Indices

घाताङ्क (Indices)

Algebra is used to solve various real-life problems. For example, if the growth of bacteria is such that its number doubles every 20 minutes, what will be the number of bacteria after 60 minutes? In such situations, we apply the concept of exponential growth using algebra.

🦠 🦠 (\(2^1\))

🦠🦠 🦠🦠 (\(2^2\))

🦠🦠🦠🦠 🦠🦠🦠🦠 🦠🦠🦠🦠 🦠🦠🦠🦠 (\(2^4\))

\(2^4\) (4 is exponent, 2 is base)

Exponent (or power or indices) is a number that indicates how many times a base number is multiplied by itself. It is written as a small number (the exponent) to the upper right of a base number. If we have a base \(2\) and an exponent \(n\), then
\(2^n\)
means that \(2\) is multiplied by itself \(n\) times.

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Product law of indices

Let \( a \neq 0 \) and \( m, n \in \mathbb{Z} \), then \(a^m \cdot a^n = a^{m+n}\)

Product law

Proof
By definition
\(a^m = \underbrace{a \cdot a \cdot \ldots \cdot a}_{m \text{ times}}\) and \(a^n = \underbrace{a \cdot a \cdot \ldots \cdot a}_{n \text{ times}}\)
Therefore
\(a^m \cdot a^n = \underbrace{a \cdot a \cdot \ldots \cdot a}_{m \text{ times}} \cdot \underbrace{a \cdot a \cdot \ldots \cdot a}_{n \text{ times}}=\underbrace{a \cdot a \cdot \ldots \cdot a}_{m+n \text{ times}} = a^{m+n}\)

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Quotient law of indices

Let \( a \neq 0 \) and \( m, n \in \mathbb{Z} \), then \(\frac{a^m}{a^n} = a^{m-n}\)

Product law

Proof
By definition
\(a^m = \underbrace{a \cdot a \cdot \ldots \cdot a}_{m \text{ times}}\) and \(a^n = \underbrace{a \cdot a \cdot \ldots \cdot a}_{n \text{ times}}\)
Therefore
\(\frac{a^m}{a^n} = \frac{\underbrace{a \cdot a \cdot \ldots \cdot a}_{m \text{ times}}}{ \underbrace{a \cdot a \cdot \ldots \cdot a}_{n \text{ times}}} =\underbrace{a \cdot a \cdot \ldots \cdot a}_{m-n \text{ times}} = a^{m-n}\)

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Power of a Power law of indices

Let \( a \neq 0 \) and \( m, n \in \mathbb{Z} \), then \((a^m)^n = a^{mn}\)

Power of a Power

Proof
By definition
\((a^m)^n = \underbrace{a^m \cdot a^m \cdot \ldots \cdot a^m}_{n \text{ times}} = a^{m + m + \cdots + m} = a^{mn}\)

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Power of product law of indices

Let \( a \neq 0 \) and \( m, n \in \mathbb{Z} \), then \((ab)^m =a^mb^n\)

4. Power of a Product

Proof
By definition
\((ab)^m = \underbrace{(ab) \cdot (ab) \cdot \ldots \cdot (ab)}_{m \text{ times}} = \underbrace{a \cdot a \cdot \ldots \cdot a}_{m \text{ times}} \cdot \underbrace{b \cdot b \cdot \ldots \cdot b}_{m \text{ times}} = a^m \cdot b^m\)

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Power of quotient law of indices

Let \( a \neq 0 \) and \( m, n \in \mathbb{Z} \), then \(\left( \frac{a}{b} \right)^m= \frac{a^m}{b^m}\)

Power of a Quotient

Proof
By definition
\(\left( \frac{a}{b} \right)^m = \underbrace{\frac{a}{b} \cdot \frac{a}{b} \cdot \ldots \cdot \frac{a}{b}}_{m \text{ times}} = \frac{a^m}{b^m}\)

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Zero Exponent law of indices

Let \( a \neq 0 \) then \(a^0=1\)

Zero Exponent Rule

Proof
By definition
\(a^0 = a^{m - m} = 1\)

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Negative Exponent law of indices

Let \( a \neq 0 \) then \(a^{-m} = \frac{1}{a^m}\)

Negative Exponent Rule

Proof
By definition
\(a^0 = a^m \cdot a^{-m} = 1 \quad \Rightarrow \quad a^{-m} = \frac{1}{a^m}\)

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For Q.No.6 (a) in BLE Examination

1. मान निकाल्नुहोस् (Find the value of): \( (27)^{-\frac{2}{3}} \) [1U]
1. मान निकाल्नुहोस् (Find the value of): \( (27)^{-\frac{2}{3}} \)
   The value is
   \(\left(3^3\right)^{-\frac{2}{3}}\)
   
   or\(3^{\cancel{3} \times \left(-\frac{2}{\cancel{3}}\right)}\)
   
   or\(3^{-2}\)
   
   or\(\dfrac{1}{3^2}\)
   
   or\(\dfrac{1}{9}\)
2. सरल गर्नुहोस् (Simplify): \( x^{a-b} \times x^{b-c} \times x^{c-a} \) [1K]
1. सरल गर्नुहोस् (Simplify): \( x^{a-b} \times x^{b-c} \times x^{c-a} \)
   The simplified form is
   \(x^{(a-b) + (b-c) + (c-a)}\)
   
   or\(x^{a - b + b - c + c - a}\)
   
   or\(x^{\cancel{a} - \cancel{b} + \cancel{b} - \cancel{c} + \cancel{c} - \cancel{a}}\)
   
   or\(x^{0}\)
   
   or\(1\)
3. मान पत्ता लगाउनुहोस् (Find the value of): \( x^0 + 3 \) [1K]
The simplified form is
\( x^0 + 3 \)

or\( 1+ 3 \)

or\(4\)
4. घाताङ्कका नियम प्रयोग गरेर सरल गर्नुहोस् (Simplify using laws of indices): \( \dfrac{a^{x+2}}{a^{x-2}} \) [1K]
The simplified form is
\( \dfrac{a^{x+2}}{a^{x-2}} \)

\(a^{(x+2)-(x-2)} \)

or\( a^{\cancel{x}+2-\cancel{x}+2} \)

or\(a^4\)
5. घाताङ्कका नियमहरू प्रयोग गरेर सरल गर्नुहोस् (Simplify using laws of indices): \( x^{3m-2} \div x^{3m-4} \) [1K]
The simplified form is
\( \dfrac{x^{3m-2}}{x^{3m-4}} \)

\(x^{(3m-2)-(3m-4)}\)

or\(x^{\cancel{3m}-2-\cancel{3m}+4}\)

or\(x^{2}\)
6. घाताङ्कका नियम प्रयोग गरी मान पत्ता लगाउनुहोस् (Find the value using laws of indices): \( \left( \dfrac{27}{8} \right)^{\frac{2}{3}} \) [1K]
The value is
\( \left( \dfrac{27}{8} \right)^{\frac{2}{3}} \)

\(= \left( \dfrac{3^3}{2^3} \right)^{\frac{2}{3}} \)

or\(= \left( \dfrac{3}{2} \right)^{3 \times \frac{2}{3}} \)

or\(= \left( \dfrac{3}{2} \right)^{\cancel{3} \times \frac{2}{\cancel{3}}} \)

or\(= \left( \dfrac{3}{2} \right)^2 \)

or\(= \dfrac{9}{4} \)
7. घाताङ्कको नियम प्रयोग गरी मान निकाल्नुहोस् (Find the value by using law of indices): \( 27 \times 9^{-1} \) [1U]
The value is
\( 27 \times 9^{-1} \)

or\(3^3 \times (3^2)^{-1} \)

or\(3^3 \times 3^{-2} \)

or\(3^{3 -2} \)

or\(3^1 = 3 \)
8. सरल गर्नुहोस् (Simplify): \( (8^x)^{-1} \times (2^3)^x \) [1U]
The simplified form is
\( (8^x)^{-1} \times (2^3)^x \)

\(= ( (2^3)^x )^{-1} \times (2^3)^x \)

or\(= (2^{3x})^{-1} \times 2^{3x} \)

or\(= 2^{-3x} \times 2^{3x} \)

or\(= 2^{-3x + 3x} \)

or\(= 2^{0} = 1 \)
9. घाताङ्कका नियम प्रयोग गरी सरल गर्नुहोस् (Simplify using laws of indices): \( (x^3 y^{-2})^3 \cdot (-2 x^{-2} y^3)^4 \) [1U]
The simplified form is
\( (x^3 y^{-2})^3 \cdot (-2 x^{-2} y^3)^4 \)

or\( x^9 y^{-6} \cdot (-2)^4 x^{-8} y^{12} \)

or\( x^9 y^{-6} \cdot 16 x^{-8} y^{12} \)

or\( 16 x^{9 -8} y^{-6 + 12} \)

or\( 16 x^1 y^6 \)

or\( 16 x y^6 \)
10. x को घाताङ्क कति हुँदा त्यसको मान \( \dfrac{1}{x^2} \) हुन्छ ?
    What should be the index of x so that its value will be \( \dfrac{1}{x^2} \)? [1U]
The index of x should be \(-2\) so that its value will be \( \dfrac{1}{x^2} \).
11. x को घाताङ्क कति हुँदा त्यसको मान \( \dfrac{1}{x^4} \) हुन्छ ?
    What should be the index of x so that its value will be \( \dfrac{1}{x^4} \)? [1U]
The index of x should be \(-4\) so that its value will be \( \dfrac{1}{x^4} \).
12. यदि \( a = -3 \) र \( b = 2 \) भए \( a^b \) को मान कति होला ?
    If \( a = -3 \) and \( b = 2 \), what will be the value of \( a^b \)? [1U]
The value is
\( a^b \)

or\( (-3)^2 \)

or\( 9 \)
13. मान निकाल्नुहोस् (Find the value of): \( (x + y)^0 + (xy)^0 \) [1K]
The value is
\( (x + y)^0 + (xy)^0 \)

or\( 1+ 1 \)

or\( 2 \)
14. यदि \( x = 3 \) भए \( \dfrac{27}{3^x} \) को मान कति होला ?
    If \( x = 3 \), what will be the value of \( \dfrac{27}{3^x} \)? [1U]
The value is
\( \dfrac{27}{3^x} \)

or\( \dfrac{27}{3^3} \)

or\( \dfrac{27}{27} \)
or\(1\)
15. यदि \( x \ne 0 \), \( y \ne 0 \) भए \( (3x)^0 + (4y)^0 \) को मान कति हुन्छ ?
    If \( x \ne 0 \), \( y \ne 0 \) what will be the value of \( (3x)^0 + (4y)^0 \)? [1K]
The value is
\( (3x)^0 + (4y)^0 \)

or\( 1 + 1 \)

or\( 2 \)
16. मान पत्ता लगाउनुहोस् (Find the value of): \( (a + b)^0 \) [1K]
The value is
\( (a + b)^0 \)

or\( 1 \)
17. मान पत्ता लगाउनुहोस् (Find the value of): \( x^0 \times 9 \) [1K]
The value is
\( x^0 \times 9 \)

or\( 1 \times 9 \)
or\( 9 \)
18. यदि \( x + y = 0 \) भए, \( (a + b)^{x+y} \) को मान कति हुन्छ ?
    If \( x + y = 0 \), what will be the value of \( (a + b)^{x+y} \)? [1K]
The value is
\( (a + b)^{x+y} \)

or\( (a + b)^{0} \)
or\( 1 \)

For Q.No.6 (b) in BLE Examination

1. मान निकाल्नुहोस् (Find the value of): \( \left( \dfrac{27}{125} \right)^{\frac{2}{3}} \times \left( \dfrac{9}{25} \right)^{-\frac{3}{2}} \) [2U]
The simplified form is
\( \left( \dfrac{27}{125} \right)^{\frac{2}{3}} \times \left( \dfrac{9}{25} \right)^{-\frac{3}{2}} \)Rewrite: \( 27 = 3^3 \), \( 125 = 5^3 \), \( 9 = 3^2 \), \( 25 = 5^2 \)

or\( \left( \dfrac{3^3}{5^3} \right)^{\frac{2}{3}} \times \left( \dfrac{3^2}{5^2} \right)^{-\frac{3}{2}} \)

or\( \dfrac{3^{3 \cdot \frac{2}{3}}}{5^{3 \cdot \frac{2}{3}}} \times \dfrac{5^{2 \cdot \frac{3}{2}}}{3^{2 \cdot \frac{3}{2}}} \)Use: \( \left( \dfrac{a}{b} \right)^n = \dfrac{a^n}{b^n} \), \( (a^m)^n = a^{m \cdot n} \), \( a^{-n} = \dfrac{1}{a^n} \)

or\( \dfrac{3^2}{5^2} \times \dfrac{5^3}{3^3} \)

or\( \dfrac{3^2 \cdot 5^3}{5^2 \cdot 3^3} \)

or\( 3^{2 - 3} \cdot 5^{3 - 2} \)

or\( 3^{-1} \cdot 5^1 \)

or\( \dfrac{5}{3} \)
2. मान निकाल्नुहोस् (Find the value of): \( (a + b)^{-1} (a^{-1} + b^{-1}) \) [2U]
The value is
\(\dfrac{1}{a + b} \left( \dfrac{1}{a} + \dfrac{1}{b} \right)\)

or\(\dfrac{1}{a + b} \left( \dfrac{b + a}{ab} \right)\)

or\(\dfrac{1}{a + b} \cdot \dfrac{a + b}{ab}\)

or\(\dfrac{\cancel{a + b}}{ab \cdot \cancel{(a + b)}}\)

or\(\dfrac{1}{ab}\)
3. घाताङ्कको नियम प्रयोग गरी सरल गर्नुहोस् (Simplify by using the law of indices): \( a^{3x+5} \div a^{x-2} \) [2U]
The simplified form is
\(\dfrac{a^{3x+5}}{a^{x-2}}\)

or\(a^{(3x+5) - (x - 2)}\)

or\(a^{3x + 5 - x + 2}\)

or\(a^{2x + 7}\)
4. घाताङ्कको नियम प्रयोग गरी सरल गर्नुहोस् (Simplify by using the law of indices): \( (3x^3y)^{-1} \times (6x^5y^7) \) [2U]
The simplified form is
\(\dfrac{1}{3x^3y} \times 6x^5y^7\)

or\(\dfrac{6}{3} \cdot \dfrac{x^5}{x^3} \cdot \dfrac{y^7}{y}\)

or\(2 \cdot x^{5-3} \cdot y^{7-1}\)


5. घाताङ्कको नियम प्रयोग गरी सरल गर्नुहोस् (Simplify by using the law of indices): \( (a^3b^{-2}c)^2 \times a^{-5}b^4c^{-2} \) [2U]
The simplified form is
\((a^3)^2 (b^{-2})^2 (c)^2 \times a^{-5} b^4 c^{-2}\)

or\(a^{6} b^{-4} c^{2} \times a^{-5} b^{4} c^{-2}\)

or\(a^{6 + (-5)} \cdot b^{-4 + 4} \cdot c^{2 + (-2)}\)

or\(a^{1} \cdot b^{0} \cdot c^{0}\)

or\(a \cdot 1 \cdot 1\)

or\(a\)
6. घाताङ्कको नियम प्रयोग गरी सरल गर्नुहोस् (Simplify by using the law of indices): \( (-2x^3y)^2 \times (x^3y^2)^{-2} \) [2U]
The simplified form is
\((-2)^2 \cdot (x^3)^2 \cdot y^2 \times (x^3)^{-2} \cdot (y^2)^{-2}\)

or\(4 \cdot x^{6} \cdot y^{2} \times x^{-6} \cdot y^{-4}\)

or\(4 \cdot x^{6 -6} \cdot y^{2 -4}\)

or\(4 \cdot x^{0} \cdot y^{-2}\)

or\(4 \cdot 1 \cdot \dfrac{1}{y^{2}}\)

or\(\dfrac{4}{y^{2}}\)
7. घाताङ्कको नियम प्रयोग गरी सरल गर्नुहोस् (Simplify by using the law of indices): \( 3a^{-n-1}b^m \times (2a^{2n+1}b^{-2}) \) [2U]
The simplified form is
\( 3a^{-n-1}b^m \times (2a^{2n+1}b^{-2}) \)

\(= 3 \times 2 \cdot a^{(-n-1) + (2n+1)} \cdot b^{m -2} \)

or\(= 6 \cdot a^{-n - 1 + 2n + 1} \cdot b^{m - 2} \)

or\(= 6 \cdot a^{-n-\cancel{1} + 2n +\cancel{1} } \cdot b^{m - 2} \)

or\(= 6a^n b^{m-2} \)
8. घाताङ्कको नियम प्रयोग गरी सरल गर्नुहोस् (Simplify by using the law of indices): \( (a^2b)^2 \times (ab)^{-1} \times (a^2b^3)^{-1} \) [2U]
The simplified form is
\( (a^2b)^2 \times (ab)^{-1} \times (a^2b^3)^{-1} \)

or\(a^{4}b^{2} \times a^{-1}b^{-1} \times a^{-2}b^{-3} \)

or\(a^{4 -1-2} \cdot b^{2 -1-3} \)

or\(a^{1} \cdot b^{-2} \)

or\(\dfrac{a}{b^{2}} \)
9. घाताङ्कको नियम प्रयोग गरी सरल गर्नुहोस् (Simplify by using the law of indices): \( (3x^2y)^2 \div 9x^3y^2 \) [2U]
The simplified form is
\( (3x^2y)^2 \div 9x^3y^2 \)

or\(\dfrac{(3x^2y)^2}{9x^3y^2} \)

or\(\dfrac{9x^4y^2}{9x^3y^2} \)

or\(\dfrac{\cancel{9}}{\cancel{9}} \cdot \dfrac{x^4}{x^3} \cdot \dfrac{y^2}{y^2} \)

or\(1 \cdot x^{4-3} \cdot y^{2-2} \)

or\(x^{1} \cdot y^{0} \)

or\(x \)
10. घाताङ्कको नियम प्रयोग गरी सरल गर्नुहोस् (Simplify by using the law of indices): \( \left( \dfrac{a^2b^3}{ab^2} \right)^3 \) [2U]
The simplified form is
\( \left( \dfrac{a^2b^3}{ab^2} \right)^3 \)

\(= \left( a^{2-1} b^{3-2} \right)^3 \)

or\(= \left( a^{1} b^{1} \right)^3 \)

or\(= (ab)^3 \)

or\(= a^3 b^3 \)
11. घाताङ्कको नियम प्रयोग गरी सरल गर्नुहोस् (Simplify by using the law of indices): \( \dfrac{x^{a+b}}{x^{c+b}} \times \dfrac{x^{c+d}}{x^{d+a}} \) [2U]
The simplified form is
\( \dfrac{x^{a+b}}{x^{c+b}} \times \dfrac{x^{c+d}}{x^{d+a}} \)

\(= x^{(a+b)-(c+b)} \times x^{(c+d)-(d+a)} \)

\(= x^{a+b-c-b} \times x^{c+d-d-a} \)

\(= x^{a-c} \times x^{c-a} \)

\(= x^{(a-c)+(c-a)} \)

\(= x^{a-c+c-a} \)

\(= x^0 \)

\(= 1 \)
12. घाताङ्कको नियम प्रयोग गरी सरल गर्नुहोस् (Simplify by using the law of indices): \( y^{a(b-c)} \times y^{b(c-a)} \times y^{c(a-b)} \) [2U]
The simplified form is
\( y^{a(b-c)} \times y^{b(c-a)} \times y^{c(a-b)} \)

\(= y^{a(b-c) + b(c-a) + c(a-b)} \)

\(= y^{ab-ac + bc-ab + ca-cb} \)

\(= y^{(ab-ab) + (bc-cb) + (ca-ac)} \)

\(= y^0 \)

\(= 1 \)
13. घाताङ्कको नियम प्रयोग गरी सरल गर्नुहोस् (Simplify by using the law of indices): \( \dfrac{x^{a+b+2} \times x^{a-b+4}}{x^{2a+6}} \) [2U]
The simplified form is
\( \dfrac{x^{a+b+2} \times x^{a-b+4}}{x^{2a+6}} \)

or\( \dfrac{x^{(a+b+2)+(a-b+4)}}{x^{2a+6}} \)

or\( \dfrac{x^{a+b+2+a-b+4}}{x^{2a+6}} \)

or\( \dfrac{x^{2a+6}}{x^{2a+6}} \)

or\( x^{(2a+6)-(2a+6)} \)

or\( x^0 \)

or\( 1 \)
14. सरल गर्नुहोस् (Simplify): \( (x^{2m+n} \times x^{n-m}) \div x^{m-2n} \) [2U]
The simplified form is
\( (x^{2m+n} \times x^{n-m}) \div x^{m-2n} \)

or\( \dfrac{x^{2m+n} \times x^{n-m}}{x^{m-2n}} \)

or\( \dfrac{x^{(2m+n)+(n-m)}}{x^{m-2n}} \)

or\( \dfrac{x^{2m+n+n-m}}{x^{m-2n}} \)

or\( \dfrac{x^{m+2n}}{x^{m-2n}} \)

or\( x^{(m+2n)-(m-2n)} \)

or\( x^{m+2n-m+2n} \)

or\( x^{4n} \)
15. सरल गर्नुहोस् (Simplify): \( \dfrac{3^{x+1} + 3^x}{2 \times 3^x} \) [2U]
The simplified form is
\( \dfrac{3^{x+1} + 3^x}{2 \times 3^x} \)

or\( \dfrac{3^x \cdot 3^1 + 3^x}{2 \cdot 3^x} \)

or\( \dfrac{3^x (3^1 + 1)}{2 \cdot 3^x} \)

or\( \dfrac{3^x (3 + 1)}{2 \cdot 3^x} \)

or\( \dfrac{3^x \cdot 4}{2 \cdot 3^x} \)

or\( \dfrac{4}{2} \)

or\( 2 \)
16. सरल गर्नुहोस् (Simplify): \( \dfrac{2^{x+1} + 2^x}{3 \times 2^x} \) [2U]
The simplified form is
\( \dfrac{2^{x+1} + 2^x}{3 \times 2^x} \)

or\( \dfrac{2^x \cdot 2^1 + 2^x}{3 \cdot 2^x} \)

or\( \dfrac{2^x (2^1 + 1)}{3 \cdot 2^x} \)

or\( \dfrac{2^x (2 + 1)}{3 \cdot 2^x} \)

or\( \dfrac{2^x \cdot 3}{3 \cdot 2^x} \)

or\( \dfrac{3}{3} \)

or\( 1 \)
17. सरल गर्नुहोस् (Simplify): \( \dfrac{3^{x+2} - 3^{x+1}}{6 \times 3^x} \) [2U]
The simplified form is
\( \dfrac{3^{x+2} - 3^{x+1}}{6 \times 3^x} \)

or\( \dfrac{3^x \cdot 3^2 - 3^x \cdot 3^1}{6 \cdot 3^x} \)

or\( \dfrac{3^x (3^2 - 3^1)}{6 \cdot 3^x} \)

or\( \dfrac{3^x (9 - 3)}{6 \cdot 3^x} \)

or\( \dfrac{3^x \cdot 6}{6 \cdot 3^x} \)

or\( \dfrac{6}{6} \)

or\( 1 \)
18. सरल गर्नुहोस् (Simplify): \( \left( \dfrac{1}{x^2} \right)^2 \left( \dfrac{x^{-1}}{x^{-3}} \right)^{-1} \) [2U]
The simplified form is
\( \left( \dfrac{1}{x^2} \right)^2 \left( \dfrac{x^{-1}}{x^{-3}} \right)^{-1} \)

or\( \left( x^{-2} \right)^2 \left( \dfrac{x^{-1}}{x^{-3}} \right)^{-1} \)

or\( x^{-4} \left( x^{-1 + 3} \right)^{-1} \)

or\( x^{-4} \left( x^2 \right)^{-1} \)

or\( x^{-4} \cdot x^{-2} \)

or\( x^{-4 - 2} \)

or\( x^{-6} \)

or\( \dfrac{1}{x^6} \)
19. मान निकाल्नुहोस् (Find the value of): \( (32)^{-\frac{7}{5}} \times (64)^{\frac{5}{6}} \) [2U]
The value is
\( (32)^{-\frac{7}{5}} \times (64)^{\frac{5}{6}} \)

or\( (2^5)^{-\frac{7}{5}} \times (2^6)^{\frac{5}{6}} \)

or\( 2^{5 \cdot (-\frac{7}{5})} \times 2^{6 \cdot (\frac{5}{6})} \)

or\( 2^{-7} \times 2^5 \)

or\( 2^{-7 + 5} \)

or\( 2^{-2} \)

or\( \dfrac{1}{2^2} \)

or\( \dfrac{1}{4} \)
20. मान निकाल्नुहोस् (Evaluate): \( \left( \dfrac{8}{27} \right)^{\frac{2}{3}} \times \left( \dfrac{81}{16} \right)^{\frac{1}{4}} \) [2U]
The value is
\( \left( \dfrac{8}{27} \right)^{\frac{2}{3}} \times \left( \dfrac{81}{16} \right)^{\frac{1}{4}} \)

or\( \left( \dfrac{2^3}{3^3} \right)^{\frac{2}{3}} \times \left( \dfrac{3^4}{2^4} \right)^{\frac{1}{4}} \)

or\( \left( \dfrac{2}{3} \right)^{3 \times \frac{2}{3}} \times \left( \dfrac{3}{2} \right)^{4 \times \frac{1}{4}} \)

or\( \left( \dfrac{2}{3} \right)^2 \times \left( \dfrac{3}{2} \right)^1 \)

or\( \left( \dfrac{2}{3} \right)^2 \times \left( \dfrac{2}{3} \right)^{-1} \)

or\( \left( \dfrac{2}{3} \right)^{2-1} \)

or\( \left( \dfrac{2}{3} \right)^1 \)

or\( \dfrac{2}{3}\ \)


21. मान पत्ता लगाउनुहोस् (Find the value of): \( \dfrac{5^9 \times 3^5}{9^2 \times 25^3} \) [2U]
The value is
\( \dfrac{5^9 \times 3^5}{9^2 \times 25^3} \)

or\( \dfrac{5^9 \times 3^5}{(3^2)^2 \times (5^2)^3} \)

or\( \dfrac{5^9 \times 3^5}{3^4 \times 5^6} \)

or\( 5^{9-6} \times 3^{5-4} \)

or\( 5^3 \times 3^1 \)

or\( 125 \times 3 \)

or\( 375 \)
22. यदि \( x = 4 \), \( y = 2 \), \( m = 1 \) र \( n = 2 \) भए \( \dfrac{x^{m+n} \times y^{m-n}}{x^{m-n} \times y^{m+n}} \) को मान पत्ता लगाउनुहोस्।
    If \( x = 4 \), \( y = 2 \), \( m = 1 \) and \( n = 2 \), then find the value of \( \dfrac{x^{m+n} \times y^{m-n}}{x^{m-n} \times y^{m+n}} \). [2A]
The value is
\( \dfrac{x^{m+n} \times y^{m-n}}{x^{m-n} \times y^{m+n}} \)

or\( x^{(m+n)-(m-n)} \times y^{(m-n)-(m+n)} \)

or\( x^{m+n-m+n} \times y^{m-n-m-n} \)

or\( x^{2n} \times y^{-2n} \)

or\( \dfrac{x^{2n}}{y^{2n}} \)

or\( \left( \dfrac{x}{y} \right)^{2n} \)

Substituting \( x = 4 \), \( y = 2 \), \( m = 1 \), \( n = 2 \):
or\( \left( \dfrac{4}{2} \right)^{2 \cdot 2} \)

or\( 2^4 \)

or\( 16 \)
23. यदि \( a = 3 \), \( b = -2 \) भए \( 3a^{-2}b^2 \) को मान पत्ता लगाउनुहोस्।
    If \( a = 3 \), \( b = -2 \), find the value of \( 3a^{-2}b^2 \). [2A]
The value is
\( 3a^{-2}b^2 \)

or\( 3 \cdot \dfrac{1}{a^2} \cdot b^2 \)

Substituting \( a = 3 \), \( b = -2 \):
or\( 3 \cdot \dfrac{1}{3^2} \cdot (-2)^2 \)

or\( 3 \cdot \dfrac{1}{9} \cdot 4 \)

or\( \dfrac{3 \cdot 4}{9} \)

or\( \dfrac{12}{9} \)

or\( \dfrac{4}{3} \)
24. यदि \( x = 2 \), \( y = 3 \), \( m = 3 \) र \( n = 2 \) भए \( \dfrac{x^{mn} \times y^{m+n}}{x^{m+n} \times y^{m-n}} \) को मान निकाल्नुहोस्।
    If \( x = 2 \), \( y = 3 \), \( m = 3 \) and \( n = 2 \), find the value of \( \dfrac{x^{mn} \times y^{m+n}}{x^{m+n} \times y^{m-n}} \). [2A]
The value is
\( \dfrac{x^{mn} \times y^{m+n}}{x^{m+n} \times y^{m-n}} \)

or\( x^{mn - (m+n)} \times y^{(m+n) - (m-n)} \)

or\( x^{mn - m - n} \times y^{m + n - m + n} \)

or\( x^{mn - m - n} \times y^{2n} \)

Substituting \( x = 2 \), \( y = 3 \), \( m = 3 \), \( n = 2 \):
or\( 2^{3 \cdot 2 - 3 - 2} \times 3^{2 \cdot 2} \)

or\( 2^{6 - 3 - 2} \times 3^4 \)

or\( 2^1 \times 3^4 \)

or\( 2 \times 81 \)

or\( 162 \)
25. यदि \( x = 8y = 2m = 2 \) र \( n = 3 \) भए, \( 2^{m+n} \cdot x^{-mn} \cdot y^{m-n} \) को मान निकाल्नुहोस्।
    If \( x = 8y = 2m = 2 \) and \( n = 3 \), find the value of \( 2^{m+n} \cdot x^{-mn} \cdot y^{m-n} \). [2A]
Solve for \( x \), we get
\( x=2 \)

Solve for \( y \), we get
\( 8y = 2 \implies y = \frac{1}{4} \)

Solve for \( m \), we get
\( 2m = 2 \implies m=1\)

Now, the simplified form is
\( 2^{m+n} \cdot x^{-mn} \cdot y^{m-n} \)

or\( 2^{1+3} \cdot 2^{-1 \cdot 3} \cdot \left( \dfrac{1}{4} \right)^{1-3} \)

or\( 2^4 \cdot 2^{-3} \cdot \left( \dfrac{1}{4} \right)^{-2} \)

or\( 2^4 \cdot 2^{-3} \cdot \left( 4^2 \right) \)

or\( 2^4 \cdot 2^{-3} \cdot 2^{4} \)

or\( 2^{4 - 3 + 4} \)

or\( 2^5 \)

or\( 32 \)
26. यदि \( a = 2 \), \( b = 0 \), \( c = -1 \) र \( d = 2 \) भए \( a^c + d^b \) को मान निकाल्नुहोस्।
    If \( a = 2 \), \( b = 0 \), \( c = -1 \) and \( d = 2 \), find the value of \( a^c + d^b \). [2A]
The value is
\( a^c + d^b \)

Substituting \( a = 2 \), \( b = 0 \), \( c = -1 \), \( d = 2 \):
or\( 2^{-1} + 2^0 \)

or\( \dfrac{1}{2} + 1 \)

or\( \dfrac{1}{2} + \dfrac{2}{2} \)

or\( \dfrac{3}{2} \)
27. यदि \( a = 2 \) र \( b = 3 \) भए \( (a + b)^{-1} \cdot (a^{-1} + b^{-1}) \) को मान निकाल्नुहोस्।
    If \( a = 2 \) and \( b = 3 \), find the value of \( (a + b)^{-1} \cdot (a^{-1} + b^{-1}) \). [2A]
The value is
\( (a + b)^{-1} \cdot (a^{-1} + b^{-1}) \)

or\( \dfrac{1}{a + b} \cdot \left( \dfrac{1}{a} + \dfrac{1}{b} \right) \)

or\( \dfrac{1}{a + b} \cdot \dfrac{b + a}{ab} \)

or\( \dfrac{a + b}{(a + b) \cdot ab} \)

or\( \dfrac{1}{ab} \)

Substituting \( a = 2 \), \( b = 3 \):
or\( \dfrac{1}{2 \cdot 3} \)

or\( \dfrac{1}{6} \)
28. यदि \( ab + bc + ac = 0 \) भए, प्रमाणित गर्नुहोस्।
    If \( ab + bc + ac = 0 \), prove that: \( x^{\frac{1}{a}} x^{\frac{1}{b}} x^{\frac{1}{c}} =1 \) [2A]
The solution is
\( x^{\frac{1}{a}} x^{\frac{1}{b}} x^{\frac{1}{c}} \)

or\( x^{\frac{1}{a} +\frac{1}{b} +\frac{1}{c}} \)

or\( x^{\frac{bc+ca+ab}{abc}} \)

or\( x^{0}\)

or\( 1\)