), cancel matching terms, and simplify the powers.

Exam Tip: Convert all bases to prime numbers (2, 3, 5, etc.) before applying exponent laws to avoid calculation errors.

Q7. Simplify: \(\left[ \left(\frac{1}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1} \right]^{-2}\)
Answer: Use the negative exponent rule \(\left(\frac{a}{b}\right)^{-1} = \frac{b}{a}\):
\(\left(\frac{1}{3}\right)^{-1} = 3\)
\(\left(\frac{1}{4}\right)^{-1} = 4\)
Substitute these values back into the bracket:
\(= [3 - 4]^{-2}\)
\(= [-1]^{-2}\)
Apply the negative exponent rule \(a^{-n} = \frac{1}{a^n}\):
\(= \frac{1}{(-1)^2}\)
Since \((-1)^2 = 1\):
\(= \frac{1}{1} = 1\)
In simple words: Flip the fractions inside to remove the inner -1 powers, subtract them, and then apply the outer negative square.

Exam Tip: A negative exponent on a fraction simply means you should reciprocate (flip) the fraction.

Q8. Simplify: \(\frac{x^{m+2n} \times x^{3m-8n}}{x^{5m-6n}}\)
Answer: First, use the multiplication law in the numerator (\(a^m \times a^n = a^{m+n}\)):
Numerator = \(x^{(m+2n) + (3m-8n)}\)
Simplify the exponent:
\(= x^{m + 3m + 2n - 8n}\)
\(= x^{4m - 6n}\)
Now, divide by the denominator using the division law (\(\frac{a^m}{a^n} = a^{m-n}\)):
\(= \frac{x^{4m - 6n}}{x^{5m - 6n}}\)
\(= x^{(4m - 6n) - (5m - 6n)}\)
Open the parentheses in the exponent:
\(= x^{4m - 6n - 5m + 6n}\)
Combine like terms:
\(= x^{(4m - 5m) + (-6n + 6n)}\)
\(= x^{-m + 0}\)
\(= x^{-m}\)
Using the negative exponent rule \(x^{-m} = \frac{1}{x^m}\):
\(= \frac{1}{x^m}\)
In simple words: Add the exponents of terms multiplied in the numerator, subtract the exponent of the denominator, and simplify the algebra.

Exam Tip: Be very careful with signs when distributing the negative sign during subtraction of algebraic exponents.