NCERT Solutions for Class 7 Maths Chapter 11: Exponents and Powers — Free PDF

Exponent (Power): In $a^n$, the number $a$ is the base and $n$ is the exponent (or power or index). It means $a$ multiplied by itself $n$ times: $a^n = \underbrace{a \times a \times \cdots \times a}_{n \text{ times}}$.

Important special cases:
- $a^1 = a$ (any number to the power $1$ is itself)
- $a^0 = 1$ for any $a \neq 0$ (any non-zero number to the power $0$ is $1$)
- $1^n = 1$ for any $n$ ($1$ raised to any power is $1$)
- $(-1)^{\text{even}} = 1$ and $(-1)^{\text{odd}} = -1$

For a non-zero base $a$ and integers $m$, $n$:

Key restriction: The first three laws require the same base. You cannot use $a^m \times a^n = a^{m+n}$ if the bases are different.

Standard form: A number written as $a \times 10^n$ where $1 \leq a < 10$ and $n$ is an integer.

For large numbers: Move the decimal left and make $n$ positive.
$384000 = 3.84 \times 10^5$ (decimal moved 5 places left).

For small numbers: Move the decimal right and make $n$ negative.
$0.00045 = 4.5 \times 10^{-4}$ (decimal moved 4 places right).

Q1. Express in exponential form:
- $5 \times 5 \times 5 \times 5 = 5^4$
- $(-3) \times (-3) \times (-3) = (-3)^3 = -27$
- $t \times t \times t \times t \times t = t^5$

Q2. Evaluate:
- $2^6 = 64$
- $(-2)^4 = 16$ (even power of negative = positive)
- $(-2)^5 = -32$ (odd power of negative = negative)

Key rule: Even power of a negative number is positive; odd power is negative.

**Q3. Express $729$ as a power of $3$.**
$729 \div 3 = 243$, $243 \div 3 = 81$, $81 \div 3 = 27$, $27 \div 3 = 9$, $9 \div 3 = 3$, $3 \div 3 = 1$.
We divided by $3$ exactly $6$ times, so $729 = 3^6$.

**Q4. Which is greater: $2^3$ or $3^2$?**
$2^3 = 8$, $3^2 = 9$. So $3^2 > 2^3$.

Express each as a product of prime factors in exponential form:

• $108 = 2^2 \times 3^3$ (since $108 = 4 \times 27 = 2^2 \times 3^3$)
  - $360 = 2^3 \times 3^2 \times 5$ (since $360 = 8 \times 9 \times 5$)
  - $1600 = 2^6 \times 5^2$ (since $1600 = 64 \times 25$)

Method: Divide repeatedly by the smallest prime factor (2, then 3, then 5, etc.) and count the occurrences.

**Q1. Simplify $2^3 \times 2^5$.**
Using $a^m \times a^n = a^{m+n}$: $2^3 \times 2^5 = 2^8 = 256$.

**Q2. Simplify $\dfrac{5^7}{5^3}$.**
Using $a^m \div a^n = a^{m-n}$: $5^{7-3} = 5^4 = 625$.

**Q3. Simplify $(3^2)^4$.**
Using $(a^m)^n = a^{mn}$: $3^{2 \times 4} = 3^8 = 6561$.

**Q4. Evaluate $7^0 + 3^0 + 5^0$.**
$1 + 1 + 1 = 3$. Any non-zero number raised to power $0$ equals $1$.

**Q5. Simplify $\dfrac{2^5 \times 3^5}{6^3}$.**

**Q6. Simplify $\dfrac{3^4 \times 3^2}{3^5}$.**
$\dfrac{3^{4+2}}{3^5} = \dfrac{3^6}{3^5} = 3^1 = 3$.

**Q7. Simplify $\dfrac{2^4 \times 5^2 \times 7}{8 \times 25}$.**
Convert to prime bases: $8 = 2^3$, $25 = 5^2$.
$\dfrac{2^4 \times 5^2 \times 7}{2^3 \times 5^2} = 2^{4-3} \times 5^{2-2} \times 7 = 2 \times 1 \times 7 = 14$.

**Q8. Find $x$: $2^x = 32$.**
$32 = 2^5$, so $2^x = 2^5 \implies x = 5$.

**Q1. Express $3{,}84{,}000$ in standard form.**
$3{,}84{,}000 = 3.84 \times 10^5$ (moved decimal 5 places left).

**Q2. Express $4.7 \times 10^4$ in usual form.**
$4.7 \times 10^4 = 47{,}000$ (moved decimal 4 places right).

**Q3. Express $5{,}00{,}00{,}000$ in standard form.**
$5{,}00{,}00{,}000 = 5 \times 10^7$.

Problem: The speed of light is $3 \times 10^8$ m/s. The distance from the Sun to Earth is $1.5 \times 10^{11}$ m. How long does light take to travel from the Sun to Earth?

Solution:

This is about $8$ minutes and $20$ seconds.

**Q5. Simplify and express in standard form: $25 \times 10^5 \times 4 \times 10^3$.**
$25 \times 4 \times 10^{5+3} = 100 \times 10^8 = 1 \times 10^{10}$.

Note: $100 = 1 \times 10^2$, so $100 \times 10^8 = 10^{10}$.

**Q6. Express $0.00045$ in standard form.**
$0.00045 = 4.5 \times 10^{-4}$ (moved decimal 4 places right, so exponent is negative).

Problem: Simplify $\dfrac{(-2)^5 \times (-2)^3}{(-2)^6}$.

Solution:

Problem: Find $x$ if $\left(\dfrac{3}{4}\right)^x = \dfrac{27}{64}$.

Solution:
$\dfrac{27}{64} = \dfrac{3^3}{4^3} = \left(\dfrac{3}{4}\right)^3$

So $x = 3$.

Problem: Is $2^3 \times 3^2 = 6^5$? Justify.

Solution:
$2^3 \times 3^2 = 8 \times 9 = 72$.
$6^5 = 7776$.
Clearly $72 \neq 7776$, so $2^3 \times 3^2 \neq 6^5$.

The law $a^m \times b^m = (ab)^m$ requires the same exponent. Here the exponents are different ($3$ and $2$), so this law does not apply.

Problem: The mass of the Earth is approximately $5.97 \times 10^{24}$ kg and the mass of the Moon is approximately $7.35 \times 10^{22}$ kg. How many times heavier is the Earth?

Solution:

The Earth is approximately $81$ times heavier than the Moon.

Problem: Simplify $\dfrac{2^5 \times 3^3 \times 5^2}{4^2 \times 9 \times 25}$.

Solution:
Convert to prime bases: $4 = 2^2$, $9 = 3^2$, $25 = 5^2$.

Question: Simplify $\dfrac{(-3)^4 \times (-3)^2}{(-3)^5}$.

Answer: $\dfrac{(-3)^{4+2}}{(-3)^5} = \dfrac{(-3)^6}{(-3)^5} = (-3)^1 = -3$.

Question: Find $x$ if $4^x = 256$.

Answer: $256 = 4^4$ (since $4 \times 4 \times 4 \times 4 = 256$). So $x = 4$.
Alternatively: $4^x = (2^2)^x = 2^{2x}$ and $256 = 2^8$, so $2x = 8$, $x = 4$.

Question: Express $0.0000067$ in standard form.

Answer: $0.0000067 = 6.7 \times 10^{-6}$ (decimal moved 6 places right).

Question: Express $3600$ in exponential form using prime factors.

Answer: $3600 = 36 \times 100 = 6^2 \times 10^2 = (2 \times 3)^2 \times (2 \times 5)^2 = 2^4 \times 3^2 \times 5^2$.