08 Mathematics

Integers as Exponents

An exponent tells us how many times a number (called the base) is multiplied by itself.

Exponent वह number है जो यह बताता है कि base को अपने आप से कितनी बार multiply करना है।

For example :

\( 3 \times 3 \times 3 \times 3 \) can be written as \( 3^4 \).
\( 3 \times 3 \times 3 \times 3 \) को हम लिख सकते हैं \( 3^4 \) के रूप में।
Here, 3 is the base and 4 is the exponent or power.
यहाँ 3 है base और 4 है exponent।
It is read as “3 raised to the power 4”.
इसे पढ़ते हैं “3 raised to the power 4”

This works for any rational number, not just whole numbers.
यह किसी भी rational number के लिए काम करता है।

\( \frac{3}{8} \times \frac{3}{8} \times \frac{3}{8} \times \frac{3}{8} = \left( \frac{3}{8} \right)^4 \)
\( \frac{3}{8} \) को 4 बार multiply करने पर : \( \left( \frac{3}{8} \right)^4 \)
\( \left( -\frac{3}{5} \right) \times \left( -\frac{3}{5} \right) \times \left( -\frac{3}{5} \right) = \left( -\frac{3}{5} \right)^3 \)
\( -\frac{3}{5} \) को 3 बार multiply करने पर : \( \left( -\frac{3}{5} \right)^3 \)

In general:
If \( \frac{a}{b} \) is any rational number and \( n \) is a natural number, then :
\( \left( \frac{a}{b} \right)^n = \frac{a}{b} \times \frac{a}{b} \times \frac{a}{b} \times … \text{(n times)} \)

किसी भी non-zero number \( x \) के लिए,
\( x^{-n} = \frac{1}{x^n} \)
और
\( \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n} \)

Positive Integral Exponents

When the exponent (n) is a positive integer (like 1, 2, 3, 4…), it is called a positive integral exponent. It simply means repeated multiplication.

जब exponent एक positive integer होता है (जैसे 1, 2, 3…), तो उसे Positive Integral Exponent कहते हैं। इसका मतलब सिर्फ repeated multiplication है।

Negative Integral Exponents

A negative exponent means we are talking about the reciprocal of the base with a positive exponent.
Negative Exponent का मतलब होता है reciprocal लेना।

For example:

The reciprocal of ( 9 ) is \( \frac{1}{9} \).
9 का reciprocal है \( \frac{1}{9} \)
We know \( 9 = 3^2 \), so \( \frac{1}{9} = \frac{1}{3^2} = 3^{-2} \).
\( 9 = 3^2 \) , इसलिए \( \frac{1}{9} = \frac{1}{3^2} = 3^{-2} \)
\( 3^{-2} \) is read as “3 raised to the power -2”.
\( 3^{-2} \) को पढ़ते हैं “3 raised to the power -2”

In general:
For any non-zero rational number \( x \),
\( x^{-n} = \frac{1}{x^n} \)
and
\( \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n} \)

Examples:

\( \frac{1}{2^5} = 2^{-5} \)
\( \frac{1}{2^{-5}} = 2^{5} \)
\( 3^5 = \frac{1}{3^{-5}} \)
\( 3^{-4} = \frac{1}{3^4} \)

Working Rules

To convert between rational numbers and exponential form, remember these key rules:

\( x^m = x \times x \times x … \) (m times)
Exponential form में convert करने के लिए : \( x^m = x \times x \times x … \) m times
\( x^{-m} = \frac{1}{x^m} \)
Negative exponent को positive करने के लिए: \( x^{-m} = \frac{1}{x^m} \)
When a Negative Rational Number is Multiplied to itself (a) Odd Times , then the Answer has Negative sign (b) Even Times, then the Answer has Positive sign.

जब कोई Negative Rational Number अपने आप से (a) Odd times multiply होता है, तो Answer में Negative sign होती है (b) Even times multiply होता है, तो Answer में Positive sign आ जाती है .

Solved Examples at a Glance

Example 1
Express the following power notations as rational numbers:
( निम्नलिखित power notations को rational numbers के रूप में express करो : )
(i) \( \left( -\frac{3}{2} \right)^4 \)
(ii) \( (-1)^7 \)

Solution :
(i) \( \left( -\frac{3}{2} \right)^4 = \left( -\frac{3}{2} \right) \times \left( -\frac{3}{2} \right) \times \left( -\frac{3}{2} \right) \times \left( -\frac{3}{2} \right) = \frac{81}{16} \)
\( \left( -\frac{3}{2} \right)^4 \) का मतलब है \( -\frac{3}{2} \) को अपने आप से 4 बार multiply करना।
जब हम multiply करते हैं, तो negative signs cancel हो जाती हैं क्योंकि 4 एक even number है।
\( \frac{3 \times 3 \times 3 \times 3}{2 \times 2 \times 2 \times 2} = \frac{81}{16} \)

(ii) \( (-1)^7 = (-1) \times (-1) \times (-1) \times (-1) \times (-1) \times (-1) \times (-1) = -1 \)
\( (-1)^7 \) का मतलब है -1 को 7 बार multiply करना।
7 एक odd number है, इसलिए answer negative होगा।
\( (-1) \times (-1) \times (-1) \times (-1) \times (-1) \times (-1) \times (-1) = -1 \)

Example 2

Express each of the following rational numbers in power notation:
(i) \( \frac{-27}{64} \)
(ii) \( \frac{-1}{32} \)

Solution:
(i) \( -27 = (-3)^3 \) and \( 64 = 4^3 \)
So, \( \frac{-27}{64} = \frac{(-3)^3}{4^3} = \left( -\frac{3}{4} \right)^3 \)

(ii) \( -1 = (-1)^5 \) and \( 32 = 2^5 \)
So, \( \frac{-1}{32} = \frac{(-1)^5}{2^5} = \left( -\frac{1}{2} \right)^5 \)

निम्नलिखित rational numbers को power notation में express करो:
(i) \( \frac{-27}{64} \)
(ii) \( \frac{-1}{32} \)

Solution:
(i) पहले numerator और denominator को किसी number की power के रूप में लिखो।
\( -27 \) है \( (-3) \times (-3) \times (-3) = (-3)^3 \)
\( 64 \) है \( 4 \times 4 \times 4 = 4^3 \)
इसलिए, \( \frac{-27}{64} = \frac{(-3)^3}{4^3} = \left( -\frac{3}{4} \right)^3 \)

(ii) \( -1 \) को \( (-1)^5 \) लिख सकते हैं।
\( 32 \) है \( 2 \times 2 \times 2 \times 2 \times 2 = 2^5 \)
इसलिए, \( \frac{-1}{32} = \frac{(-1)^5}{2^5} = \left( -\frac{1}{2} \right)^5 \)

Example 3
Find the value of:
(i) \( (3^2 – 2^2) \div \left( \frac{1}{5} \right)^2 \)
(ii) Reciprocal of \( \left[ \left( \frac{1}{2} \right)^2 – \left( \frac{1}{4} \right)^3 \right] \times 2^3 \)

Solution:
(i) \( (9 – 4) \div \frac{1}{25} = 5 \div \frac{1}{25} = 5 \times 25 = 125 \)

(ii) First, simplify inside the brackets:
\( \left( \frac{1}{4} – \frac{1}{64} \right) \times 8 = \left( \frac{16}{64} – \frac{1}{64} \right) \times 8 = \frac{15}{64} \times 8 = \frac{15}{8} \)
The reciprocal of \( \frac{15}{8} \) is \( \frac{8}{15} \).

निम्नलिखित का मान ज्ञात करो:
(i) \( (3^2 – 2^2) \div \left( \frac{1}{5} \right)^2 \)
(ii) \( \left[ \left( \frac{1}{2} \right)^2 – \left( \frac{1}{4} \right)^3 \right] \times 2^3 \) का reciprocal

Solution:
(i) पहले brackets को solve करो: \( 3^2 = 9 \), \( 2^2 = 4 \), so \( 9 – 4 = 5 \)
अब, \( \left( \frac{1}{5} \right)^2 = \frac{1}{25} \)
Division by a fraction is multiplication by its reciprocal : \( 5 \div \frac{1}{25} = 5 \times 25 = 125 \)

(ii) पहले brackets के अंदर solve करो :
\( \left( \frac{1}{2} \right)^2 = \frac{1}{4} ), ( \left( \frac{1}{4} \right)^3 = \frac{1}{64} \)
So, \( \frac{1}{4} – \frac{1}{64} = \frac{16}{64} – \frac{1}{64} = \frac{15}{64} \)
अब, \( \frac{15}{64} \times 2^3 = \frac{15}{64} \times 8 = \frac{15}{8} \)
Reciprocal means अंश (Numerator) और हर (Denominator) को बदलना: \( \frac{8}{15} \)

Example 4
The value of \( \left( \frac{1}{3} \right)^{-3} \) is :
(a) \( \frac{1}{27} \) (b) \( -\frac{1}{27} \) (c) -27 (d) 27

Solution :
\( \left( \frac{1}{3} \right)^{-3} = \frac{1}{\left( \frac{1}{3} \right)^3} = \frac{1}{\frac{1}{27}} = 1 \times 27 = 27 \)
So, the correct option is (d).

\( \left( \frac{1}{3} \right)^{-3} \) का मान है :
(a) \( \frac{1}{27} \) (b) \( -\frac{1}{27} \) (c) -27 (d) 27

Solution:
Negative exponent rule : \( \left( \frac{1}{3} \right)^{-3} = \frac{1}{\left( \frac{1}{3} \right)^3} \)
अब, \( \left( \frac{1}{3} \right)^3 = \frac{1}{27} \)
So, \( \frac{1}{\frac{1}{27}} = 1 \times 27 = 27 \)
सही option है (d) 27

Example 5
The value of \( (-5)^{-4} \) is :
(a) \( \frac{1}{625} \) (b) -125 (c) \( -\frac{1}{12} \) (d) -625

Solution:
\( (-5)^{-4} = \frac{1}{(-5)^4} = \frac{1}{(-5) \times (-5) \times (-5) \times (-5)} = \frac{1}{625} \)
So, the correct option is (a).

\( (-5)^{-4} \) का मान है :
(a) \( \frac{1}{625} \) (b) -125 (c) \( -\frac{1}{12} \) (d) -625

Solution :
Negative exponent rule : \( (-5)^{-4} = \frac{1}{(-5)^4} \)
अब, \( (-5)^4 = (-5) \times (-5) \times (-5) \times (-5) \)
Negative signs cancel out (even exponent), so answer positive: ( 5 \times 5 \times 5 \times 5 = 625 )
So, \( \frac{1}{625} \)
सही option है (a) \( \frac{1}{625} \)

Of course! Here are the solutions for Exercise 2.1, Question 1 (MCQ) with detailed explanations.

Exercise 2.1

Question 1 (i)
The value of \( \left( \frac{2}{5} \right)^{-4} \) is :
(a) \( \frac{16}{625} \)
(b) \( -\frac{625}{16} \)
(c) \( \frac{625}{16} \)
(d) \( -\frac{16}{625} \)

Solution:
We use the rule for negative exponents : \( \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n} \)

So,\( \left( \frac{2}{5} \right)^{-4} = \left( \frac{5}{2} \right)^{4}\)
Now, calculate \( \left( \frac{5}{2} \right)^{4} \) :
\( \left( \frac{5}{2} \right)^{4} = \frac{5^4}{2^4} = \frac{5 \times 5 \times 5 \times 5}{2 \times 2 \times 2\times 2} = \frac{625}{16}\)

Therefore, the correct option is (c) \( \frac{625}{16} \) .

\( \left( \frac{2}{5} \right)^{-4} \) का मान (value) है :
(a) \( \frac{16}{625} \)
(b) \( -\frac{625}{16} \)
(c) \( \frac{625}{16} \)
(d) \( -\frac{16}{625} \)

Solution:
हम negative exponent के rule का use करेंगे :
\( \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n} \)

इसलिए,
\( \left( \frac{2}{5} \right)^{-4} = \left( \frac{5}{2} \right)^{4} \)
अब, \( \left( \frac{5}{2} \right)^{4} \) की value निकालते हैं:
\(\left( \frac{5}{2} \right)^{4} = \frac{5^4}{2^4} = \frac{5 \times 5 \times 5 \times 5}{2 \times 2 \times 2 \times 2} = \frac{625}{16} \)

इसलिए, सही option है (c) \( \frac{625}{16} \)

Question 1 (ii)

Left Side (Simple English):
The reciprocal of \( \left[ \left( \frac{1}{3} \right)^2 – \left( \frac{1}{3} \right)^3 \right] \times 3^3 \) is :
(a) \( \frac{1}{2} \)
(b) 2
(c) \( -\frac{1}{2} \)
(d) -2

Solution:
First, simplify the expression inside the brackets.

Step 1: Simplify the terms inside
\( \left( \frac{1}{3} \right)^2 = \frac{1}{9} \) \( \left( \frac{1}{3} \right)^3 = \frac{1}{27} \) \( \ left( \frac{1}{3}
\right)^2 – \left( \frac{1}{3} \right)^3 = \frac{1}{9} – \frac{1}{27} \)
Find a common denominator (which is 27) :
\( \frac{1}{9} = \frac{3}{27}, \quad \text{so} \quad \frac{3}{27} – \frac{1}{27} = \frac{2}{27} \)
So, the expression inside the brackets becomes \( \frac{2}{27} \).

Step 2: Multiply by \( 3^3 \)
\(3^3 = 27 \) \( \left[ \frac{2}{27} \right] \times 27 = 2 \)
The value of the expression is 2.

Step 3: Find the reciprocal
The reciprocal of 2 is \( \frac{1}{2} \).

Therefore, the correct option is (a) \( \frac{1}{2} \).

\( \left[ \left( \frac{1}{3} \right)^2 – \left( \frac{1}{3} \right)^3 \right] \times 3^3 \)
का reciprocal है :
(a) \( \frac{1}{2} \)
(b) 2
(c) \( -\frac{1}{2} \)
(d) -2

Solution:
सबसे पहले brackets के अंदर के expression को simplify करते हैं।

Step 1: Brackets के अंदर simplify करो
\( \left( \frac{1}{3} \right)^2 = \frac{1}{9} \) \( \left( \frac{1}{3} \right)^3
= \frac{1}{27} \) \( \left( \frac{1}{3} \right)^2 – \left( \frac{1}{3} \right)^3 = \frac{1}{9} – \frac{1}{27} \)
Common denominator 27 लो :
\( \frac{1}{9} = \frac{3}{27}, \quad \text{so} \quad \frac{3}{27} – \frac{1}{27} = \frac{2}{27} \)
तो, brackets के अंदर की value है \( \frac{2}{27} \).

Step 2: \( 3^3 \) से multiply करो
\(3^3 = 27\) \( \left[ \frac{2}{27} \right] \times 27 = 2 \)
Pure expression की value है 2.

Step 3: Reciprocal निकालो
2 का reciprocal है \( \frac{1}{2} \).

इसलिए, सही option है (a) \( \frac{1}{2}\)

Question 2 (i)

Express the following power of a rational number as a rational number :
\( \left( \frac{-4}{5} \right)^3 \)

Solution:
The expression \( \left( \frac{-4}{5} \right)^3 \) means \( \frac{-4}{5} \) multiplied by itself 3 times.
\(
\left( \frac{-4}{5} \right)^3 = \left( \frac{-4}{5} \right) \times \left( \frac{-4}{5} \right) \times \left( \frac{-4}{5} \right) \)

Step 1: Multiply the numerators :
\( (-4) \times (-4) \times (-4) \)
First, \( (-4) \times (-4) = +16 \)
Then, \( 16 \times (-4) = -64 \)
So, the numerator is -64.

Step 2: Multiply the denominators :
\( 5 \times 5 \times 5 = 125 \)
So, the denominator is 125.

Step 3: Combine the results :
\( \left( \frac{-4}{5} \right)^3 = \frac{-64}{125}\)

Final Answer : \( \frac{-64}{125} \)

निम्नलिखित power of a rational number को एक rational number के रूप में express करो :
\( \left( \frac{-4}{5} \right)^3 \)

Solution :
\( \left( \frac{-4}{5} \right)^3 \) का मतलब है \( \frac{-4}{5} \) को अपने आप से 3 बार multiply करना।
\( \left( \frac{-4}{5} \right)^3 = \left( \frac{-4}{5} \right) \times \left( \frac{-4}{5} \right) \times \left( \frac{-4}{5} \right \)

Step 1 : Numerators को multiply करो :
\( (-4) \times (-4) \times (-4) \)
पहले, \( (-4) \times (-4) = +16 \)
फिर, \( 16 \times (-4) = -64 \)
तो, numerator है -64

Step 2 : Denominators को multiply करो :
\( 5 \times 5 \times 5 = 125 \)
तो, denominator है 125

Step 3 : Results को combine करो :
\( \left( \frac{-4}{5} \right)^3 = \frac{-64}{125} \)

Final Answer : \( \frac{-64}{125} \)

Question 2 (ii)

Express the following power of a rational number as a rational number :
\( (-1)^{12} \)

Solution:
The expression \( (-1)^{12} \) means (-1) multiplied by itself 12 times.
\((-1)^{12} = \underbrace{(-1) \times (-1) \times (-1) \times \dots \times (-1)}_{12\ \text{times}} \)

Key Concept:

When (-1) is raised to an even power (like 2, 4, 6, … 12), the result is +1.
When (-1) is raised to an odd power (like 1, 3, 5, …), the result is -1.

Since 12 is an even number, the negative signs will all cancel out, and the final result will be positive.

Therefore :
\((-1)^{12} = +1\)

Final Answer: ( 1 )

निम्नलिखित power of a rational number को एक rational number के रूप में express करो :
\( (-1)^{12} \)

Solution:
\( (-1)^{12} \) का मतलब है (-1) को अपने आप से 12 बार multiply करना
\((-1)^{12} = \underbrace{(-1) \times (-1) \times (-1) \times \dots \times (-1)}_{12\ \text{बार}} \)

Important Rule:

जब (-1) को किसी even power (जैसे 2, 4, 6, … 12) पर raise किया जाता है, तो result +1 होता है
जब (-1) को किसी odd power (जैसे 1, 3, 5, …) पर raise किया जाता है, तो result -1 होता है

क्योंकि 12 एक even number है, सारे negative signs cancel हो जाएंगे, और final result positive होगा

इसलिए:
\((-1)^{12} = +1 \)

Final Answer: ( 1 )

Question 3 (i)

Express the rational number in power notation :
\( -\frac{1}{125} \)

Solution:
We need to write both the numerator and the denominator as powers.

The numerator is -1. We can write \( -1 = (-1)^3 \).
The denominator is 125. We know \( 125 = 5 \times 5 \times 5 = 5^3 \).

Therefore :
\( -\frac{1}{125} = \frac{-1}{5^3} = \frac{(-1)^3}{5^3} = \left( -\frac{1}{5} \right)^3\)

Final Answer : \( \left( -\frac{1}{5} \right)^3 \)

Rational number को power notation में express करो :
\( -\frac{1}{125} \)

Solution :
हमें numerator और denominator दोनों को powers के रूप में लिखना है

Numerator है -1. हम लिख सकते हैं \( -1 = (-1)^3 \).
Denominator है 125. हम जानते हैं \( 125 = 5 \times 5 \times 5 = 5^3 \).

इसलिए :
\( -\frac{1}{125} = \frac{-1}{5^3} = \frac{(-1)^3}{5^3} = \left( -\frac{1}{5} \right)^3 \)

Final Answer: \( \left( -\frac{1}{5} \right)^3 \)

Question 3 (ii)

Express the rational number in power notation:
\( -\frac{27}{8} \)

Solution:
We need to write both the numerator and the denominator as powers.

The numerator is -27. We know \( 27 = 3 \times 3 \times 3 = 3^3 ), so ( -27 = (-3)^3 \).
The denominator is 8. We know \( 8 = 2 \times 2 \times 2 = 2^3 \).

Therefore:
\( -\frac{27}{8} = \frac{(-3)^3}{2^3} = \left( -\frac{3}{2} \right)^3 \)

Final Answer: \( \left( -\frac{3}{2} \right)^3 \)

Rational number को power notation में express करो :
\( -\frac{27}{8} \)

Solution:
हमें numerator और denominator दोनों को powers के रूप में लिखना है

Numerator है -27. हम जानते हैं \( 27 = 3 \times 3 \times 3 = 3^3 ), इसलिए ( -27 = (-3)^3 \).
Denominator है 8. हम जानते हैं \( 8 = 2 \times 2 \times 2 = 2^3 \).

इसलिए :
\( -\frac{27}{8} = \frac{(-3)^3}{2^3} = \left( -\frac{3}{2} \right)^3 \)

Final Answer : \(\left( -\frac{3}{2} \right)^3 \)

Question 3 (iii)

Left Side (Simple English):
Express the rational number in power notation:
\( \frac{9}{25} \times \frac{64}{25} \)

Solution:
We can write each number as a square.

\( 9 = 3^2 ), so ( \frac{9}{25} = \frac{3^2}{5^2} = \left( \frac{3}{5} \right)^2 \)
\( 64 = 8^2 ), so ( \frac{64}{25} = \frac{8^2}{5^2} = \left( \frac{8}{5} \right)^2 \)

Now, the expression becomes :
\( \left( \frac{3}{5} \right)^2 \times \left( \frac{8}{5} \right)^2 \)
Using the law of exponents: \( a^m \times b^m = (a \times b)^m \)
\( \left( \frac{3}{5} \times \frac{8}{5} \right)^2 = \left( \frac{3 \times 8}{5 \times 5} \right)^2 = \left( \frac{24}{25} \right)^2 \)

Final Answer: \( \left( \frac{24}{25} \right)^2 \)

Rational number को power notation में express करो :
\( \frac{9}{25} \times \frac{64}{25} \)

Solution:
हम प्रत्येक number को एक square के रूप में लिख सकते हैं

\( 9 = 3^2 ), इसलिए ( \frac{9}{25} = \frac{3^2}{5^2} = \left( \frac{3}{5} \right)^2 \)
\( 64 = 8^2 ), इसलिए ( \frac{64}{25} = \frac{8^2}{5^2} = \left( \frac{8}{5} \right)^2 \)

अब, expression बन जाता है :
\( \left( \frac{3}{5} \right)^2 \times \left( \frac{8}{5} \right)^2 \)
Exponents के law का use करते हुए : \( a^m \times b^m = (a \times b)^m \)
\( \left( \frac{3}{5} \times \frac{8}{5} \right)^2 = \left( \frac{3 \times 8}{5 \times 5} \right)^2 = \left( \frac{24}{25} \right)^2 \)

Final Answer : \( \left( \frac{24}{25} \right)^2 \)

Question 3 (iv)

Express the rational number in power notation :
\( \frac{49}{64} \)

Solution :
We need to write both the numerator and the denominator as squares.

The numerator is 49. We know \( 49 = 7 \times 7 = 7^2 \).
The denominator is 64. We know \( 64 = 8 \times 8 = 8^2 \).

Therefore :
\( \frac{49}{64} = \frac{7^2}{8^2} = \left( \frac{7}{8} \right)^2 \)

Final Answer : \( \left( \frac{7}{8} \right)^2 \)

Rational number को power notation में express करो :
\( \frac{49}{64} \)

Solution:
हमें numerator और denominator दोनों को squares के रूप में लिखना है

Numerator है 49. हम जानते हैं \( 49 = 7 \times 7 = 7^2 \).
Denominator है 64. हम जानते हैं \( 64 = 8 \times 8 = 8^2 \).

इसलिए :
\( \frac{49}{64} = \frac{7^2}{8^2} = \left( \frac{7}{8} \right)^2 \)

Final Answer : \( \left( \frac{7}{8} \right)^2 \)

Question 3 (v)

Express the rational number in power notation :
\( 0.49 \)

Solution :
First, convert the decimal into a fraction.
\( 0.49 = \frac{49}{100} \)
Now, write both the numerator and the denominator as squares.

The numerator is 49. We know \( 49 = 7 \times 7 = 7^2 \).
The denominator is 100. We know \( 100 = 10 \times 10 = 10^2 \).

Therefore :
\( 0.49 = \frac{49}{100} = \frac{7^2}{10^2} = \left( \frac{7}{10} \right)^2 \)

Final Answer: \( \left( \frac{7}{10} \right)^2 \)

Rational number को power notation में express करो:
\( 0.49 \)

Solution:
सबसे पहले, decimal को fraction में convert करो।
\( 0.49 = \frac{49}{100} \)
अब, numerator और denominator दोनों को squares के रूप में लिखो

Numerator है 49. हम जानते हैं \( 49 = 7 \times 7 = 7^2 \) .
Denominator है 100. हम जानते हैं \( 100 = 10 \times 10 = 10^2 \).

इसलिए :
\( 0.49 = \frac{49}{100} = \frac{7^2}{10^2} = \left( \frac{7}{10} \right)^2 \)

Final Answer: \( \left( \frac{7}{10} \right)^2 \)

Question 4 Convert negative exponents to positive exponents:

Question 4 (i) \( \left( \frac{4}{5} \right)^{-5} \)

We use the rule for negative exponents : \( \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n} \)

Applying the rule :
\( \left( \frac{4}{5} \right)^{-5} = \left( \frac{5}{4} \right)^{5} \)

Final Answer : \( \left( \frac{5}{4} \right)^{5} \)

Negative exponent को positive बनाने का rule है : \( \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n} \)

इस rule को apply करते हैं :
\( \left( \frac{4}{5} \right)^{-5} = \left( \frac{5}{4} \right)^{5} \)

Final Answer : \( \left( \frac{5}{4} \right)^{5} \)

Question 4 (ii) \( \left( -\frac{8}{9} \right)^{-10} \)

We use the rule : \( \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n} \)

Applying the rule :
\( \left( -\frac{8}{9} \right)^{-10} = \left( -\frac{9}{8} \right)^{10} \)

Important Note : The exponent 10 is even. A negative base raised to an even power results in a positive number.
So, \( \left( -\frac{9}{8} \right)^{10} = \left( \frac{9}{8} \right)^{10} \) .

Final Answer : \( \left( \frac{9}{8} \right)^{10} \)

Rule use करते हैं : \( \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n} \)

Apply करने पर :
\( \left( -\frac{8}{9} \right)^{-10} = \left( -\frac{9}{8} \right)^{10} \)

Important Note : Exponent 10 एक even number है। Negative base को even power पर raise करने पर result positive होता है।
इसलिए, \( \left( -\frac{9}{8} \right)^{10} = \left( \frac{9}{8} \right)^{10} \) .

Final Answer: \( \left( \frac{9}{8} \right)^{10} \)

Question 4 (iii) \( \left( \frac{7}{8} \right)^9 \)

This expression already has a positive exponent. There is no negative exponent to convert. The expression remains unchanged.

Final Answer : \( \left( \frac{7}{8} \right)^9 \)

इस expression में पहले से ही एक positive exponent है। इसमें कोई negative exponent नहीं है जिसे convert करना हो। Expression वैसा का वैसा ही रहेगा

Final Answer : \( \left( \frac{7}{8} \right)^9 \)

Question 4 (iv) \( \left( -\frac{6}{5} \right)^{-11} \)

We use the rule : \( \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n} \)

Applying the rule :
\( \left( -\frac{6}{5} \right)^{-11} = \left( -\frac{5}{6} \right)^{11} \)

Important Note : The exponent 11 is odd. A negative base raised to an odd power remains negative. So, the expression stays as \( \left( -\frac{5}{6} \right)^{11} \).

Final Answer: \( \left( -\frac{5}{6} \right)^{11} \)

Rule use करते हैं: \( \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n} \)

Apply करने पर :
\( \left( -\frac{6}{5} \right)^{-11} = \left( -\frac{5}{6} \right)^{11} \)

Important Note : Exponent 11 एक odd number है। Negative base को odd power पर raise करने पर result negative ही रहता है।
इसलिए, expression \( \left( -\frac{5}{6} \right)^{11} \) के रूप में ही रहेगा

Final Answer : \( \left( -\frac{5}{6} \right)^{11} \)

Question 5 (i)

Simplify:
\(\left[ \left( -\frac{2}{3} \right)^3 + \frac{4}{9} \right] + \left( \frac{3}{5} \right)^{-3}\)

Step 1: Simplify the first bracket.
\( \left( -\frac{2}{3} \right)^3 = \frac{(-2)^3}{(3)^3} = \frac{-8}{27} \)
So, the bracket becomes :
\(\left[ \frac{-8}{27} + \frac{4}{9} \right] \)
Find a common denominator (27) :
\( \frac{4}{9} = \frac{12}{27} \)
\( \frac{-8}{27} + \frac{12}{27} = \frac{4}{27} \)

Step 2 : Simplify the second term.
\( \left( \frac{3}{5} \right)^{-3} = \left( \frac{5}{3} \right)^{3} = \frac{125}{27} \)

Step 3 : Add the results.
\( \frac{4}{27} + \frac{125}{27} = \frac{129}{27} = \frac{43}{9} \)

Final Answer: \(\frac{43}{9}\)

Step 1: पहले हम brackets को simplify करेंगे.
\( \left( -\frac{2}{3} \right)^3 = \frac{(-2)^3}{(3)^3} = \frac{-8}{27} \)
तो , Bracket की Value होगी :
\( \left[ \frac{-8}{27} + \frac{4}{9} \right] \)
Common denominator (27) लेते है :
\( \frac{4}{9} = \frac{12}{27} \)
\( \frac{-8}{27} + \frac{12}{27} = \frac{4}{27} \)

Step 2: दूसरे term ko simplify करेंगे.
\( \left( \frac{3}{5} \right)^{-3} = \left( \frac{5}{3} \right)^{3} = \frac{125}{27} \)

Step 3: दोनों results को add करेंगे .
\( \frac{4}{27} + \frac{125}{27} = \frac{129}{27} = \frac{43}{9} \)

Final Answer: \(\frac{43}{9}\)

Question 5 (ii)

Simplify:
\((4)^{-3} + \left( -\frac{1}{2} \right)^3 \)

Step 1: Simplify each term.
\( (4)^{-3} = \frac{1}{4^3} = \frac{1}{64} \)
\( \left( -\frac{1}{2} \right)^3 = \frac{(-1)^3}{(2)^3} = \frac{-1}{8} \)

Step 2: Add the results.
Find a common denominator (64) :
\( \frac{-1}{8} = \frac{-8}{64} \)
\( \frac{1}{64} + \left( \frac{-8}{64} \right) = \frac{1 – 8}{64} = \frac{-7}{64} \)

Final Answer: \(\frac{-7}{64})\

Step 1: हर term को हम simplify करेंगे.
\( (4)^{-3} = \frac{1}{4^3} = \frac{1}{64} \)
\( \left( -\frac{1}{2} \right)^3 = \frac{(-1)^3}{(2)^3} = \frac{-1}{8} \)

Step 2: Results को add करेंगे.
Common denominator (64) लेते है :
\( \frac{-1}{8} = \frac{-8}{64} \)
[ \( \frac{1}{64} + \left( \frac{-8}{64} \right) = \frac{1 – 8}{64} = \frac{-7}{64} \)

Final Answer: \(\frac{-7}{64}\)

Question 5 (iii)

Simplify :
\(\left( \frac{3}{7} \right)^2 \times \frac{25}{7} \times \left( -\frac{7}{5} \right)^2 \)

Step 1: Write all terms with exponents.
\( \left( \frac{3}{7} \right)^2 = \frac{3^2}{7^2} = \frac{9}{49} \)
\( \left( -\frac{7}{5} \right)^2 = \frac{(-7)^2}{(5)^2} = \frac{49}{25} \)
The middle term is \(\frac{25}{7} \).

Step 2: Multiply all terms together.
\( \frac{9}{49} \times \frac{25}{7} \times \frac{49}{25} \)
Notice that 49 and 25 cancel out:
\( \frac{9}{\cancel{49}} \times \frac{\cancel{25}}{7} \times \frac{\cancel{49}}{\cancel{25}} = \frac{9}{7} \)

Final Answer: \( \frac{9}{7} \)

Step 1: सभी terms को Exponents के साथ लिखेंगे .
\( \left( \frac{3}{7} \right)^2 = \frac{3^2}{7^2} = \frac{9}{49} \)
\( \left( -\frac{7}{5} \right)^2 = \frac{(-7)^2}{(5)^2} = \frac{49}{25} \)
बीच वाले terms है \(\frac{25}{7}\).

Step 2: सभी terms को multiply करेंगे.
\( \frac{9}{49} \times \frac{25}{7} \times \frac{49}{25} \)
यहाँ 49 और 25 cancel हो जाते है :
\( \frac{9}{\cancel{49}} \times \frac{\cancel{25}}{7} \times \frac{\cancel{49}}{\cancel{25}} = \frac{9}{7} \)

Final Answer: \(\frac{9}{7}\)

Question 5 (iv)

Simplify:
\(\left( -\frac{3}{4} \right)^3 \times \left[ \left( \frac{3}{4} \right)^{-3} – \left( \frac{2}{3} \right)^3 \right]\)

Step 1: Simplify each exponent.
\( \left( -\frac{3}{4} \right)^3 = \frac{-27}{64} \)
\( \left( \frac{3}{4} \right)^{-3} = \left( \frac{4}{3} \right)^{3} = \frac{64}{27} \)
\( \left( \frac{2}{3} \right)^3 = \frac{8}{27} \)

Step 2: Simplify the expression inside the bracket.
\( \left[ \frac{64}{27} – \frac{8}{27} \right] = \frac{56}{27} \)

Step 3: Multiply the results.
\( \frac{-27}{64} \times \frac{56}{27} \)
Cancel 27 and simplify 56/64 :
\( \frac{-\cancel{27}}{\cancel{64}} \times \frac{\cancel{56}^7}{\cancel{27}} = \frac{-7}{8} \)

Final Answer: \(\frac{-7}{8}\)

Step 1: हर Exponent को Simplify करेंगे.
\( \left( -\frac{3}{4} \right)^3 = \frac{-27}{64} \)
\( \left( \frac{3}{4} \right)^{-3} = \left( \frac{4}{3} \right)^{3} = \frac{64}{27} \)
\( \left( \frac{2}{3} \right)^3 = \frac{8}{27} \)

Step 2: Bracket ke andar wale expression ko simplify karenge.
\( \left[ \frac{64}{27} – \frac{8}{27} \right] = \frac{56}{27} \)

Step 3: Results ko multiply karenge.
\( \frac{-27}{64} \times \frac{56}{27} \)
27 cancel हो जायेगा और 56 & 64 को simplify करेंगे :
\( \frac{-\cancel{27}}{\cancel{64}} \times \frac{\cancel{56}^7}{\cancel{27}} = \frac{-7}{8} \)

Final Answer: \( \frac{-7}{8} \)

Question 5 (v)

Simplify:
\(\left( \frac{2}{3} \right)^4 \times \left( \frac{3}{2} \right)^5 \times \left( \frac{5}{3} \right)^{-2} \times \left( \frac{3}{5} \right)^{-3} \)

Step 1: Convert all negative exponents to positive.
\( \left( \frac{5}{3} \right)^{-2} = \left( \frac{3}{5} \right)^{2} \)
\( \left( \frac{3}{5} \right)^{-3} = \left( \frac{5}{3} \right)^{3} \)

Step 2: Rewrite the entire expression.
\( \left( \frac{2}{3} \right)^4 \times \left( \frac{3}{2} \right)^5 \times \left( \frac{3}{5} \right)^{2} \times \left( \frac{5}{3} \right)^{3} \)

Step 3: Group the same bases together.
\( = \frac{2^4}{3^4} \times \frac{3^5}{2^5} \times \frac{3^2}{5^2} \times \frac{5^3}{3^3} \)

Step 4: Combine the exponents for each base.

For base 2 : \(4 – 5 = -1) → (\frac{1}{2^1} \)
For base 3 : \(-4 + 5 + 2 – 3 = 0) → (3^0 = 1 \)
For base 5 : \(-2 + 3 = 1) → (5^1 \)

Step 5: Write the final expression.
\( \frac{1}{2} \times 1 \times 5 = \frac{5}{2} \)

Final Answer: \( \frac{5}{2} \)

Step 1: सभी Negative Exponents को Positive बनाएंगे .
\( \left( \frac{5}{3} \right)^{-2} = \left( \frac{3}{5} \right)^{2} \)
\( \left( \frac{3}{5} \right)^{-3} = \left( \frac{5}{3} \right)^{3} \)

Step 2: Pure expression को फिर से लिखेंगे.
\( \left( \frac{2}{3} \right)^4 \times \left( \frac{3}{2} \right)^5 \times \left( \frac{3}{5} \right)^{2} \times \left( \frac{5}{3} \right)^{3} \)

Step 3: Same bases को group करेंगे.
\( = \frac{2^4}{3^4} \times \frac{3^5}{2^5} \times \frac{3^2}{5^2} \times \frac{5^3}{3^3} \)

Step 4: हर Base के Exponents को Combine करेंगे .

Base 2: \(4 – 5 = -1) → (\frac{1}{2^1} \)
Base 3: \(-4 + 5 + 2 – 3 = 0) → (3^0 = 1 \)
Base 5: \(-2 + 3 = 1) → (5^1 \)

Step 5: Final expression लिखेंगे.
\( \frac{1}{2} \times 1 \times 5 = \frac{5}{2} \)

Final Answer: \(\frac{5}{2} \)

Question 6

Life Skills Problem: A gardener plants in a pattern :

Row 1 : 1 plant
Row 2 : 3 plants
Row 3 : 9 plants
If this pattern continues, how many plants will be in the 9th row? Give your answer in exponential form.

Step 1: Identify the Pattern

Row 1 : \( 3^0 = 1 \) plant
Row 2 : \( 3^1 = 3 \) plants
Row 3 : \( 3^2 = 9 \) plants

The number of plants in each row follows powers of 3.

Step 2: Find the Formula
For the n-th row, the number of plants is :
\( 3^{(n-1)} \)

Step 3: Apply for 9th Row
For the 9th row \((n = 9) \) : \( 3^{(9-1)} = 3^8 \)

Final Answer in Exponential Form: \(3^8 \)

Step 1: Pattern को समझेंगे :

Row 1 : 1 plant → \(3^0 = 1 \)
Row 2 : 3 plants → \(3^1 = 3 \)
Row 3 : 9 plants → \(3^2 = 9 \)

हर Row में Plants का Number 3 के Power के हिसाब से Increase हो रहा है.

Step 2: Formula बनाएं
n-th row में plants = \(3^{(n-1)} \)

Step 3: 9th row के लिए calculate करें
9th row के लिए \((n = 9)\ ) : \( 3^{(9-1)} = 3^8 \)

Final Answer : \(3^8 \)

2.2 LAWS OF EXPONENTS WITH INTEGRAL POWERS

Law 1: Product of Powers with Same Base

If you multiply two powers with the same base, you keep the base and add the exponents.
Formula : \( x^m \times x^n = x^{m+n} \)
Example : \( 2^3 \times 2^4 = 2^{3+4} = 2^7 \)

Law 2: Power of a Power

When you raise a power to another power, you keep the base and multiply the exponents.
Formula : \( (x^m)^n = x^{m \times n} \)
Example : \( (3^2)^4 = 3^{2 \times 4} = 3^8 \)

Law 3: Quotient of Powers with Same Base

When you divide two powers with the same base, you keep the base and subtract the exponents.
Formula : \( x^m \div x^n = x^{m-n} \)
Example : \( 5^7 \div 5^2 = 5^{7-2} = 5^5 \)

Law 4: Zero Exponent Rule

Any non-zero number raised to the power of zero is always equal to 1.
Formula : \( x^0 = 1 ) (where ( x \neq 0 )\)
Example : \( 7^0 = 1 ), ( (-4)^0 = 1 ), ( \left(\frac{2}{3}\right)^0 = 1 \)

Solved Example 6

Find the value of : \( \left( -\frac{2}{3} \right)^{-3} \times \left( -\frac{2}{3} \right)^6 \)

Solution:

Use the law : \( x^m \times x^n = x^{m+n} \)
Here, base = \( -\frac{2}{3} \), exponents = -3 and 6
Add exponents: -3 + 6 = 3
So : \( \left( -\frac{2}{3} \right)^3 = \frac{-8}{27} \)

Final Answer : \( -\frac{8}{27} \)

Solved Example 7

Express as a single exponential form:

(i) \( \left( -\frac{4}{5} \right)^{-10} \times \left( -\frac{4}{5} \right)^{15} \div \left( -\frac{4}{5} \right)^9 \)

Multiply = add exponents, Divide = subtract exponents
Combine : \( (-10) + 15 – 9 = -4 \)
So : \( \left( -\frac{4}{5} \right)^{-4} \)

(ii) \( \left( \frac{3}{2} \right)^{33} \div \left( \frac{2}{3} \right)^{-35} \)

First : \( \left( \frac{2}{3} \right)^{-35} = \left( \frac{3}{2} \right)^{35} \)
Now : \( \left( \frac{3}{2} \right)^{33} \div \left( \frac{3}{2} \right)^{35} = \left( \frac{3}{2} \right)^{-2} \)

Final Answers:
(i) \( \left( -\frac{4}{5} \right)^{-4} \)
(ii) \( \left( \frac{3}{2} \right)^{-2} \)

Solved Example 8

Simplify :

(i) \( \left( 4^{-1} + 8^{-1} \right) \div \frac{1}{\left( \frac{3}{2} \right)^{-2}} \)

\( 4^{-1} = \frac{1}{4} ), ( 8^{-1} = \frac{1}{8} \)
Sum : \( \frac{1}{4} + \frac{1}{8} = \frac{3}{8} \)
\( \frac{1}{\left( \frac{3}{2} \right)^{-2}} = \left( \frac{3}{2} \right)^2 = \frac{9}{4} \)
Now : \( \frac{3}{8} \div \frac{9}{4} = \frac{3}{8} \times \frac{4}{9} = \frac{1}{6} \)

(ii) \( \left( 5^{-1} \times 5^{-19} \right)^2 \times \left( -\frac{5}{2} \right)^{40} \)

\( 5^{-1} \times 5^{-19} = 5^{-20} \)
\( (5^{-20})^2 = 5^{-40} \)
\( \left( -\frac{5}{2} \right)^{40} = \frac{5^{40}}{2^{40}} \) (since exponent is even)
Multiply : \( 5^{-40} \times \frac{5^{40}}{2^{40}} = \frac{1}{2^{40}} = \left( \frac{1}{2} \right)^{40} \)

(iii) \( \left[ \left( -\frac{3}{2} \right)^{20} \right]^{20} \)

Power of a power: multiply exponents : \( 20 \times 20 = 400 \)
So : \( \left( -\frac{3}{2} \right)^{400} \)
Since 400 is even : \( \left( \frac{3}{2} \right)^{400} \)

Final Answers:
(i) \( \frac{1}{6} \)
(ii) \( \left( \frac{1}{2} \right)^{40} \)
(iii) \( \left( \frac{3}{2} \right)^{400} \)

Solved Example 9

By what number should \( (-9)^{-1} ) be multiplied to get ( (-12)^{-1} \) ?

Let the number be \( x \)
Equation : \( (-9)^{-1} \times x = (-12)^{-1} \)
So : \( x = (-12)^{-1} \div (-9)^{-1} \)
\( x = \left( -\frac{1}{12} \right) \div \left( -\frac{1}{9} \right) = \left( -\frac{1}{12} \right) \times (-9) = \frac{9}{12} = \frac{3}{4} \)

Final Answer : \( \frac{3}{4} \)

Solved Example 10

If \( \left( \frac{3}{4} \right)^{-6} \times \left( \frac{4}{3} \right)^{-8} = \left( \frac{3}{4} \right)^n \), find \(n\)

\( \left( \frac{4}{3} \right)^{-8} = \left( \frac{3}{4} \right)^{8} \)
So : \( \left( \frac{3}{4} \right)^{-6} \times \left( \frac{3}{4} \right)^{8} = \left( \frac{3}{4} \right)^n \)
Add exponents : \( -6 + 8 = 2 \)
So : \( \left( \frac{3}{4} \right)^{2} = \left( \frac{3}{4} \right)^n \)
Therefore : \( n = 2 \)

Final Answer : \( n = 2 \)

Exercise 2.2

Question 1 (i) Find the value of :
\(\left[ \left( \left( -\frac{1}{2} \right)^2 \right)^{-2} \right]^{-1} \)

Step-by-step solution using Power of a Power Law :

Law Used : \((x^m)^n = x^{m \times n} \)

1. Multiply all exponents :
\( 2 \times (-2) \times (-1) = 4 \)

2. Raise the base to the resulting power : \( \left( -\frac{1}{2} \right)^4 = \frac{(-1)^4}{2^4} = \frac{1}{16} \)

\( \left( -\frac{1}{2} \right)^4 = \frac{(-1)^4}{2^4} = \frac{1}{16} \)

Final Answer : \(\frac{1}{16} \)

Question 1 (ii)

Find the value of :
\( left{ \left( \frac{1}{4} \right)^{-3} – \left( \frac{1}{2} \right)^{-3} \right} \div \left( \frac{1}{4} \right)^{-3} \)

Step-by-step solution:

Simplify each term using \( a^{-m} = \frac{1}{a^m} \) :
\( \left( \frac{1}{4} \right)^{-3} = 4^3 = 64 \)
\( \left( \frac{1}{2} \right)^{-3} = 2^3 = 8 \)
Substitute into the expression:
\( \left{ 64 – 8 \right} \div 64 = 56 \div 64 \)
Simplify the division:
\( \frac{56}{64} = \frac{7}{8} \)

Final Answer: \(\frac{7}{8} \)

Exercise 2.2 Question 1 (ii)

Find the value of : \( \left{ \left( \frac{1}{4} \right)^{-3} – \left( \frac{1}{2} \right)^{-3} \right} \div \left( \frac{1}{4} \right)^{-3} \)

Step-by-step solution :

Simplify each term using \( a^{-m} = \frac{1}{a^m} \) :
\( \left( \frac{1}{4} \right)^{-3} = 4^3 = 64 \)
\( \left( \frac{1}{2} \right)^{-3} = 2^3 = 8 \)
Substitute into the expression :
\( \left{ 64 – 8 \right} \div 64 = 56 \div 64 \)
Simplify the division :
\( \frac{56}{64} = \frac{7}{8} \)

Final Answer : \(\frac{7}{8}\)

Question 1 (iii)

By what number should \( \left( \frac{1}{-2} \right)^{-1} \) be multiplied so that the product equals \( \left( \frac{4}{7} \right)^{-1} \) ?

Step-by-step solution :

Let the required number be \( x \).

Set up the equation :
\( \left( \frac{1}{-2} \right)^{-1} \times x = \left( \frac{4}{7} \right)^{-1} \)
Simplify both sides :
\( \left( \frac{1}{-2} \right)^{-1} = (-2)^1 = -2 \)
\( \left( \frac{4}{7} \right)^{-1} = \frac{7}{4} \)
Substitute into the equation :
\( -2 \times x = \frac{7}{4} \)
Solve for \( x \) :
\( x = \frac{7}{4} \div (-2) = \frac{7}{4} \times \left( -\frac{1}{2} \right) = -\frac{7}{8} \)

Final Answer : \( -\frac{7}{8} \)

Question 2 (i)

Express the following into a single exponential form :
\( \left( \frac{2}{3} \right)^{-5} \times \left( \frac{2}{3} \right)^6 \times \left( \frac{2}{3} \right)^2 \)

Step-by-step solution :

Law Used : \( x^m \times x^n \times x^p = x^{m+n+p} \)

Identify the base : \( \frac{2}{3} \)
Add the exponents :
\( -5 + 6 + 2 = 3 \)
Write in single exponential form :
\( \left( \frac{2}{3} \right)^3 \)

Final Answer : \( \left( \frac{2}{3} \right)^3 \)

Question 2 (ii)

Express the following into a single exponential form :
\( \left( \frac{-6}{7} \right)^{10} \times \left( \frac{-6}{7} \right)^{-15} \times \left( \frac{-6}{7} \right)^2 \)

Step-by-step solution :

Law Used : \( x^m \times x^n \times x^p = x^{m+n+p} \)

Identify the base : \( \frac{-6}{7} \)
Add the exponents : \(10 + (-15) + 2 = -3 \)
Write in single exponential form : \( \left( \frac{-6}{7} \right)^{-3} \)

Final Answer : \( \left( \frac{-6}{7} \right)^{-3} \)

Question 3 (i)

Express the following into a single exponential form :
\( \left( -\frac{2}{3} \right)^{10} \times \left( -\frac{2}{3} \right)^{-15} \div \left( -\frac{2}{3} \right)^{25} \)

Step-by-step solution :

Laws Used :

Multiplication : \( x^m \times x^n = x^{m+n} \)
Division : \( x^m \div x^n = x^{m-n} \)

Combine all operations :
\( \left( -\frac{2}{3} \right)^{10 + (-15) – 25} \)
Simplify the exponent :
\( 10 – 15 – 25 = -30 \)
Write in single exponential form :
\( \left( -\frac{2}{3} \right)^{-30} \)

Final Answer : \( \left( -\frac{2}{3} \right)^{-30} \)