Exponents

(i) 7⁴

7⁴ = 7 x 7 x 7 x 7 = 2401

Hence, the answer is 2401

(ii) (-5)

(-5)³ = (-5) x (-5) x (-5) = -125

Hence, the answer is -125

(iii) $\Big(\frac{3}{4}\Big)^5$

We know that

$\Big(\frac{a}{b}\Big)^n = \frac{a^n}{b^n} \quad \text{[Power of division rule]}$

$\therefore \Big(\frac{3}{4}\Big)^5 = \frac{3^5}{4^5} = \frac{3 \times 3 \times 3 \times 3 \times 3}{4 \times 4 \times 4 \times 4 \times 4} = \frac{243}{1024}$

Hence, the answer is $\frac{243}{1024}$

(iv) $\Big(\frac{-5}{2}\Big)^2$

$\Big(\frac{a}{b}\Big)^n = \frac{a^n}{b^n} \quad \text{[Power of division rule]}$

$\Big(\frac{-5}{2}\Big)^2 = \frac{(-5)^2}{2^2} = \frac{(-5) \times (-5)}{2 \times 2} = \frac{25}{4} = 6\frac{1}{4}$

Hence, the answer is $6\frac{1}{4}$

(i) 10000

10000 = 10 x 10 x 10 x 10 = 10⁴.

Hence, the answer is 10⁴

(ii) One Crore

One Crore is written as 1,00,00,000, which has 7 zeros.

1,00,00,000 = 10 x 10 x 10 x 10 x 10 x 10 x 10 = 10⁷.

Hence, the answer is 10⁷

(iii) One million

One million is written as 1,000,000, which has 6 zeros.

1,000,000 = 10 x 10 x 10 x 10 x 10 x 10 = 10⁶.

Hence, the answer is 10⁶

(i) $\Big(\frac{-7}{13}\Big) \times \Big(\frac{-7}{13}\Big) \times \Big(\frac{-7}{13}\Big)$

The rational number $\Big(\frac{-7}{13}\Big)$ is being multiplied by itself 3 times.

Hence, the answer is $\Big(\frac{-7}{13}\Big)^3$

(ii) $\Big(\frac{-8}{3}\Big) \times \Big(\frac{-8}{3}\Big) \times \Big(\frac{-8}{3}\Big) \times \Big(\frac{-8}{3}\Big)$

The rational number $\Big(\frac{-8}{3}\Big)$ is being multiplied by itself 4 times.

Hence, the answer is $\Big(\frac{-8}{3}\Big)^4$

(i) $\frac{343}{512}$

We have:

343 = 7 x 7 x 7 = 7³

512 = 8 x 8 x 8 = 8³

∴ $\frac{343}{512} = \frac{7^3}{8^3} = \Big(\frac{7}{8}\Big)^3$

Hence, the answer is $\Big(\frac{7}{8}\Big)^3$

(ii) $\frac{-32}{243}$

We have:

-32 = (-2) x (-2) x (-2) x (-2) x (-2) = (-2)⁵

243 = 3 x 3 x 3 x 3 x 3 = 3⁵

∴ $\frac{-32}{243} = \frac{(-2)^5}{3^5} = \Big(\frac{-2}{3}\Big)^5$

Hence, the answer is $\Big(\frac{-2}{3}\Big)^5$

(iii) $\frac{-1}{128}$

We have denominator 128:

∴ 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁷

-1 = (-1)⁷ $\quad$ (Since any odd power of -1 is -1)

∴ $\frac{-1}{128} = \frac{(-1)^7}{2^7} = \Big(\frac{-1}{2}\Big)^7$

Hence, the answer is $\Big(\frac{-1}{2}\Big)^7$

(iv) $\frac{729}{64}$

We have:

729 = 3 x 3 x 3 x 3 x 3 x 3 = 3⁶

64 = 2 x 2 x 2 x 2 x 2 x 2 = 2⁶

∴ $\frac{729}{64} = \Big(\frac{3}{2}\Big)^6$

Hence, the answer is $\Big(\frac{3}{2}\Big)^6$

(i) $\Big(\frac{5}{21}\Big)^3 \times \Big(\frac{5}{21}\Big)^8$

Using the multiplication rule, we add the exponents: 3 + 8 = 11.

$$\begin{aligned}\therefore \Big(\frac{5}{21}\Big)^3 \times \Big(\frac{5}{21}\Big)^8 \\[1em] = \Big(\frac{5}{21}\Big)^{3+8} \\[1em] = \Big(\frac{5}{21}\Big)^{11}\end{aligned}$$

Hence, the answer is $\Big(\frac{5}{21}\Big)^{11}$

(ii) $\Big(\frac{-7}{3}\Big)^{11} \times \Big(\frac{-7}{3}\Big)^{13}$

Using the multiplication rule, we add the exponents: 11 + 13 = 24.

$$\begin{aligned}\therefore \Big(\frac{-7}{3}\Big)^{11} \times \Big(\frac{-7}{3}\Big)^{13} \\[1em] = \Big(\frac{-7}{3}\Big)^{11+13} \\[1em] = \Big(\frac{-7}{3}\Big)^{24}\end{aligned}$$

Hence, the answer is $\Big(\frac{-7}{3}\Big)^{24}$

(iii) $\Big(\frac{13}{43}\Big)^7 ÷ \Big(\frac{13}{43}\Big)^2$

Using the division rule, we subtract the exponents: 7 - 2 = 5.

$$\begin{aligned}\therefore \Big(\frac{13}{43}\Big)^7 ÷ \Big(\frac{13}{43}\Big)^2 \\[1em] = \Big(\frac{13}{43}\Big)^{7-2} \\[1em] = \Big(\frac{13}{43}\Big)^5\end{aligned}$$

Hence, the answer is $\Big(\frac{13}{43}\Big)^5$

(iv) $\Big(\frac{-16}{35}\Big)^{16} ÷ \Big(\frac{-16}{35}\Big)^3$

Using the division rule, we subtract the exponents: 16 - 3 = 13.

$$\begin{aligned}\therefore \Big(\frac{-16}{35}\Big)^{16} ÷ \Big(\frac{-16}{35}\Big)^3 \\[1em] = \Big(\frac{-16}{35}\Big)^{16-3} \\[1em] = \Big(\frac{-16}{35}\Big)^{13}\end{aligned}$$

Hence, the answer is $\Big(\frac{-16}{35}\Big)^{13}$

(v) $\Big(\frac{-7}{15}\Big)^{12} ÷ \Big(\frac{-7}{15}\Big)^{15}$

Using the division rule, we subtract the exponents: 12 - 15 = -3.

$$\begin{aligned}\therefore \Big(\frac{-7}{15}\Big)^{12} ÷ \Big(\frac{-7}{15}\Big)^{15} \\[1em] = \Big(\frac{-7}{15}\Big)^{12-15} \\[1em] = \Big(\frac{-7}{15}\Big)^{-3} \quad \text{[Power is negative, take reciprocal]} \\[1em] = \Big(\frac{-15}{7}\Big)^3\end{aligned}$$

Hence, the answer is $\Big(\frac{-15}{7}\Big)^3$

(vi) $\Big(\frac{1}{24}\Big)^{13} ÷ \Big(\frac{1}{24}\Big)^{16}$

Using the division rule, we subtract the exponents: 13 - 16 = -3.

$$\begin{aligned}\therefore \Big(\frac{1}{24}\Big)^{13} ÷ \Big(\frac{1}{24}\Big)^{16} \\[1em] = \Big(\frac{1}{24}\Big)^{13-16} \\[1em] = \Big(\frac{1}{24}\Big)^{-3} \quad \text{[Power is negative, take the reciprocal]} \\[1em] = 24^3\end{aligned}$$

Hence, the answer is 24³

(i) We have:

$$\begin{aligned}\phantom{=} \Big(\frac{6}{5}\Big)^3 \times \Big(\frac{5}{2}\Big)^2 \\[1em] = \frac{6^3}{5^3} \times \frac{5^2}{2^2} \\[1em] = \frac{6^3 \times 5^2}{5^3 \times 2^2} \\[1em] = \frac{6^3 \times 5^{2-3}}{2^2}\quad \text{[Applying exponent law]} \\[1em] = \frac{6^3 \times 5^{-1}}{2^2} \\[1em] = \frac{6 \times 6 \times 6}{5 \times 2 \times 2} \quad {[5^{-1} = \frac{1}{5}]} \\[1em] = \frac{216}{20} \\[1em] = \frac{54}{5} = 10\frac{4}{5}\end{aligned}$$

Hence, the answer is $10\frac{4}{5}$

(ii) We have:

$$\begin{aligned}\phantom{=} \Big(\frac{3}{4}\Big)^2 \times \Big(\frac{-1}{2}\Big)^5 \times 2^3 \\[1em] = \frac{3^2}{4^2} \times \frac{(-1)^5}{2^5} \times 2^3 \\[1em] = \frac{3^2 \times (-1)^5 \times 2^3}{4^2 \times 2^5} \\[1em] = \frac{3^2 \times (-1)^5 \times 2^{3-5}}{4^2} \\[1em] = \frac{3^2 \times (-1)^5 \times 2^{-2}}{4^2} \\[1em] = \frac{(3 \times 3) \times (-1)}{4 \times (4 \times 4)} \quad [2^{-2} = \frac{1}{2^2} = \frac{1}{4}] \\[1em] = \frac{-9}{64}\end{aligned}$$

Hence, the answer is $\frac{-9}{64}$

(iii) We have:

$$\begin{aligned}\phantom{=} \Big(\frac{5}{4}\Big)^2 \times \Big(\frac{2}{3}\Big)^2 \times \Big(\frac{-3}{5}\Big)^3 \\[1em] \frac{5^2}{4^2} \times \frac{2^2}{3^2} \times \frac{(-3)^3}{5^3} \\[1em] = \frac{5^2 \times 2^2 \times (-3)^3}{4^2 \times 3^2 \times 5^3} \\[1em] = \frac{2^2 \times (-3)^{3-2} \times 5^{2-3}}{4^2} \\[1em] = \frac{2^2 \times (-3)^1 \times 5^{-1}}{4^2} \\[1em] = \frac{(2 \times 2) \times (-3)}{(4 \times 4) \times 5} \quad[5^{-1} = \frac{1}{5}] \\[1em] = \frac{-12}{80} = \frac{-3}{20}\end{aligned}$$

Hence, the answer is $\frac{-3}{20}$

(iv) We have:

$$\begin{aligned}\phantom{=} \Big(\frac{-3}{4}\Big)^3 \times \Big(\frac{-5}{2}\Big)^3 \times \Big(\frac{2}{3}\Big)^5 \\[1em] = \frac{(-3)^3}{4^3} \times \frac{(-5)^3}{2^3} \times \frac{2^5}{3^5} \\[1em] = \frac{(-3)^3 \times (-5)^3 \times 2^5}{4^3 \times 2^3 \times 3^5} \\[1em] = \frac{(-3)^3 \times (-5)^3 \times 2^{5-3}}{4^3 \times 3^5} \\[1em] = \frac{(-3)^3 \times (-5)^3 \times 2^2}{4^3 \times 3^5} \\[1em] = \frac{(-3 \times -3 \times -3) \times (-5 \times -5 \times -5) \times (2 \times 2)}{(4 \times 4 \times 4) \times (3 \times 3 \times 3 \times 3 \times 3)} \\[1em] = \frac{(-27) \times -125 \times 4}{64 \times 243} \\[1em] = \frac{-1 \times -125 \times 1}{16 \times 9} \\[1em] = \frac{125}{144} \\[1em]\end{aligned}$$

Hence, the answer is $\frac{125}{144}$

(v) We have:

$$\begin{aligned}\phantom{=} \Big(\frac{7}{11}\Big)^6 ÷ \Big(\frac{7}{11}\Big)^3 \\[1em] = \Big(\frac{7}{11}\Big)^{6-3} \quad \text{[By division rule]} \\[1em] = \Big(\frac{7}{11}\Big)^3 \\[1em] = \frac{7^3}{11^3} \\[1em] = \frac{7 \times 7 \times 7}{11 \times 11 \times 11} \\[1em] = \frac{343}{1331}\end{aligned}$$

Hence, the answer is $\frac{343}{1331}$

(vi) We have:

$$\begin{aligned}\phantom{=} \Big(\frac{-4}{3}\Big)^8 ÷ \Big(\frac{-4}{3}\Big)^{12} \\[1em] = \Big(\frac{-4}{3}\Big)^{8-12} \quad \text{[By division rule]} \\[1em] = \Big(\frac{-4}{3}\Big)^{-4} \\[1em] = \Big(\frac{3}{-4}\Big)^{4} \quad \text{[By reciprocal rule]} \\[1em] = \frac{3 \times 3 \times 3 \times 3}{(-4) \times (-4) \times (-4) \times (-4)} \\[1em] = \frac{81}{256}\end{aligned}$$

Hence, the answer is $\frac{81}{256}$

(i) We have:

$$\begin{aligned}\phantom{=} \frac{10^2 \times 15^3}{2^2 \times 3 \times 5^5 \times 6^4} \\[1em] = \frac{(2 \times 5)^2 \times (3 \times 5)^3}{2^2 \times 3^1 \times 5^5 \times (2 \times 3)^4} \\[1em] = \frac{2^2 \times 5^2 \times 3^3 \times 5^3}{2^2 \times 3^1 \times 5^5 \times 2^4 \times 3^4} \\[1em] = \frac{2^2 \times 3^3 \times 5^{2+3}}{2^{2+4} \times 3^{1+4} \times 5^5} \\[1em] = \frac{2^2 \times 3^3 \times 5^5}{2^6 \times 3^5 \times 5^5} \\[1em] = \frac{2^{2-6}}{3^{5-3} \times 5^{5-5}} \\[1em] = \frac{2^{-4}}{3^2 \times 5^0} \quad \text{[Using law of exponents]} \\[1em] = \frac{1}{2^4 \times 3^2 \times 1} \quad [2^{-4} = \frac{1}{2^4}] \\[1em] = \frac{1}{16 \times 9} \\[1em] = \frac{1}{144}\end{aligned}$$

Hence, the answer is $\frac{1}{144}$

(ii) We have:

$$\begin{aligned}\phantom{=} \frac{3^5 \times 25 \times 10^5}{5^7 \times 6^5} \\[1em] = \frac{3^5 \times 5^2 \times (2 \times 5)^5}{5^7 \times (2 \times 3)^5} \\[1em] = \frac{3^5 \times 5^2 \times 2^5 \times 5^5}{5^7 \times 2^5 \times 3^5} \\[1em] = \frac{2^5 \times 3^5 \times 5^{2+5}}{2^5 \times 3^5 \times 5^7} \\[1em] = \frac{2^5 \times 3^5 \times 5^7}{2^5 \times 3^5 \times 5^7} \\[1em] = 2^{5-5} \times 3^{5-5} \times 5^{7-7} \quad \text{[law of exponents]} \\[1em] = 2^0 \times 3^0 \times 5^0 \\[1em] = 1 \times 1 \times 1 \\[1em] = 1\end{aligned}$$

Hence, the answer is 1

Given:

Distance between the Earth and the Moon = 384000 km.

We know that 1 km = 1000 m = 10³ m.

Distance in metres = 384000 x 10³ m.

We can write 384000 as 384 x 1000 = 384 x 10³.

Total distance = (384 x 10³) x 10³ m.

Using the law of exponents (a^m x aⁿ = a^m+n), we add the powers: 3+3 = 6.

Distance = 384 x 10⁶ m.

Hence, the distance is 384 x 10⁶ m.

Given:

Total RAM = 8 gigabytes.

Value of 1 gigabyte = 10⁹ bytes.

RAM in bytes = Total RAM x Value of 1 gigabyte

Substituting the values in above, we get:

RAM in bytes = 8 x 10⁹ bytes.

Hence, the RAM is 8 x 10⁹ bytes.

Given:

Initial number of players = 128.

In a knockout round, the number of players is reduced to half $\Big(\frac{1}{2}\Big)$.

After 1st round, players left = $\frac{1}{2} \times 128$.

After 2nd round, players left = $\frac{1}{2} \times \frac{1}{2} \times 128 = \Big(\frac{1}{2}\Big)^2 \times 128$.

After 3rd round, players left = $\Big(\frac{1}{2}\Big)^3 \times 128$.

After 4th round, players left = $\Big(\frac{1}{2}\Big)^4 \times 128$ $$\begin{aligned}= \frac{1^4}{2^4} \times 128 \\[1em] = \frac{128}{2^4} \\[1em] = \frac{128}{2 \times 2 \times 2 \times 2} \\[1em] = \frac{128}{16} \\[1em] = 8\end{aligned}$$

Hence, 8 players moved to the next round.

Exponential notation in initial no. of players = $\frac{\bold{128}}{\bold{2}^\bold{4}}$

(i) 98 hm

1 hm = 100 m and 1 m = 100 cm.

Therefore, 1 hm = 100 x 100 cm = 10000 cm.

In exponential form, 10000 = 10⁴ cm.

98 hm = 98 x 10⁴ cm.

Hence, the answer is 98 x 10⁴ cm.

(ii) 156 km

1 km = 1000 m and 1 m = 100 cm.

Therefore, 1 km = 1000 x 100 cm = 100000 cm.

In exponential form, 100000 = 10⁵ cm.

156 km = 156 x 10⁵ cm.

Hence, the answer is 156 x 10⁵ cm.

(iii) 371 m

1 m = 100 cm.

In exponential form, 100 = 10² cm.

371 m = 371 x 10² cm.

Hence, the answer is 371 x 10² cm.

3² = 3 x 3 = 9

2³ = 2 x 2 x 2 = 8.

3² + 2³ = 9 + 8 = 17.

Hence, option 3 is the correct option.

We have:

(7 - 5)⁵

⇒ (7 - 5) = 2

2⁵ = 2 x 2 x 2 x 2 x 2 = 32

Hence, option 4 is the correct option.

Concept:

(-5)⁴ = (-5) x (-5) x (-5) x (-5) = 625.

Hence, option 3 is the correct option.

If the base is negative and the exponent is odd, the result is negative.

$$\begin{aligned}\Big(\frac{-3}{5}\Big)^5 \\[1em] = \frac{(-3)^5}{5^5} \\[1em] = \frac{(-3) \times (-3) \times (-3) \times (-3) \times (-3)}{5 \times 5 \times 5 \times 5 \times 5} \\[1em] = \frac{-243}{3125}\end{aligned}$$

Hence, option 1 is the correct option.

Given:

$\Big(\frac{3}{7}\Big)^3 \times \Big(\frac{3}{7}\Big)^{-8} = \Big(\frac{3}{7}\Big)^{2x + 3}$

LHS = $\Big(\frac{3}{7}\Big)^3 \times \Big(\frac{3}{7}\Big)^{-8}$

$$\begin{aligned}= \frac{3^3}{7^3} \times \frac{3^{-8}}{7^{-8}} \\[1em] = \frac{3^3 \times 3^{-8}}{7^3 \times 7^{-8}} \\[1em] = \frac{3^{3+(-8)}}{7^{3+(-8)}} = \frac{3^{3-8}}{7^{3-8}} \\[1em] = \frac{3^{-5}}{7^{-5}} \\[1em] = \Big(\frac{3}{7}\Big)^{-5} \\[1em]\end{aligned}$$

Now, LHS = $\Big(\frac{3}{7}\Big)^{-5}$

As the base of both LHS and RHS is same, let us compare the exponents:

-5 = 2x + 3
2x = -5 - 3
2x = -(5 + 3)
2x = -8
x = $\frac{-8}{2}$
x = -4

Hence, option 1 is the correct option.

Given:

$\Big[\Big\lbrace\Big(\frac{-1}{3}\Big)^{-2}\Big\rbrace^{-1}\Big]^{2}$

Multiply the exponents: (-2) x (-1) x 2 = 4. $\quad$ [Power of power rule]

∴ $\Big(\frac{-1}{3}\Big)^4 = \frac{(-1)^4}{3^4} = \frac{1}{81}$

Hence, option 3 is the correct option.

Move the decimal 7 places to the left: 3.4613 x 10⁷.

Verification: 3.4613 x 10,000,000 = 34613000.

Hence, option 2 is the correct option.

(i) 2³ or 3²

2³ = 2 x 2 x 2 = 8

3² = 3 x 3 = 9

Clearly, 8 < 9.

Hence, 3² is larger.

(ii) 2⁵ or 5²

2⁵ = 2 x 2 x 2 x 2 x 2 = 32

5² = 5 x 5 = 25

Clearly, 32 > 25.

Hence, 2⁵ is larger.

(iii) 3⁷ or 7³.

3⁷ = 3 x 3 x 3 x 3 x 3 x 3 x 3 = 2187

7³ = 7 x 7 x 7 = 343

Clearly, 2187 > 343.

Hence, 3⁷ is larger.

(i) In an exponential form x^m; x is called the base.

(ii) We have : $\frac{a^m}{a^n} = \frac{1}{a^{n-m}}$ if n > m.

(iii) Any number to the power 0 is equal to 1.

(iv) The exponential form is also called power notation.

(v) The reciprocal of x^a is equal to x to the power -a.

(i) False
Reason — The power of a power law states that (a^m)ⁿ = a^{m x n}. The expression a^m+n is the result of multiplying powers with the same base (a^m x aⁿ).

(ii) True
Reason — According to the power of a power rule, $\Big\lbrace\left(\frac{a}{b}\right)^m\Big\rbrace^n = \left(\frac{a}{b}\right)^{mn}$. By applying the reciprocal rule $\left(\frac{a}{b}\right)^{mn} = \left(\frac{b}{a}\right)^{-mn}$, the statement is mathematically correct.

(iii) True
Reason — This is the product law of exponents, which states that for any non-zero base x, x^m x xⁿ = x^(m+n).

(iv) True
Reason — By definition, exponential notation is a shorthand for repeated multiplication; if p is multiplied n times, it is written as pⁿ.

(v) True
Reason — In the notation xⁿ, the number n is commonly referred to as the exponent, power, or index.

(1) Given:

Initial Value = ₹72,900

Loss in value every year = $\frac{2}{3}$

Value remaining every year = $1 - \frac{2}{3} = \frac{1}{3}$ of the previous year's value.

Every year, the value is multiplied by $\frac{1}{3}$. After n years, the value is $72900 \times (\frac{1}{3})^n$, which is $\frac{72900}{3^n}$.

Hence, option 3 is the correct option.

(2) Value of the computer after 3 years = ?

Value of the computer after n years = $\frac{72900}{3^n} \quad$ [From previous step]

By replacing the value of 'n' with 3, we get:

₹ $\frac{72900}{3^3} = ₹ \frac{72900}{27}$

= ₹2700

Hence, option 4 is the correct option.

(3) We know at 3 years, value = ₹2700. $\quad$ [From previous step]

Value remaining every year = $1 - \frac{2}{3} = \frac{1}{3}$ of the previous year's value. $\quad$ [From step 1]

∴ Value after 4 years = $\frac{1}{3} \times 2700 = \frac{2700}{3} = ₹ 900$

Value after 5 years = $\frac{1}{3} \times 900 = \frac{900}{3} = ₹ 300$

Value after 6 years = $\frac{1}{3} \times 300 = \frac{300}{3} = ₹ 100$

Since 100 < 200, it takes 6 years.

Hence, option 1 is the correct option.

(4) Value of the computer reduced in 4 years = ?

Value after 4 years = ₹ $\frac{72900}{3^4} = ₹ \frac{72900}{81} = ₹ 900$

Value of the computer reduced in 4 years = Initial value - Value after 4 years

Substituting the values in above, we get:

Value of the computer reduced in 4 years = ₹72900 - ₹ 900 = ₹ 72000

Hence, option 4 is the correct option.

(1) Given:

Initial Population = 50

Growth Rate = Doubles every hour (x2)

The population starts at 50 and multiplies by 2 for every hour.

General formula after n hours:

Population = 50 × 2ⁿ

Hence, option 3 is the correct option.

(2) Population size after 3 hours = ?

Population size after n hours = 50 × 2ⁿ $\quad$ [From step 1]

By replacing the value of 'n' with 3, we get:

Population size after 3 hours = 50 × 2³ = 50 x 8 = 400

Hence, option 3 is the correct option.

(3) Bacteria in the culture after 1 day = ?

We know that 1 day has 24 hours

Population size after n hours = 50 × 2ⁿ $\quad$ [From step 1]

By replacing the value of 'n' with 24, we get:

Bacteria in the culture after 1 day = 50 × 2²⁴

Hence, option 2 is the correct option.

(4) After how many hours will the population be larger than 1000?

Let's test hours (n):

n = 4: 50 x 2⁴ = 50 x 16 = 800

n = 5: 50 x 2⁵ = 50 x 32 = 1600

Since 1600 > 1000, it happens at 5 hours.

Hence, option 1 is the correct option.

Assertion (A) is false but Reason (R) is true.

Explanation

Assertion is false because, the expression $\frac{x^4}{y^3}$ represents repeated multiplication, not addition. Specifically, $x^4 = x \times x \times x \times x$ and $y^3 = y \times y \times y$. Adding the variables as shown in the assertion is incorrect.

Reason is true as it is the correct mathematical definition for the components of an exponential term.

Hence, option 4 is the correct option.

Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Explanation

Assertion is true because, it is the Power of a Quotient rule. It states that when two different bases are divided and raised to the same power, the power can be applied to the quotient.

Reason is true because, it is the Power of a Product rule. It correctly states that a^m x b^m = (ab)^m.

While both are valid laws of exponents, the rule for multiplication (Reason) does not explain the rule for division (Assertion).

Hence, option 2 is the correct option.

Assertion (A) is false but Reason (R) is true.

Explanation

Assertion is false.

Let's evaluate both sides using the zero exponent rule:

LHS: 2⁰ + 3⁰ + 4⁰ = 1 + 1 + 1 = 3.

RHS: (2 + 3 + 4)⁰ = (9)⁰ = 1

Since $3 \neq 1$, the assertion is false.

Reason is true because, the rule x⁰ = 1 (for any non-zero x) is a fundamental law of exponents.

Hence, option 4 is the correct option.

Given expression: 2³ + 2³ + 2³ + 2³

Since 2³ is being added 4 times, we can write it as:

= 4 × 2³

= 2² × 2³ $\hspace{2cm}$[∵ 4 = 2²]

= 2^{2 + 3} $\hspace{2cm}$[∵ a^m × aⁿ = a^{m + n}]

= 2⁵

∴ 2³ + 2³ + 2³ + 2³ = 2⁵.

Hence, option 1 is the correct option.

Step 1: Simplify the first fraction

$$\begin{aligned}\frac{\left(-\frac{1}{2}\right)^5}{\left(-\frac{1}{2}\right)^4} \\[1em] = \left(-\frac{1}{2}\right)^{5 - 4} \quad [∵ \frac{a^m}{a^n} = a^{m - n}, \text{ if } m \gt n] \\[1em] = \left(-\frac{1}{2}\right)^1 \\[1em] = -\frac{1}{2}\end{aligned}$$

Step 2: Simplify the second fraction

$$\begin{aligned}\frac{\left(-\frac{1}{8}\right)}{\left(-\frac{1}{4}\right)} = \left(-\frac{1}{8}\right) \times \left(-\frac{4}{1}\right) \quad \text{[Reciprocal of } -\frac{1}{4} \text{ is } -\frac{4}{1}] \\[1em] = \frac{4}{8} \\[1em] = \frac{1}{2}\end{aligned}$$

Step 3: Divide the two results

$$\begin{aligned}-\frac{1}{2} \div \frac{1}{2} = -\frac{1}{2} \times \frac{2}{1} \quad \text{[Reciprocal of } \frac{1}{2} \text{ is } \frac{2}{1}] \\[1em] = -\frac{2}{2} \\[1em] = -1\end{aligned}$$

Hence, option 4 is the correct option.

Given expression: $\left(\frac{a^p}{a^q}\right) \times \left(\frac{a^q}{a^r}\right) \times \left(\frac{a^r}{a^p}\right)$

Rearranging the factors,

= $\frac{a^p \times a^q \times a^r}{a^q \times a^r \times a^p}$

= 1

for a ≠ 0.

∴ The value of the given expression is 1.

Hence, option 1 is the correct option.

Given:

a = 3 and b = 7

Substituting the values of a and b in the given expression, we get:

a^b − b^a = 3⁷ − 7³

Step 1: Evaluate 3⁷

3⁷ = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 2187

Step 2: Evaluate 7³

7³ = 7 × 7 × 7 = 343

Step 3: Subtract

3⁷ − 7³ = 2187 − 343 = 1844

∴ The value of a^b − b^a is 1844.

Hence, option 3 is the correct option.

Step 1: Evaluate the first term

$\frac{3^4}{2^3} = \frac{3 \times 3 \times 3 \times 3}{2 \times 2 \times 2} = \frac{81}{8}$

Step 2: Evaluate the numerator of the second term

$\left(\frac{1}{9} \times 4\right)^2 = \left(\frac{4}{9}\right)^2 = \frac{4^2}{9^2} = \frac{16}{81}$

Step 3: Evaluate the denominator of the second term

$$\begin{aligned}\left(\frac{-1}{3}\right)^3 = \frac{(-1)^3}{3^3} = \frac{-1}{27} \\[1em] \left(\frac{1}{-2}\right)^3 = \frac{1^3}{(-2)^3} = \frac{-1}{8} \\[1em] \left(\frac{-1}{3}\right)^3 \div \left(\frac{1}{-2}\right)^3 = \frac{-1}{27} \div \frac{-1}{8} \\[1em] = \frac{-1}{27} \times \frac{8}{-1} \quad \text{[Reciprocal of } \frac{-1}{8} \text{ is } \frac{8}{-1}] \\[1em] = \frac{8}{27}\end{aligned}$$

Step 4: Simplify the second term

$\frac{\frac{16}{81}}{\frac{8}{27}} = \left(\frac{16}{81} \times \frac{27}{8}\right) = \left(\frac{2}{3} \times \frac{1}{1}\right) = \frac{2}{3}$

Step 5: Add the two terms

$$\begin{aligned}\frac{81}{8} + \frac{2}{3} = \frac{81 \times 3 + 2 \times 8}{24} \quad \text{[LCM of 8 and 3 is 24]} \\[1em] = \frac{243 + 16}{24} \\[1em] = \frac{259}{24}\end{aligned}$$

∴ The value of the given expression is $\frac{259}{24}$.

Hence, option 2 is the correct option.

The given expression has (−8) multiplied by itself 10 times.

So, (−8) × (−8) × (−8) × (−8) × (−8) × (−8) × (−8) × (−8) × (−8) × (−8) = (−8)¹⁰

Since 10 is an even number, the result is positive.
[∵ A negative number raised to an even power is positive]

(−8)¹⁰ = 8¹⁰ = (2³)¹⁰ = 2³⁰ $\hspace{2cm}$[∵ (a^m)ⁿ = a^mn]

Now, let us check each option:

Option 1: −2(8)⁵ = −2 × 2¹⁵ = −2¹⁶ (negative value)

Option 2: −(2)¹⁶ (negative value)

Option 3: −(4)⁸ = −(2²)⁸ = −2¹⁶ (negative value)

Since 2³⁰ is positive and all the given options are negative, none of them is equal to (−8)¹⁰.

∴ The value of the given expression does not match any of options 1, 2 or 3.

Hence, option 4 is the correct option.

Given expression: (p^x × p^y) ÷ (p^x ÷ p^y)

Rewrite the division in the second bracket as a fraction:

= $(p^x \times p^y) \div \left(\frac{p^x}{p^y}\right)$

Dividing by a fraction means multiplying by its reciprocal:

= $p^x \times p^y \times \frac{p^y}{p^x}$

Cancel p^x from the numerator and denominator:

= p^y × p^x

= p^{y + y} $\quad$ [using a^m × aⁿ = a^{m + n}]

= p^2y

∴ The value of the given expression is p^2y.

Hence, option 4 is the correct option.

Given:

x = 6 and y = 2

Substituting the values of x and y in the given expression, we get:

x^y − y^x + xy = 6² − 2⁶ + (6 × 2)

Step 1: Evaluate 6²

6² = 6 × 6 = 36

Step 2: Evaluate 2⁶

2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

Step 3: Evaluate 6 × 2

6 × 2 = 12

Step 4: Substitute and simplify

6² − 2⁶ + 6 × 2 = 36 − 64 + 12

= 48 − 64

= −16

∴ The value of x^y − y^x + xy is −16.

Hence, option 3 is the correct option.

Given:

3^x = 500

We need to find the value of 3^{x − 2}.

Using the quotient law of exponents:

3^{x − 2} = $\frac{3^x}{3^2}$ $\hspace{2cm}$[∵ a^{m − n} = $\frac{a^m}{a^n}$]

Substituting the value of 3^x, we get:

3^{x − 2} = $\frac{500}{3^2}$

= $\frac{500}{9}$ $\hspace{2cm}$[∵ 3² = 9]

∴ The value of 3^{x − 2} is $\frac{500}{9}$.

Hence, option 1 is the correct option.

Rajat claims, “A negative number raised to a power is always less than the number itself.”

To check the statement, consider a negative number raised to an even power.

For example,

(−2)² = (−2) × (−2) = 4

But,

4 > −2

So, a negative number raised to a power is not always less than the number itself.

∴ Rajat's statement is false.

Hence, option 2 is the correct option.

Given expression: {(9²)⁴ × 9⁵} ÷ 9⁸

Step 1: Simplify (9²)⁴

(9²)⁴ = 9^{2 × 4} = 9⁸ $\hspace{2cm}$[∵ (a^m)ⁿ = a^mn]

Step 2: Simplify the bracket

9⁸ × 9⁵ = 9^{8 + 5} = 9¹³ $\hspace{2cm}$[∵ a^m × aⁿ = a^{m + n}]

Step 3: Divide by 9⁸

9¹³ ÷ 9⁸ = 9^{13 − 8} = 9⁵ $\hspace{2cm}$[∵ a^m ÷ aⁿ = a^{m − n}]

∴ The value of the given expression is 9⁵.

Hence, option 3 is the correct option.