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| Example 1 to 7 (Before Exercise 10.1) | Exercise 10.1 | Example 8 & 9 (Before Exercise 10.2) |
| Exercise 10.2 | ||
Chapter 10 Exponents and Powers
Welcome to the solutions guide for Chapter 10, "Exponents and Powers," featured in the latest Class 8 NCERT mathematics textbook for the academic session 2024-25. This chapter significantly extends the concepts introduced in Class 7, equipping students with the ability to handle exponents that are negative integers and mastering the use of standard form (scientific notation) for representing both extremely large and extremely small numbers. These skills are fundamental in various scientific disciplines, engineering, and higher mathematics for concise representation and efficient calculation. This resource provides comprehensive, step-by-step solutions designed to clarify concepts and build computational fluency.
While reinforcing the laws of exponents for positive integer powers learned previously, the core advancement in this chapter is the introduction and manipulation of negative exponents. The solutions meticulously explain the fundamental rule governing negative integer exponents: for any non-zero rational number 'a' and positive integer 'n', the expression $a^{-n}$ is defined as the multiplicative inverse (reciprocal) of $a^n$. That is, $a^{-n} = \frac{1}{a^n}$. This simple yet powerful rule allows us to work seamlessly with powers that represent division or fractions. For instance, $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$. The solutions provide numerous worked examples demonstrating how to evaluate expressions involving negative exponents and simplify complex terms by converting negative powers to positive ones by moving them between the numerator and denominator of a fraction (e.g., $\frac{x^{-2}}{y^{-3}} = \frac{y^3}{x^2}$).
A crucial aspect is applying the standard Laws of Exponents consistently, now extended to encompass all integer exponents (positive, negative, and zero). The solutions thoroughly illustrate the application of these laws in simplifying intricate exponential expressions:
- Product Rule: $a^m \times a^n = a^{m+n}$ (e.g., $2^5 \times 2^{-3} = 2^{5+(-3)} = 2^2$)
- Quotient Rule: $\frac{a^m}{a^n} = a^{m-n}$ (where $a \neq 0$) (e.g., $\frac{5^2}{5^{-1}} = 5^{2-(-1)} = 5^3$)
- Power of a Power Rule: $(a^m)^n = a^{mn}$ (e.g., $(3^{-2})^4 = 3^{-8}$)
- Power of a Product Rule: $(ab)^m = a^m b^m$
- Power of a Quotient Rule: $(\frac{a}{b})^m = \frac{a^m}{b^m}$ (where $b \neq 0$)
- Zero Exponent Rule: $a^0 = 1$ (where $a \neq 0$)
A significant practical application of exponents is expressing numbers in Standard Form (Scientific Notation). This format, $k \times 10^n$, where $1 \le k < 10$ and $n$ is an integer, provides a compact way to write very large and, importantly introduced here, very small numbers. While positive values of $n$ are used for large numbers (e.g., $345,000,000 = 3.45 \times 10^8$), this chapter emphasizes using negative values of $n$ for numbers smaller than 1. The solutions demonstrate step-by-step how to convert small decimal numbers into standard form by shifting the decimal point appropriately (e.g., $0.000072 = 7.2 \times 10^{-5}$). Conversely, converting numbers from standard form (with negative $n$) back to their usual decimal form is also clearly explained (e.g., $1.9 \times 10^{-6} = 0.0000019$). Furthermore, the solutions address problems involving the comparison of very large and very small numbers, often facilitated by first expressing them in standard form.
Regarding the rationalized syllabus for 2024-25, Chapter 10, "Exponents and Powers," in the Class 8 NCERT textbook retains its core focus on extending laws of exponents to negative integers and expressing both large and small numbers in standard form. The rationalization primarily involved streamlining the exercises, ensuring a strong grasp of applying these rules and standard form conversions without excessive repetition or overly complex nested simplifications. By diligently working through these detailed solutions, students can achieve fluency in handling negative exponents, master the universal application of exponent laws, and become proficient in using standard form for representing and comparing numbers across vast scales.
Example 1 to 7 (Before Exercise 10.1)
Example 1: Find the value of
(i) 2–3
(ii) $\frac{1}{3^{-2}}$
Answer:
Solution:
We need to find the value of the given expressions using the laws of exponents.
(i) Find the value of $2^{-3}$.
Using the law of exponents, $a^{-n} = \frac{1}{a^n}$, where $a$ is any non-zero number and $n$ is a positive integer, we have:
$2^{-3} = \frac{1}{2^3}$
Now, we calculate the value of $2^3$ which is $2 \times 2 \times 2 = 8$.
So, $2^{-3} = \frac{1}{8}$.
Therefore, the value of $2^{-3}$ is $\frac{1}{8}$.
(ii) Find the value of $\frac{1}{3^{-2}}$.
Using the law of exponents, $\frac{1}{a^{-n}} = a^n$, where $a$ is any non-zero number and $n$ is a positive integer, we have:
$\frac{1}{3^{-2}} = 3^2$
Now, we calculate the value of $3^2$ which is $3 \times 3 = 9$.
So, $\frac{1}{3^{-2}} = 9$.
Therefore, the value of $\frac{1}{3^{-2}}$ is $9$.
Example 2: Simplify
(i) (– 4)5 × (– 4)–10
(ii) 25 ÷ 2– 6
Answer:
Solution:
We need to simplify the given expressions using the laws of exponents.
(i) Simplify $(-4)^5 \times (-4)^{-10}$.
Using the law of exponents, $a^m \times a^n = a^{m+n}$, where $a$ is any non-zero number and $m$, $n$ are integers, we have:
$(-4)^5 \times (-4)^{-10} = (-4)^{5 + (-10)}$
$ = (-4)^{5 - 10}$
$ = (-4)^{-5}$
Now, using the law $a^{-n} = \frac{1}{a^n}$, we get:
$(-4)^{-5} = \frac{1}{(-4)^5}$
We calculate $(-4)^5$: $(-4) \times (-4) \times (-4) \times (-4) \times (-4)$.
Since the exponent is odd, the result will be negative.
$4^5 = 4 \times 4 \times 4 \times 4 \times 4 = 1024$.
So, $(-4)^5 = -1024$.
Therefore, $(-4)^{-5} = \frac{1}{-1024} = -\frac{1}{1024}$.
The simplified value is $-\frac{1}{1024}$.
(ii) Simplify $2^5 \div 2^{-6}$.
Using the law of exponents, $a^m \div a^n = a^{m-n}$, where $a$ is any non-zero number and $m$, $n$ are integers, we have:
$2^5 \div 2^{-6} = 2^{5 - (-6)}$
$ = 2^{5 + 6}$
$ = 2^{11}$
Now, we calculate the value of $2^{11}$.
$2^{11} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2048$.
Therefore, the value of $2^5 \div 2^{-6}$ is $2048$.
Example 3: Express 4–3 as a power with the base 2.
Answer:
Solution:
We are asked to express $4^{-3}$ as a power with base 2.
We know that the base 4 can be written as a power of 2, i.e., $4 = 2^2$.
Substitute this into the expression $4^{-3}$:
$4^{-3} = (2^2)^{-3}$
Using the law of exponents, $(a^m)^n = a^{mn}$, where $a$ is any non-zero number and $m$, $n$ are integers, we have:
$(2^2)^{-3} = 2^{2 \times (-3)}$
$ = 2^{-6}$
Thus, $4^{-3}$ expressed as a power with the base 2 is $2^{-6}$.
Example 4: Simplify and write the answer in the exponential form.
(i) (25 ÷ 28)5 × 2–5
(ii) (–4)–3 × (5)–3 × (–5)–3
(iii) $\frac{1}{8}$ × (3)-3
(iv) (-3)4 × $\left( \frac{5}{3} \right)^{4}$
Answer:
Solution:
We need to simplify the given expressions and write the answer in exponential form using the laws of exponents.
(i) Simplify $(2^5 \div 2^8)^5 \times 2^{-5}$.
First, simplify the expression inside the parenthesis using the law $a^m \div a^n = a^{m-n}$:
$2^5 \div 2^8 = 2^{5-8} = 2^{-3}$
Now, the expression becomes $(2^{-3})^5 \times 2^{-5}$.
Simplify $(2^{-3})^5$ using the law $(a^m)^n = a^{mn}$:
$(2^{-3})^5 = 2^{-3 \times 5} = 2^{-15}$
The expression is now $2^{-15} \times 2^{-5}$.
Finally, use the law $a^m \times a^n = a^{m+n}$:
$2^{-15} \times 2^{-5} = 2^{-15 + (-5)} = 2^{-15 - 5} = 2^{-20}$
The simplified expression in exponential form is $2^{-20}$.
(ii) Simplify $(-4)^{-3} \times (5)^{-3} \times (-5)^{-3}$.
We can use the law $a^m \times b^m = (ab)^m$ repeatedly.
$(-4)^{-3} \times (5)^{-3} \times (-5)^{-3} = ((-4) \times 5)^{-3} \times (-5)^{-3}$
$ = (-20)^{-3} \times (-5)^{-3}$
$ = ((-20) \times (-5))^{-3}$
$ = (100)^{-3}$
We can also write 100 as a power of 10:
$100 = 10^2$
So, the expression becomes $(10^2)^{-3}$.
Using the law $(a^m)^n = a^{mn}$:
$(10^2)^{-3} = 10^{2 \times (-3)} = 10^{-6}$
The simplified expression in exponential form is $10^{-6}$.
(iii) Simplify $\frac{1}{8} \times (3)^{-3}$.
First, express $\frac{1}{8}$ as a power of 2.
$8 = 2^3$, so $\frac{1}{8} = \frac{1}{2^3}$.
Using the law $\frac{1}{a^n} = a^{-n}$:
$\frac{1}{2^3} = 2^{-3}$
Next, express $(3)^{-3}$ using the law $a^{-n} = \frac{1}{a^n}$:
$(3)^{-3} = \frac{1}{3^3}$
The expression is now $2^{-3} \times 3^{-3}$.
Using the law $a^m \times b^m = (ab)^m$:
$2^{-3} \times 3^{-3} = (2 \times 3)^{-3} = 6^{-3}$
The simplified expression in exponential form is $6^{-3}$.
(iv) Simplify $(-3)^4 \times \left( \frac{5}{3} \right)^{4}$.
Using the law $a^m \times b^m = (ab)^m$:
$(-3)^4 \times \left( \frac{5}{3} \right)^{4} = \left( (-3) \times \frac{5}{3} \right)^4$
Calculate the product inside the parenthesis:
$(-3) \times \frac{5}{3} = -\cancel{3} \times \frac{5}{\cancel{3}} = -5$
The expression becomes $(-5)^4$.
The simplified expression in exponential form is $(-5)^4$.
Alternatively, we can note that $(-5)^4 = (-5) \times (-5) \times (-5) \times (-5) = 625$ and $5^4 = 5 \times 5 \times 5 \times 5 = 625$. So, $(-5)^4 = 5^4$. However, the question asks for the answer in exponential form, and $(-5)^4$ is a valid exponential form derived directly from the calculation.
Example 5: Find m so that (–3)m + 1 × (–3)5 = (–3)7
Answer:
Solution:
We are given the equation: $(–3)^{m + 1} \times (–3)^{5} = (–3)^{7}$.
We will use the law of exponents which states that for any non-zero number $a$ and integers $m$ and $n$, $a^m \times a^n = a^{m+n}$.
Applying this law to the left side of the given equation:
$(-3)^{m + 1} \times (-3)^{5} = (-3)^{(m + 1) + 5}$
$ = (-3)^{m + 6}$
Now, the equation becomes:
$(-3)^{m + 6} = (-3)^{7}$
Since the bases are the same and non-zero, we can equate the exponents.
$m + 6 = 7$
To find the value of $m$, subtract 6 from both sides of the equation.
$m = 7 - 6$
$m = 1$
Thus, the value of $m$ is $\mathbf{1}$.
Verification:
Substitute $m=1$ back into the original equation:
$(-3)^{1 + 1} \times (-3)^{5} = (-3)^{2} \times (-3)^{5}$
Using the law $a^m \times a^n = a^{m+n}$:
$(-3)^{2} \times (-3)^{5} = (-3)^{2+5} = (-3)^7$
This matches the right side of the original equation, so the value $m=1$ is correct.
Example 6: Find the value of $\left( \frac{2}{3} \right)^{-2}$
Answer:
Solution:
We need to find the value of the given expression $\left( \frac{2}{3} \right)^{-2}$.
We use the law of exponents which states that for any non-zero rational number $\frac{a}{b}$ and positive integer $n$, $\left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n}$.
Applying this law to the given expression:
$\left( \frac{2}{3} \right)^{-2} = \left( \frac{3}{2} \right)^{2}$
Now, we use the law $\left( \frac{a}{b} \right)^{n} = \frac{a^n}{b^n}$.
$\left( \frac{3}{2} \right)^{2} = \frac{3^2}{2^2}$
Calculate the values of $3^2$ and $2^2$:
$3^2 = 3 \times 3 = 9$
$2^2 = 2 \times 2 = 4$
So, $\frac{3^2}{2^2} = \frac{9}{4}$.
Therefore, the value of $\left( \frac{2}{3} \right)^{-2}$ is $\mathbf{\frac{9}{4}}$.
Example 7: Simplify
(i) $\left\{ \left( \frac{1}{3} \right)^{-2} - \left( \frac{1}{2} \right)^{-3}\right\}$ ÷ $\left( \frac{1}{4} \right)^{-2}$
(ii) $\left( \frac{5}{8} \right)^{-7}$ × $\left( \frac{8}{5} \right)^{-5}$
Answer:
Solution:
We need to simplify the given expressions using the laws of exponents.
(i) Simplify $\left\{ \left( \frac{1}{3} \right)^{-2} - \left( \frac{1}{2} \right)^{-3}\right\}$ ÷ $\left( \frac{1}{4} \right)^{-2}$.
First, evaluate the terms with negative exponents using the property $\left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n}$.
$\left( \frac{1}{3} \right)^{-2} = \left( \frac{3}{1} \right)^{2} = 3^2 = 9$
$\left( \frac{1}{2} \right)^{-3} = \left( \frac{2}{1} \right)^{3} = 2^3 = 8$
$\left( \frac{1}{4} \right)^{-2} = \left( \frac{4}{1} \right)^{2} = 4^2 = 16$
Now substitute these values back into the expression:
$\{9 - 8\} \div 16$
Perform the subtraction inside the curly braces:
$1 \div 16$
Perform the division:
$1 \div 16 = \frac{1}{16}$
The simplified value is $\frac{1}{16}$.
(ii) Simplify $\left( \frac{5}{8} \right)^{-7}$ × $\left( \frac{8}{5} \right)^{-5}$.
We can rewrite the second term $\left( \frac{8}{5} \right)^{-5}$ using the property $\left( \frac{b}{a} \right)^{-n} = \left( \frac{a}{b} \right)^{n}$.
$\left( \frac{8}{5} \right)^{-5} = \left( \frac{5}{8} \right)^{5}$
Now the expression becomes:
$\left( \frac{5}{8} \right)^{-7}$ × $\left( \frac{5}{8} \right)^{5}$
Using the law of exponents $a^m \times a^n = a^{m+n}$, where $a = \frac{5}{8}$, $m = -7$, and $n = 5$:
$\left( \frac{5}{8} \right)^{-7 + 5} = \left( \frac{5}{8} \right)^{-2}$
Finally, use the property $\left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n}$ again:
$\left( \frac{5}{8} \right)^{-2} = \left( \frac{8}{5} \right)^{2}$
Now, calculate the value:
$\left( \frac{8}{5} \right)^{2} = \frac{8^2}{5^2} = \frac{64}{25}$
The simplified value is $\frac{64}{25}$.
Exercise 10.1
Question 1. Evaluate.
(i) 3–2
(ii) (– 4)–2
(iii) $\left( \frac{1}{2} \right)^{-5}$
Answer:
Solution:
We need to evaluate the given expressions using the laws of exponents.
(i) Evaluate $3^{-2}$.
Using the law of exponents, $a^{-n} = \frac{1}{a^n}$, where $a$ is any non-zero number and $n$ is a positive integer, we have:
$3^{-2} = \frac{1}{3^2}$
Now, calculate the value of $3^2$:
$3^2 = 3 \times 3 = 9$
So, $3^{-2} = \frac{1}{9}$.
The value of $3^{-2}$ is $\mathbf{\frac{1}{9}}$.
(ii) Evaluate $(-4)^{-2}$.
Using the law of exponents, $a^{-n} = \frac{1}{a^n}$, where $a$ is any non-zero number and $n$ is a positive integer, we have:
$(-4)^{-2} = \frac{1}{(-4)^2}$
Now, calculate the value of $(-4)^2$:
$(-4)^2 = (-4) \times (-4) = 16$
So, $(-4)^{-2} = \frac{1}{16}$.
The value of $(-4)^{-2}$ is $\mathbf{\frac{1}{16}}$.
(iii) Evaluate $\left( \frac{1}{2} \right)^{-5}$.
Using the law of exponents, $\left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n}$, where $\frac{a}{b}$ is a non-zero rational number and $n$ is a positive integer, we have:
$\left( \frac{1}{2} \right)^{-5} = \left( \frac{2}{1} \right)^{5}$
$\left( \frac{2}{1} \right)^{5} = 2^5$
Now, calculate the value of $2^5$:
$2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$
So, $\left( \frac{1}{2} \right)^{-5} = 32$.
The value of $\left( \frac{1}{2} \right)^{-5}$ is $\mathbf{32}$.
Question 2. Simplify and express the result in power notation with positive exponent.
(i) (– 4)5 ÷ (– 4)8
(ii) $\left( \frac{1}{2^3} \right)^{2}$
(iii) (-3)4 × $\left( \frac{5}{3} \right)^{4}$
(iv) (3–7 ÷ 3–10) × 3–5
(v) 2–3 × (–7)–3
Answer:
Solution:
We need to simplify the given expressions and express the result in power notation with a positive exponent using the laws of exponents.
(i) Simplify $(-4)^5 \div (-4)^8$.
Using the law $a^m \div a^n = a^{m-n}$:
$(-4)^5 \div (-4)^8 = (-4)^{5-8}$
$ = (-4)^{-3}$
To express with a positive exponent, use the law $a^{-n} = \frac{1}{a^n}$:
$(-4)^{-3} = \frac{1}{(-4)^3}$
The result in power notation with positive exponent is $\mathbf{\frac{1}{(-4)^3}}$.
(ii) Simplify $\left( \frac{1}{2^3} \right)^{2}$.
First, rewrite $\frac{1}{2^3}$ using $a^{-n} = \frac{1}{a^n}$ or $\frac{1}{a^n} = a^{-n}$:
$\frac{1}{2^3} = 2^{-3}$
Now the expression is $(2^{-3})^2$.
Using the law $(a^m)^n = a^{mn}$:
$(2^{-3})^2 = 2^{-3 \times 2}$
$ = 2^{-6}$
To express with a positive exponent, use the law $a^{-n} = \frac{1}{a^n}$:
$2^{-6} = \frac{1}{2^6}$
The result in power notation with positive exponent is $\mathbf{\frac{1}{2^6}}$.
(iii) Simplify $(-3)^4 \times \left( \frac{5}{3} \right)^{4}$.
Using the law $a^m \times b^m = (ab)^m$:
$(-3)^4 \times \left( \frac{5}{3} \right)^{4} = \left( (-3) \times \frac{5}{3} \right)^{4}$
Calculate the product inside the parenthesis:
$(-3) \times \frac{5}{3} = -\cancel{3} \times \frac{5}{\cancel{3}} = -5$
The expression becomes $(-5)^4$.
The exponent is already positive.
The result in power notation with positive exponent is $\mathbf{(-5)^4}$.
(iv) Simplify $(3^{-7} \div 3^{-10}) \times 3^{-5}$.
First, simplify the expression inside the parenthesis using the law $a^m \div a^n = a^{m-n}$:
$3^{-7} \div 3^{-10} = 3^{-7 - (-10)}$
$ = 3^{-7 + 10}$
$ = 3^3$
Now, the expression is $3^3 \times 3^{-5}$.
Using the law $a^m \times a^n = a^{m+n}$:
$3^3 \times 3^{-5} = 3^{3 + (-5)}$
$ = 3^{3 - 5}$
$ = 3^{-2}$
To express with a positive exponent, use the law $a^{-n} = \frac{1}{a^n}$:
$3^{-2} = \frac{1}{3^2}$
The result in power notation with positive exponent is $\mathbf{\frac{1}{3^2}}$.
(v) Simplify $2^{-3} \times (–7)^{-3}$.
Using the law $a^m \times b^m = (ab)^m$:
$2^{-3} \times (-7)^{-3} = (2 \times (-7))^{-3}$
$ = (-14)^{-3}$
To express with a positive exponent, use the law $a^{-n} = \frac{1}{a^n}$:
$(-14)^{-3} = \frac{1}{(-14)^3}$
The result in power notation with positive exponent is $\mathbf{\frac{1}{(-14)^3}}$.
Question 3. Find the value of.
(i) (30 + 4–1) × 22
(ii) (2–1 × 4–1) ÷ 2–2
(iii) $\left( \frac{1}{2} \right)^{-2}$ + $\left( \frac{1}{3} \right)^{-2}$ + $\left( \frac{1}{4} \right)^{-2}$
(iv) (3–1 + 4–1 + 5–1)0
(v) $\left\{ \left( \frac{-2}{3} \right)^{-2} \right\}^{2}$
Answer:
Solution:
We need to find the value of the given expressions by simplifying them using the laws of exponents.
(i) Evaluate $(3^0 + 4^{-1}) \times 2^2$.
Using the laws $a^0 = 1$ and $a^{-n} = \frac{1}{a^n}$:
$3^0 = 1$
$4^{-1} = \frac{1}{4^1} = \frac{1}{4}$
$2^2 = 2 \times 2 = 4$
Substitute these values into the expression:
$(1 + \frac{1}{4}) \times 4$
First, calculate the sum inside the parenthesis:
$1 + \frac{1}{4} = \frac{4}{4} + \frac{1}{4} = \frac{4+1}{4} = \frac{5}{4}$
Now, multiply the result by 4:
$\frac{5}{4} \times 4 = 5$
The value is $\mathbf{5}$.
(ii) Evaluate $(2^{-1} \times 4^{-1}) \div 2^{-2}$.
We can use the laws $a^{-n} = \frac{1}{a^n}$ and $a^m \times a^n = a^{m+n}$ and $a^m \div a^n = a^{m-n}$.
Rewrite the expression using negative exponents for the division part.
$(2^{-1} \times 4^{-1}) \div 2^{-2}$
Express 4 as $2^2$:
$(2^{-1} \times (2^2)^{-1}) \div 2^{-2}$
Using $(a^m)^n = a^{mn}$ for $(2^2)^{-1}$:
$(2^{-1} \times 2^{2 \times (-1)}) \div 2^{-2}$
$(2^{-1} \times 2^{-2}) \div 2^{-2}$
Using $a^m \times a^n = a^{m+n}$ for the product in parenthesis:
$2^{-1 + (-2)} \div 2^{-2}$
$2^{-3} \div 2^{-2}$
Using $a^m \div a^n = a^{m-n}$:
$2^{-3 - (-2)} = 2^{-3 + 2} = 2^{-1}$
Finally, using $a^{-n} = \frac{1}{a^n}$:
$2^{-1} = \frac{1}{2^1} = \frac{1}{2}$
The value is $\mathbf{\frac{1}{2}}$.
(iii) Evaluate $\left( \frac{1}{2} \right)^{-2}$ + $\left( \frac{1}{3} \right)^{-2}$ + $\left( \frac{1}{4} \right)^{-2}$.
Using the law $\left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n}$:
$\left( \frac{1}{2} \right)^{-2} = \left( \frac{2}{1} \right)^{2} = 2^2 = 4$
$\left( \frac{1}{3} \right)^{-2} = \left( \frac{3}{1} \right)^{2} = 3^2 = 9$
$\left( \frac{1}{4} \right)^{-2} = \left( \frac{4}{1} \right)^{2} = 4^2 = 16$
Substitute these values into the expression:
$4 + 9 + 16$
Perform the addition:
$4 + 9 + 16 = 13 + 16 = 29$
The value is $\mathbf{29}$.
(iv) Evaluate $(3^{-1} + 4^{-1} + 5^{-1})^0$.
Using the law $a^0 = 1$ for any non-zero number $a$.
The base of the expression is $(3^{-1} + 4^{-1} + 5^{-1})$.
Using $a^{-n} = \frac{1}{a^n}$:
$3^{-1} = \frac{1}{3}$
$4^{-1} = \frac{1}{4}$
$5^{-1} = \frac{1}{5}$
The base is $\frac{1}{3} + \frac{1}{4} + \frac{1}{5}$.
This sum is $\frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{47}{60}$.
Since the base $\frac{47}{60}$ is non-zero, any non-zero number raised to the power of 0 is 1.
$(3^{-1} + 4^{-1} + 5^{-1})^0 = \left(\frac{47}{60}\right)^0 = 1$
The value is $\mathbf{1}$.
(v) Evaluate $\left\{ \left( \frac{-2}{3} \right)^{-2} \right\}^{2}$.
Using the law $(a^m)^n = a^{mn}$:
$\left\{ \left( \frac{-2}{3} \right)^{-2} \right\}^{2} = \left( \frac{-2}{3} \right)^{-2 \times 2} = \left( \frac{-2}{3} \right)^{-4}$
Using the law $\left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n}$:
$= \left( \frac{3}{-2} \right)^{4}$
This can be written as $\left( -\frac{3}{2} \right)^{4}$.
Using the law $\left( \frac{a}{b} \right)^{n} = \frac{a^n}{b^n}$:
$= \frac{(-3)^4}{2^4}$
Calculate the values of the numerator and denominator:
$(-3)^4 = (-3) \times (-3) \times (-3) \times (-3) = 81$
$2^4 = 2 \times 2 \times 2 \times 2 = 16$
$= \frac{81}{16}$
The value is $\mathbf{\frac{81}{16}}$.
Question 4. Evaluate
(i) $\frac{8^{-1} \;\times\; 5^{3}}{2^{-4}}$
(ii) (5–1 × 2–1) × 6–1
Answer:
Solution:
We need to evaluate the given expressions using the laws of exponents.
(i) Evaluate $\frac{8^{-1} \;\times\; 5^{3}}{2^{-4}}$.
We can rewrite the expression using the law $a^{-n} = \frac{1}{a^n}$ and also by expressing 8 as a power of 2, i.e., $8 = 2^3$.
The expression is $\frac{(2^3)^{-1} \times 5^3}{2^{-4}}$.
Using the law $(a^m)^n = a^{mn}$, we simplify $(2^3)^{-1}$:
$(2^3)^{-1} = 2^{3 \times (-1)} = 2^{-3}$
So the expression becomes $\frac{2^{-3} \times 5^3}{2^{-4}}$.
Now, use the law $\frac{a^m}{a^n} = a^{m-n}$ for the terms with base 2:
$\frac{2^{-3}}{2^{-4}} = 2^{-3 - (-4)} = 2^{-3 + 4} = 2^1$
The expression simplifies to $2^1 \times 5^3$.
Now, we evaluate the powers:
$2^1 = 2$
$5^3 = 5 \times 5 \times 5 = 125$
Multiply the results:
$2 \times 125 = 250$
The value is $\mathbf{250}$.
(ii) Evaluate $(5^{-1} \times 2^{-1}) \times 6^{-1}$.
We use the law of exponents $a^m \times b^m = (ab)^m$.
Apply this law to the terms inside the parenthesis:
$(5^{-1} \times 2^{-1}) = (5 \times 2)^{-1} = 10^{-1}$
The expression becomes $10^{-1} \times 6^{-1}$.
Apply the law $a^m \times b^m = (ab)^m$ again:
$10^{-1} \times 6^{-1} = (10 \times 6)^{-1} = 60^{-1}$
Finally, use the law $a^{-n} = \frac{1}{a^n}$:
$60^{-1} = \frac{1}{60^1} = \frac{1}{60}$
The value is $\mathbf{\frac{1}{60}}$.
Question 5. Find the value of m for which 5m ÷ 5–3 = 55.
Answer:
Solution:
We are given the equation: $5^m \div 5^{-3} = 5^5$.
We will use the law of exponents which states that for any non-zero number $a$ and integers $p$ and $q$, $a^p \div a^q = a^{p-q}$.
Applying this law to the left side of the given equation:
$5^m \div 5^{-3} = 5^{m - (-3)}$
$ = 5^{m + 3}$
Now, the equation becomes:
$5^{m + 3} = 5^5$
Since the bases on both sides of the equation are the same and non-zero (which is 5), the exponents must be equal.
$m + 3 = 5$
To find the value of $m$, subtract 3 from both sides of the equation.
$m = 5 - 3$
$m = 2$
Thus, the value of $m$ is $\mathbf{2}$.
Verification:
Substitute $m=2$ back into the original equation:
$5^{2} \div 5^{-3}$
Using the law $a^p \div a^q = a^{p-q}$:
$5^{2 - (-3)} = 5^{2+3} = 5^5$
This matches the right side of the original equation, so the value $m=2$ is correct.
Question 6. Evaluate
(i) $\left\{ \left( \frac{1}{3} \right)^{-1}-\left( \frac{1}{4} \right)^{-1} \right\}^{-1}$
(ii) $\left( \frac{5}{8} \right)^{-7}$ × $\left( \frac{8}{5} \right)^{-4}$
Answer:
Solution:
We need to evaluate the given expressions using the laws of exponents.
(i) Evaluate $\left\{ \left( \frac{1}{3} \right)^{-1}-\left( \frac{1}{4} \right)^{-1} \right\}^{-1}$.
First, evaluate the terms inside the curly braces using the property $\left( \frac{a}{b} \right)^{-1} = \frac{b}{a}$.
$\left( \frac{1}{3} \right)^{-1} = \frac{3}{1} = 3$
$\left( \frac{1}{4} \right)^{-1} = \frac{4}{1} = 4$
Substitute these values back into the expression inside the curly braces:
$3 - 4 = -1$
Now, the expression becomes $(-1)^{-1}$.
Using the property $a^{-1} = \frac{1}{a}$:
$(-1)^{-1} = \frac{1}{-1} = -1$
The value is $\mathbf{-1}$.
(ii) Evaluate $\left( \frac{5}{8} \right)^{-7}$ × $\left( \frac{8}{5} \right)^{-4}$.
We can rewrite the second term $\left( \frac{8}{5} \right)^{-4}$ using the property $\left( \frac{b}{a} \right)^{-n} = \left( \frac{a}{b} \right)^{n}$.
$\left( \frac{8}{5} \right)^{-4} = \left( \frac{5}{8} \right)^{4}$
Now the expression becomes:
$\left( \frac{5}{8} \right)^{-7}$ × $\left( \frac{5}{8} \right)^{4}$
Using the law of exponents $a^m \times a^n = a^{m+n}$, where $a = \frac{5}{8}$, $m = -7$, and $n = 4$:
$\left( \frac{5}{8} \right)^{-7 + 4} = \left( \frac{5}{8} \right)^{-3}$
Finally, use the property $\left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^{n}$ again:
$\left( \frac{5}{8} \right)^{-3} = \left( \frac{8}{5} \right)^{3}$
Now, calculate the value by applying the exponent to both numerator and denominator:
$\left( \frac{8}{5} \right)^{3} = \frac{8^3}{5^3}$
Calculate $8^3$ and $5^3$:
$8^3 = 8 \times 8 \times 8 = 64 \times 8 = 512$
$5^3 = 5 \times 5 \times 5 = 25 \times 5 = 125$
So, $\frac{8^3}{5^3} = \frac{512}{125}$.
The value is $\mathbf{\frac{512}{125}}$.
Question 7. Simplify.
(i) $\frac{25 \;\times\; t^{-4}}{5^{-3} \;\times\; 10 \;\times\; t^{-8}}$ (t ≠ 0)
(ii) $\frac{3^{-5} \;\times\; 10^{-5} \;\times\; 125}{5^{-7} \;\times\; 6^{-5}}$
Answer:
Solution:
We need to simplify the given expressions using the laws of exponents.
(i) Simplify $\frac{25 \;\times\; t^{-4}}{5^{-3} \;\times\; 10 \;\times\; t^{-8}}$ (t ≠ 0).
First, express the constants as powers of their prime factors:
$25 = 5^2$
$10 = 2 \times 5$
Substitute these into the expression:
$\frac{5^2 \;\times\; t^{-4}}{5^{-3} \;\times\; (2 \times 5) \;\times\; t^{-8}}$
Simplify the denominator by combining terms with the same base:
$5^{-3} \times 2^1 \times 5^1 \times t^{-8} = 5^{-3+1} \times 2^1 \times t^{-8} = 5^{-2} \times 2^1 \times t^{-8}$
The expression becomes:
$\frac{5^2 \;\times\; t^{-4}}{5^{-2} \;\times\; 2^1 \;\times\; t^{-8}}$
Now, use the laws of exponents $\frac{a^m}{a^n} = a^{m-n}$ and $\frac{1}{a^n} = a^{-n}$:
$\frac{5^2}{5^{-2}} = 5^{2 - (-2)} = 5^{2+2} = 5^4$
$\frac{t^{-4}}{t^{-8}} = t^{-4 - (-8)} = t^{-4+8} = t^4$
$\frac{1}{2^1} = 2^{-1}$ (This term is in the denominator, so it will appear in the numerator with a negative exponent or remain in the denominator with a positive exponent).
Combining the terms:
$5^4 \times \frac{1}{2^1} \times t^4 = \frac{5^4 \;\times\; t^4}{2}$
Evaluate the constant term $5^4$:
$5^4 = 5 \times 5 \times 5 \times 5 = 625$
The simplified expression is $\mathbf{\frac{625 t^4}{2}}$.
(ii) Simplify $\frac{3^{-5} \;\times\; 10^{-5} \;\times\; 125}{5^{-7} \;\times\; 6^{-5}}$.
First, express the constants as powers of their prime factors:
$10 = 2 \times 5$
$125 = 5^3$
$6 = 2 \times 3$
Substitute these into the expression and use the law $(ab)^n = a^n b^n$:
$\frac{3^{-5} \;\times\; (2 \times 5)^{-5} \;\times\; 5^3}{5^{-7} \;\times\; (2 \times 3)^{-5}}$
$\frac{3^{-5} \;\times\; 2^{-5} \;\times\; 5^{-5} \;\times\; 5^3}{5^{-7} \;\times\; 2^{-5} \;\times\; 3^{-5}}$
Simplify the numerator by combining terms with the same base (5):
$3^{-5} \times 2^{-5} \times 5^{-5+3} = 3^{-5} \times 2^{-5} \times 5^{-2}$
The denominator is already in terms of prime factor powers:
$5^{-7} \times 2^{-5} \times 3^{-5}$
The expression is now:
$\frac{3^{-5} \;\times\; 2^{-5} \;\times\; 5^{-2}}{3^{-5} \;\times\; 2^{-5} \;\times\; 5^{-7}}$
Now, use the law of exponents $\frac{a^m}{a^n} = a^{m-n}$ for each base:
For base 3: $\frac{3^{-5}}{3^{-5}} = 3^{-5 - (-5)} = 3^{-5+5} = 3^0 = 1$
For base 2: $\frac{2^{-5}}{2^{-5}} = 2^{-5 - (-5)} = 2^{-5+5} = 2^0 = 1$
For base 5: $\frac{5^{-2}}{5^{-7}} = 5^{-2 - (-7)} = 5^{-2+7} = 5^5$
Combine the results:
$1 \times 1 \times 5^5 = 5^5$
Evaluate the constant term $5^5$:
$5^5 = 5 \times 5 \times 5 \times 5 \times 5 = 3125$
The simplified value is $\mathbf{3125}$.
Example 8 & 9 (Before Exercise 10.2)
Example 8: Express the following numbers in standard form.
(i) 0.000035
(ii) 4050000
Answer:
Solution:
We need to express the given numbers in standard form, which is of the form $a \times 10^k$, where $1 \leq |a| < 10$ and $k$ is an integer.
(i) Express 0.000035 in standard form.
To write 0.000035 in standard form, we need to move the decimal point so that there is only one non-zero digit to the left of the decimal point.
The number is 0.000035.
Move the decimal point 5 places to the right to get 3.5.
Since we moved the decimal point 5 places to the right, the power of 10 will be negative, i.e., $10^{-5}$.
So, $0.000035 = 3.5 \times 10^{-5}$.
In standard form, the number is $\mathbf{3.5 \times 10^{-5}}$.
(ii) Express 4050000 in standard form.
To write 4050000 in standard form, we need to move the decimal point so that there is only one non-zero digit to the left of the decimal point.
The number is 4050000.
The decimal point is implicitly after the last zero. We move it to the left.
Move the decimal point 6 places to the left to get 4.05.
Since we moved the decimal point 6 places to the left, the power of 10 will be positive, i.e., $10^{6}$.
So, $4050000 = 4.05 \times 10^{6}$.
In standard form, the number is $\mathbf{4.05 \times 10^{6}}$.
Example 9: Express the following numbers in usual form.
(i) 3.52 × 105
(ii) 7.54 × 10–4
(iii) 3 × 10–5
Answer:
Solution:
We need to express the given numbers from standard form to the usual form.
(i) Express $3.52 \times 10^5$ in usual form.
The exponent of 10 is positive (5), so we move the decimal point 5 places to the right.
Start with 3.52. Move the decimal point 5 places to the right.
3.52 $\rightarrow$ 35.2 (1 place) $\rightarrow$ 352. (2 places) $\rightarrow$ 3520. (3 places) $\rightarrow$ 35200. (4 places) $\rightarrow$ 352000. (5 places)
So, $3.52 \times 10^5 = 352000$.
The usual form is $\mathbf{352000}$.
(ii) Express $7.54 \times 10^{-4}$ in usual form.
The exponent of 10 is negative (-4), so we move the decimal point 4 places to the left.
Start with 7.54. Move the decimal point 4 places to the left, adding zeros as needed.
7.54 $\rightarrow$ 0.754 (1 place) $\rightarrow$ 0.0754 (2 places) $\rightarrow$ 0.00754 (3 places) $\rightarrow$ 0.000754 (4 places)
So, $7.54 \times 10^{-4} = 0.000754$.
The usual form is $\mathbf{0.000754}$.
(iii) Express $3 \times 10^{-5}$ in usual form.
The exponent of 10 is negative (-5), so we move the decimal point 5 places to the left.
Start with 3. (the decimal point is after 3). Move the decimal point 5 places to the left, adding zeros as needed.
3. $\rightarrow$ 0.3 (1 place) $\rightarrow$ 0.03 (2 places) $\rightarrow$ 0.003 (3 places) $\rightarrow$ 0.0003 (4 places) $\rightarrow$ 0.00003 (5 places)
So, $3 \times 10^{-5} = 0.00003$.
The usual form is $\mathbf{0.00003}$.
Exercise 10.2
Question 1. Express the following numbers in standard form.
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000
Answer:
Solution:
We need to express the given numbers in standard form, which is of the form $a \times 10^k$, where $1 \leq |a| < 10$ and $k$ is an integer.
(i) Express 0.0000000000085 in standard form.
To get a number between 1 and 10, we move the decimal point to the right until it is after the first non-zero digit, which is 8.
0.0000000000085 becomes 8.5.
The decimal point was moved 12 places to the right.
Since the decimal point was moved to the right, the exponent of 10 is negative.
The standard form is $\mathbf{8.5 \times 10^{-12}}$.
(ii) Express 0.00000000000942 in standard form.
To get a number between 1 and 10, we move the decimal point to the right until it is after the first non-zero digit, which is 9.
0.00000000000942 becomes 9.42.
The decimal point was moved 12 places to the right.
Since the decimal point was moved to the right, the exponent of 10 is negative.
The standard form is $\mathbf{9.42 \times 10^{-12}}$.
(iii) Express 6020000000000000 in standard form.
To get a number between 1 and 10, we move the decimal point to the left until it is after the first digit, which is 6.
The original number has an implied decimal point after the last zero: 6020000000000000.
Moving the decimal point 15 places to the left gives 6.02.
Since the decimal point was moved to the left, the exponent of 10 is positive.
The standard form is $\mathbf{6.02 \times 10^{15}}$.
(iv) Express 0.00000000837 in standard form.
To get a number between 1 and 10, we move the decimal point to the right until it is after the first non-zero digit, which is 8.
0.00000000837 becomes 8.37.
The decimal point was moved 9 places to the right.
Since the decimal point was moved to the right, the exponent of 10 is negative.
The standard form is $\mathbf{8.37 \times 10^{-9}}$.
(v) Express 31860000000 in standard form.
To get a number between 1 and 10, we move the decimal point to the left until it is after the first digit, which is 3.
The original number has an implied decimal point after the last zero: 31860000000.
Moving the decimal point 10 places to the left gives 3.186.
Since the decimal point was moved to the left, the exponent of 10 is positive.
The standard form is $\mathbf{3.186 \times 10^{10}}$.
Question 2. Express the following numbers in usual form.
(i) 3.02 × 10–6
(ii) 4.5 × 104
(iii) 3 × 10–8
(iv) 1.0001 × 109
(v) 5.8 × 1012
(vi) 3.61492 × 106
Answer:
Solution:
We need to express the given numbers from standard form to the usual form.
(i) Express $3.02 \times 10^{-6}$ in usual form.
The exponent of 10 is negative (-6). We move the decimal point 6 places to the left.
Start with 3.02. Move the decimal point 6 places to the left, adding zeros as needed:
3.02 $\rightarrow$ 0.302 (1 place) $\rightarrow$ 0.0302 (2 places) $\rightarrow$ 0.00302 (3 places) $\rightarrow$ 0.000302 (4 places) $\rightarrow$ 0.0000302 (5 places) $\rightarrow$ 0.00000302 (6 places)
The usual form is $\mathbf{0.00000302}$.
(ii) Express $4.5 \times 10^{4}$ in usual form.
The exponent of 10 is positive (4). We move the decimal point 4 places to the right.
Start with 4.5. Move the decimal point 4 places to the right, adding zeros as needed:
4.5 $\rightarrow$ 45. (1 place) $\rightarrow$ 450. (2 places) $\rightarrow$ 4500. (3 places) $\rightarrow$ 45000. (4 places)
The usual form is $\mathbf{45000}$.
(iii) Express $3 \times 10^{-8}$ in usual form.
The exponent of 10 is negative (-8). We move the decimal point 8 places to the left.
Start with 3. (the decimal point is after 3). Move the decimal point 8 places to the left, adding zeros as needed:
3. $\rightarrow$ 0.3 (1 place) $\rightarrow$ 0.03 (2 places) $\rightarrow$ 0.003 (3 places) $\rightarrow$ 0.0003 (4 places) $\rightarrow$ 0.00003 (5 places) $\rightarrow$ 0.000003 (6 places) $\rightarrow$ 0.0000003 (7 places) $\rightarrow$ 0.00000003 (8 places)
The usual form is $\mathbf{0.00000003}$.
(iv) Express $1.0001 \times 10^{9}$ in usual form.
The exponent of 10 is positive (9). We move the decimal point 9 places to the right.
Start with 1.0001. Move the decimal point 9 places to the right, adding zeros as needed:
1.0001 $\rightarrow$ 10.001 (1 place) $\rightarrow$ 100.01 (2 places) $\rightarrow$ 1000.1 (3 places) $\rightarrow$ 10001. (4 places) $\rightarrow$ 100010. (5 places) $\rightarrow$ 1000100. (6 places) $\rightarrow$ 10001000. (7 places) $\rightarrow$ 100010000. (8 places) $\rightarrow$ 1000100000. (9 places)
The usual form is $\mathbf{1000100000}$.
(v) Express $5.8 \times 10^{12}$ in usual form.
The exponent of 10 is positive (12). We move the decimal point 12 places to the right.
Start with 5.8. Move the decimal point 12 places to the right, adding zeros as needed:
5.8 $\rightarrow$ 58. (1 place). We need to move 11 more places, so we add 11 zeros.
The usual form is $\mathbf{5800000000000}$.
(vi) Express $3.61492 \times 10^{6}$ in usual form.
The exponent of 10 is positive (6). We move the decimal point 6 places to the right.
Start with 3.61492. Move the decimal point 6 places to the right:
3.61492 $\rightarrow$ 36.1492 (1 place) $\rightarrow$ 361.492 (2 places) $\rightarrow$ 3614.92 (3 places) $\rightarrow$ 36149.2 (4 places) $\rightarrow$ 361492. (5 places) $\rightarrow$ 3614920. (6 places)
The usual form is $\mathbf{3614920}$.
Question 3. Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to $\frac{1}{1000000}$ m.
(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.
(iii) Size of a bacteria is 0.0000005 m
(iv) Size of a plant cell is 0.00001275 m
(v) Thickness of a thick paper is 0.07 mm
Answer:
Solution:
We need to express the numbers in the given statements in standard form ($a \times 10^k$, where $1 \leq |a| < 10$ and $k$ is an integer).
(i) 1 micron is equal to $\frac{1}{1000000}$ m.
The number is $\frac{1}{1000000}$.
$\frac{1}{1000000} = \frac{1}{10^6} = 10^{-6}$.
In standard form, this is written as $1 \times 10^{-6}$.
So, 1 micron = $\mathbf{1 \times 10^{-6}}$ m.
(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb.
The number is 0.00000000000000000016.
Move the decimal point to the right to get 1.6. The number of places moved is 19.
Since we moved the decimal to the right, the exponent is negative.
The standard form is $\mathbf{1.6 \times 10^{-19}}$.
So, the charge of an electron is $\mathbf{1.6 \times 10^{-19}}$ coulomb.
(iii) Size of a bacteria is 0.0000005 m.
The number is 0.0000005.
Move the decimal point to the right to get 5. The number of places moved is 7.
Since we moved the decimal to the right, the exponent is negative.
The standard form is $\mathbf{5 \times 10^{-7}}$.
So, the size of a bacteria is $\mathbf{5 \times 10^{-7}}$ m.
(iv) Size of a plant cell is 0.00001275 m.
The number is 0.00001275.
Move the decimal point to the right to get 1.275. The number of places moved is 5.
Since we moved the decimal to the right, the exponent is negative.
The standard form is $\mathbf{1.275 \times 10^{-5}}$.
So, the size of a plant cell is $\mathbf{1.275 \times 10^{-5}}$ m.
(v) Thickness of a thick paper is 0.07 mm.
The number is 0.07.
Move the decimal point to the right to get 7. The number of places moved is 2.
Since we moved the decimal to the right, the exponent is negative.
The standard form is $\mathbf{7 \times 10^{-2}}$.
So, the thickness of a thick paper is $\mathbf{7 \times 10^{-2}}$ mm.
Question 4. In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack.
Answer:
Solution:
We need to find the total thickness of the stack which consists of books and paper sheets.
Given information:
Number of books = 5
Thickness of each book = 20 mm
Number of paper sheets = 5
Thickness of each paper sheet = 0.016 mm
First, calculate the total thickness of the 5 books.
Total thickness of books = Number of books $\times$ Thickness of each book
Total thickness of books = $5 \times 20$ mm
Total thickness of books = 100 mm
Next, calculate the total thickness of the 5 paper sheets.
Total thickness of paper sheets = Number of paper sheets $\times$ Thickness of each paper sheet
Total thickness of paper sheets = $5 \times 0.016$ mm
To multiply 5 by 0.016:
$5 \times 0.016 = 0.080$ mm
Total thickness of paper sheets = 0.08 mm
Finally, calculate the total thickness of the stack by adding the total thickness of the books and the total thickness of the paper sheets.
Total thickness of the stack = Total thickness of books + Total thickness of paper sheets
Total thickness of the stack = $100 + 0.08$ mm
Total thickness of the stack = 100.08 mm
The total thickness of the stack is $\mathbf{100.08}$ mm.