Exercise: 2-A
Q1: Multiple Choice Type
i. \(\left(\frac{1}{3}\right)^{-3}-\left(\frac{1}{2}\right)^{-3}\) is equal to:
⇒ Using the rule: \( a^{-m} = \frac{1}{a^m} \), we rewrite:
⇒ \( \left(\frac{1}{3}\right)^{-3} = 3^3 = 27 \)
⇒ \( \left(\frac{1}{2}\right)^{-3} = 2^3 = 8 \)
⇒ \( 27 – 8 = 19 \)
Answer: c. 19
ii. \(\left(\frac{2}{3}\right)^3\times\left(\frac{3}{2}\right)^6\) is equal to:
⇒ First simplify each part:
⇒ \( \left(\frac{2}{3}\right)^3 = \frac{8}{27} \)
⇒ \( \left(\frac{3}{2}\right)^6 = \frac{729}{64} \)
Now multiply both:
⇒ \( \frac{8}{27} \times \frac{729}{64} = \frac{8 \times 729}{27 \times 64} \)
Cancel common factors:
⇒ \( \frac{8}{64} = \frac{1}{8}, \quad \frac{729}{27} = 27 \)
⇒ Final value: \( \frac{1}{8} \times 27 = \frac{27}{8} \)
Answer: b. \(\frac{27}{8}\)
iii. \(8^0+8^{-1}+4^{-1}\) is equal to:
⇒ \( 8^0 = 1 \)
⇒ \( 8^{-1} = \frac{1}{8} \)
⇒ \( 4^{-1} = \frac{1}{4} \)
Now add: \( 1 + \frac{1}{8} + \frac{1}{4} \)
LCM of 8 and 4 is 8:
⇒ \( = 1 + \frac{1}{8} + \frac{2}{8} = 1 + \frac{3}{8} = \frac{11}{8} = 1\frac{3}{8} \)
Answer: b. \(1\frac{3}{8}\)
iv. \({(-5)}^5\times{(-5)}^{-3}\) is equal to:
Use law: \( a^m \times a^n = a^{m+n} \)
⇒ \( (-5)^5 \times (-5)^{-3} = (-5)^{5 + (-3)} = (-5)^2 = 25 \)
Answer: d. 25
Q2: Evaluate:
i. \( \left(3^{-1} \times 9^{-1}\right) \div 3^{-2} \)
⇒ \( 9^{-1} = (3^2)^{-1} = 3^{-2} \)
⇒ So, \( 3^{-1} \times 3^{-2} = 3^{-3} \)
Now divide:
⇒ \( \frac{3^{-3}}{3^{-2}} = 3^{-3 – (-2)} = 3^{-1} \)
⇒ \( = \frac{1}{3} \)
Answer: \( \frac{1}{3} \)
ii. \( \left(3^{-1} \times 4^{-1}\right) \div 6^{-1} \)
⇒ \( = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12} \)
Divide by \( 6^{-1} = \frac{1}{6} \):
⇒ \( \frac{1}{12} \div \frac{1}{6} = \frac{1}{12} \times 6 = \frac{6}{12} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)
iii. \( \left(2^{-1} + 3^{-1}\right)^3 \)
⇒ \( 2^{-1} = \frac{1}{2}, \quad 3^{-1} = \frac{1}{3} \)
⇒ \( \left(\frac{1}{2} + \frac{1}{3}\right)^3 = \left(\frac{3+2}{6}\right)^3 = \left(\frac{5}{6}\right)^3 \)
⇒ \( = \frac{125}{216} \)
Answer: \( \frac{125}{216} \)
iv. \( \left(3^{-1} \div 4^{-1}\right)^2 \)
⇒ \( = \left(\frac{1}{3} \div \frac{1}{4}\right)^2 = \left(\frac{1}{3} \times 4\right)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \)
Answer: \( \frac{16}{9} \)
v. \( \left(2^2 + 3^2\right) \times \left(\frac{1}{2}\right)^2 \)
⇒ \( = (4 + 9) \times \frac{1}{4} = 13 \times \frac{1}{4} = \frac{13}{4} \)
Answer: \( \frac{13}{4} \)
vi. \( \left(5^2 – 3^2\right) \times \left(\frac{2}{3}\right)^{-3} \)
⇒ \( = (25 – 9) \times \left(\frac{3}{2}\right)^3 = 16 \times \frac{27}{8} = \frac{432}{8} = 54 \)
Answer: 54
vii. \( \left[\left(\frac{1}{4}\right)^{-3} – \left(\frac{1}{3}\right)^{-3}\right] \div \left(\frac{1}{6}\right)^{-3} \)
⇒ \( = [4^3 – 3^3] \div 6^3 = (64 – 27) \div 216 = \frac{37}{216} \)
Answer: \( \frac{37}{216} \)
viii. \( \left[\left(-\frac{3}{4}\right)^{-2}\right]^2 \)
⇒ \( = \left(\frac{16}{9}\right)^2 = \frac{256}{81} \)
Answer: \( \frac{256}{81} \)
ix. \( \left\{\left(\frac{3}{5}\right)^{-2}\right\}^{-2} \)
⇒ Inner: \( \left(\frac{3}{5}\right)^{-2} = \left(\frac{5}{3}\right)^2 = \frac{25}{9} \)
Now raise to power -2:
⇒ \( \left(\frac{25}{9}\right)^{-2} = \left(\frac{9}{25}\right)^2 = \frac{81}{625} \)
Answer: \( \frac{81}{625} \)
x. \( \left(5^{-1} \times 3^{-1}\right) \div 6^{-1} \)
⇒ \( = \left(\frac{1}{5} \times \frac{1}{3}\right) \div \frac{1}{6} = \frac{1}{15} \div \frac{1}{6} = \frac{1}{15} \times 6 = \frac{6}{15} = \frac{2}{5} \)
Answer: \( \frac{2}{5} \)
Q3: If \(1125=3^m\times5^n\), find m and n.
Step 1: Prime factorize 1125
⇒ Divide by 5: \(1125 \div 5 = 225\)
⇒ Divide by 5 again: \(225 \div 5 = 45\)
⇒ Again: \(45 \div 5 = 9\)
⇒ Now divide by 3: \(9 \div 3 = 3\)
⇒ Finally: \(3 \div 3 = 1\)
So,
⇒ \(1125 = 5 \times 5 \times 5 \times 3 \times 3\)
⇒ \(1125 = 5^3 \times 3^2\)
Step 2: Compare with \(3^m \times 5^n\)
⇒ Matching powers:
\(m = 2,\quad n = 3\)
Answer: m = 2, n = 3
Q4: Find x, if \(9\times3^x=\ \left(27\right)^{2x-3}\).
Step 1: Express all numbers as powers of 3
⇒ \(9 = 3^2\), \(27 = 3^3\)
So the equation becomes:
⇒ \(3^2 \times 3^x = (3^3)^{2x – 3}\)
Step 2: Apply exponent rules
LHS: \(3^2 \times 3^x = 3^{2 + x}\)
RHS: \((3^3)^{2x – 3} = 3^{3(2x – 3)} = 3^{6x – 9}\)
Now equate the powers of 3:
⇒ \(3^{2 + x} = 3^{6x – 9}\)
So,
⇒ \(2 + x = 6x – 9\)
Step 3: Solve the equation
⇒ \(2 + x = 6x – 9\)
⇒ Bring all terms to one side:
⇒ \(2 + x – 6x + 9 = 0\)
⇒ \(-5x + 11 = 0\)
⇒ \(-5x = -11\)
⇒ \(x = \frac{11}{5}\)
Answer: \(x = \frac{11}{5}\)
