Exponents and Powers — Exercise 2(B)

Which of the following values are equal?

I. 1⁵    II. 5⁰    III. 0⁵    IV. 5¹

1. I and II

2. II and III

3. I and III

4. I and IV

Answer

We know that,

$a^0 = 1$ and $a^1 = a$

Solving each value,

$\Rightarrow 1^5 = 1 \times 1 \times 1 \times 1 \times 1 = 1\\[1em] \Rightarrow 5^0 = 1\\[1em] \Rightarrow 0^5 = 0 \times 0 \times 0 \times 0 \times 0 = 0\\[1em] \Rightarrow 5^1 = 5$

Since, I = II = 1

Hence, option 1 is the correct option.

If $5^x = 3125$, then the value of $5^{(x-3)}$ is

1. 25

2. 125

3. 625

4. 1625

Answer

We know that 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵

$\Rightarrow 5^x = 3125\\[1em] \Rightarrow 5^x = 5^5$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x = 5$

Substituting the value of $x$,

$\Rightarrow 5^{x-3}\\[1em] \Rightarrow 5^{5-3}\\[1em] \Rightarrow 5^2\\[1em] \Rightarrow 25$

Hence, option 1 is the correct option.

The value of $(256)^{\dfrac{5}{4}}$ is

1. 512

2. 984

3. 1024

4. 1032

Answer

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$

We know that 256 = 4⁴

Solving,

$\Rightarrow (256)^{\dfrac{5}{4}}\\[1em] \Rightarrow (4^4)^{\dfrac{5}{4}}\\[1em] \Rightarrow 4^{4 \times \dfrac{5}{4}}\\[1em] \Rightarrow 4^5\\[1em] \Rightarrow 4 \times 4 \times 4 \times 4 \times 4\\[1em] \Rightarrow 1024$

Hence, option 3 is the correct option.

The value of $2^{\dfrac{5}{4}}$ lies between

1. 0 and 1

2. 1 and 2

3. 2 and 3

4. 3 and 4

Answer

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$

Solving,

$\Rightarrow 2^{\dfrac{5}{4}}\\[1em] \Rightarrow 2^{1 + \dfrac{1}{4}}\\[1em] \Rightarrow 2^1 \times 2^{\dfrac{1}{4}}\\[1em] \Rightarrow 2 \times \sqrt[4]{2}$

Since $\sqrt[4]{2}$ lies between 1 and 2, the value of $2 \times \sqrt[4]{2}$ lies between 2 and 4.

More precisely, $\sqrt[4]{2} \approx 1.189$, so

$\Rightarrow 2 \times 1.189\\[1em] \Rightarrow 2.378$

Since, 2 < 2.378 < 3

Hence, option 3 is the correct option.

The value of $\left(\dfrac{32}{243}\right)^{-\dfrac{4}{5}}$ is

1. $\dfrac{4}{9}$

2. $\dfrac{9}{4}$

3. $\dfrac{16}{81}$

4. $\dfrac{81}{16}$

Answer

We know that,

For any non-zero rational number ($a$)

$\Big(\dfrac{a}{b}\Big)^{-m} = \Big(\dfrac{b}{a}\Big)^{m}$ and $(a^m)^n = a^{mn}$

We know that 32 = 2⁵ and 243 = 3⁵

Solving,

$\Rightarrow \Big(\dfrac{32}{243}\Big)^{\dfrac{-4}{5}}\\[1em] \Rightarrow \Big(\dfrac{243}{32}\Big)^{\dfrac{4}{5}}\\[1em] \Rightarrow \Big(\dfrac{3^5}{2^5}\Big)^{\dfrac{4}{5}}\\[1em] \Rightarrow \Big(\dfrac{3}{2}\Big)^{5 \times \dfrac{4}{5}}\\[1em] \Rightarrow \Big(\dfrac{3}{2}\Big)^{4}\\[1em] \Rightarrow \dfrac{81}{16}$

Hence, option 4 is the correct option.

The value of [2 - 3(2 - 3)^-1]^-1 is

1. 5

2. -5

3. $\dfrac{1}{5}$

4. $\dfrac{-1}{5}$

Answer

We know that,

For any non-zero rational number ($a$)

$a^{-n} = \dfrac{1}{a^n}$

Solving,

$\Rightarrow [2 - 3(2-3)^{-1}]^{-1}\\[1em] \Rightarrow [2 - 3(-1)^{-1}]^{-1}\\[1em] \Rightarrow \Big[2 - 3 \times \dfrac{1}{-1}\Big]^{-1}\\[1em] \Rightarrow [2 - (-3)]^{-1}\\[1em] \Rightarrow [2 + 3]^{-1}\\[1em] \Rightarrow 5^{-1}\\[1em] \Rightarrow \dfrac{1}{5}$

Hence, option 3 is the correct option.

The value of (8^-1 - 9^-1)^-1 ÷ (4^-1 - 9^-1)^-1 is

1. 5

2. 10

3. 14

4. 25

Answer

We know that,

For any non-zero rational number ($a$)

$a^{-n} = \dfrac{1}{a^n}$

Solving the first bracket,

$\Rightarrow (8^{-1} - 9^{-1})^{-1}\\[1em] \Rightarrow \Big(\dfrac{1}{8} - \dfrac{1}{9}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{9 - 8}{72}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{1}{72}\Big)^{-1}\\[1em] \Rightarrow 72$

Solving the second bracket,

$\Rightarrow (4^{-1} - 9^{-1})^{-1}\\[1em] \Rightarrow \Big(\dfrac{1}{4} - \dfrac{1}{9}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{9 - 4}{36}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{5}{36}\Big)^{-1}\\[1em] \Rightarrow \dfrac{36}{5}$

So,

$\Rightarrow 72 \div \dfrac{36}{5}\\[1em] \Rightarrow 72 \times \dfrac{5}{36}\\[1em] \Rightarrow \dfrac{360}{36}\\[1em] \Rightarrow 10$

Hence, option 2 is the correct option.

$(64)^{\dfrac{-1}{2}} - (-32)^{\dfrac{-4}{5}}$ = ?

1. $\dfrac{1}{8}$

2. $\dfrac{3}{8}$

3. $\dfrac{1}{16}$

4. $\dfrac{3}{16}$

Answer

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$ and $a^{-n} = \dfrac{1}{a^n}$

We know that 64 = 8² and −32 = (−2)⁵

Solving the first term,

$\Rightarrow (64)^{\dfrac{-1}{2}}\\[1em] \Rightarrow (8^2)^{\dfrac{-1}{2}}\\[1em] \Rightarrow 8^{-1}\\[1em] \Rightarrow \dfrac{1}{8}$

Solving the second term,

$\Rightarrow (-32)^{\dfrac{-4}{5}}\\[1em] \Rightarrow ((-2)^5)^{\dfrac{-4}{5}}\\[1em] \Rightarrow (-2)^{-4}\\[1em] \Rightarrow \dfrac{1}{(-2)^4}\\[1em] \Rightarrow \dfrac{1}{16}$

So,

$\Rightarrow \dfrac{1}{8} - \dfrac{1}{16}\\[1em] \Rightarrow \dfrac{2 - 1}{16}\\[1em] \Rightarrow \dfrac{1}{16}$

Hence, option 3 is the correct option.

Which of the following is the same as $\left(\dfrac{-5}{7}\right)^{-7}$?

1. $\left(\dfrac{5}{7}\right)^{-7}$

2. $-\left(\dfrac{5}{7}\right)^{-7}$

3. $\left(\dfrac{7}{5}\right)^{7}$

4. $\left(\dfrac{-7}{5}\right)^{7}$

Answer

We know that,

For any non-zero rational number ($a$)

$\Big(\dfrac{p}{q}\Big)^{-m} = \Big(\dfrac{q}{p}\Big)^{m}$

Solving,

$\Rightarrow \Big(\dfrac{-5}{7}\Big)^{-7}\\[1em] \Rightarrow \Big(\dfrac{7}{-5}\Big)^{7}\\[1em] \Rightarrow \Big(\dfrac{-7}{5}\Big)^{7}$

Hence, option 4 is the correct option.

If $3^{x+y} = 81$ and $81^{x-y} = 3^8$, then the values of $x$ and $y$ are respectively

1. -1, -3

2. -1, 3

3. 1, 3

4. 3, 1

Answer

We know that 81 = 3⁴

Solving the first equation,

$\Rightarrow 3^{x+y} = 81\\[1em] \Rightarrow 3^{x+y} = 3^4$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x + y = 4$    ..... (1)

Solving the second equation,

$\Rightarrow 81^{x-y} = 3^8\\[1em] \Rightarrow (3^4)^{x-y} = 3^8\\[1em] \Rightarrow 3^{4(x-y)} = 3^8$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 4(x - y) = 8\\[1em] \Rightarrow x - y = 2$

   ..... (2)

Adding equations (1) and (2),

$\Rightarrow (x + y) + (x - y) = 4 + 2\\[1em] \Rightarrow 2x = 6\\[1em] \Rightarrow x = 3$

Substituting $x = 3$ in equation (1),

$\Rightarrow 3 + y = 4\\[1em] \Rightarrow y = 1$

Hence, option 4 is the correct option.

Out of the following, which one is the greatest?

1. $(0.008)^{\dfrac{1}{3}}$

2. $(0.01)^{\dfrac{1}{2}}$

3. (0.2)²

4. $\dfrac{1}{100}$

Answer

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$

Solving each value,

$\Rightarrow (0.008)^{\dfrac{1}{3}} = ((0.2)^3)^{\dfrac{1}{3}} = 0.2\\[1em] \Rightarrow (0.01)^{\dfrac{1}{2}} = ((0.1)^2)^{\dfrac{1}{2}} = 0.1\\[1em] \Rightarrow (0.2)^2 = 0.2 \times 0.2 = 0.04\\[1em] \Rightarrow \dfrac{1}{100} = 0.01$

Since, 0.01 < 0.04 < 0.1 < 0.2

So, the greatest value is 0.2.

Hence, option 1 is the correct option.

(1000)¹² ÷ (10)³⁰ = ?

1. (1000)²

2. 10

3. 100

4. (100)²

Answer

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$ and $\dfrac{a^m}{a^n} = a^{m-n}$

We know that 1000 = 10³

Solving,

$\Rightarrow (1000)^{12} \div (10)^{30}\\[1em] \Rightarrow (10^3)^{12} \div 10^{30}\\[1em] \Rightarrow 10^{36} \div 10^{30}\\[1em] \Rightarrow 10^{36-30}\\[1em] \Rightarrow 10^6$

Also, (1000)² = (10³)² = 10⁶

Hence, option 1 is the correct option.

If $(0.04)^2 \div (0.008) \times (0.2)^6 = (0.2)^x$, then the value of $x$ is

1. 5

2. 6

3. 8

4. 7

Answer

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}, \dfrac{a^m}{a^n} = a^{m-n}$ and $a^m \times a^n = a^{m+n}$

We know that 0.04 = (0.2)² and 0.008 = (0.2)³

Solving L.H.S.,

$\Rightarrow (0.04)^2 \div (0.008) \times (0.2)^6\\[1em] \Rightarrow ((0.2)^2)^2 \div (0.2)^3 \times (0.2)^6\\[1em] \Rightarrow (0.2)^4 \div (0.2)^3 \times (0.2)^6\\[1em] \Rightarrow (0.2)^{4-3+6}\\[1em] \Rightarrow (0.2)^7$

So,

$\Rightarrow (0.2)^7 = (0.2)^x$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x = 7$

Hence, option 4 is the correct option.

If $(25)^{7.5} \times (5)^{2.5} \div (125)^{1.5} = 5^x$, then the value of $x$ is

1. 8.5

2. 13

3. 16

4. 17.5

Answer

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}, a^m \times a^n = a^{m+n}$ and $\dfrac{a^m}{a^n} = a^{m-n}$

We know that 25 = 5² and 125 = 5³

Solving L.H.S.,

$\Rightarrow (25)^{7.5} \times (5)^{2.5} \div (125)^{1.5}\\[1em] \Rightarrow (5^2)^{7.5} \times 5^{2.5} \div (5^3)^{1.5}\\[1em] \Rightarrow 5^{15} \times 5^{2.5} \div 5^{4.5}\\[1em] \Rightarrow 5^{15 + 2.5 - 4.5}\\[1em] \Rightarrow 5^{13}$

So,

$\Rightarrow 5^{13} = 5^x$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x = 13$

Hence, option 2 is the correct option.

If $2^{x+1} = 8^x$, then $x$ has the value

1. $\dfrac{-1}{3}$

2. $\dfrac{1}{2}$

3. 1

4. 3

Answer

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$

We know that 8 = 2³

Solving,

$\Rightarrow 2^{x+1} = 8^x\\[1em] \Rightarrow 2^{x+1} = (2^3)^x\\[1em] \Rightarrow 2^{x+1} = 2^{3x}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x + 1 = 3x\\[1em] \Rightarrow 1 = 3x - x\\[1em] \Rightarrow 2x = 1\\[1em] \Rightarrow x = \dfrac{1}{2}$

Hence, option 2 is the correct option.

Given that $9^n + 9^n + 9^n = 3^{2013}$, what is the value of $n$?

1. 1005

2. 1006

3. 2011

4. 6019

Answer

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$ and $a^m \times a^n = a^{m+n}$

We know that 9 = 3²

Solving L.H.S.,

$\Rightarrow 9^n + 9^n + 9^n\\[1em] \Rightarrow 3 \times 9^n\\[1em] \Rightarrow 3 \times (3^2)^n\\[1em] \Rightarrow 3^1 \times 3^{2n}\\[1em] \Rightarrow 3^{2n+1}$

So,

$\Rightarrow 3^{2n+1} = 3^{2013}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 2n + 1 = 2013\\[1em] \Rightarrow 2n = 2012\\[1em] \Rightarrow n = \dfrac{2012}{2}\\[1em] \Rightarrow n = 1006$

Hence, option 2 is the correct option.

If $3^{x-1} + 3^{x+1} = 90$, then $x$ is equal to

1. 0

2. 1

3. 2

4. 3

Answer

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$ and $a^{-n} = \dfrac{1}{a^n}$

Solving L.H.S.,

$\Rightarrow 3^{x-1} + 3^{x+1}\\[1em] \Rightarrow 3^x \times 3^{-1} + 3^x \times 3^1\\[1em] \Rightarrow 3^x\Big(\dfrac{1}{3} + 3\Big)\\[1em] \Rightarrow 3^x\Big(\dfrac{1 + 9}{3}\Big)\\[1em] \Rightarrow 3^x \times \dfrac{10}{3}$

So,

$\Rightarrow 3^x \times \dfrac{10}{3} = 90\\[1em] \Rightarrow 3^x = 90 \times \dfrac{3}{10}\\[1em] \Rightarrow 3^x = 27\\[1em] \Rightarrow 3^x = 3^3$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x = 3$

Hence, option 4 is the correct option.

If $\left(\dfrac{x}{y}\right)^{n-1} = \left(\dfrac{y}{x}\right)^{n-3}$ then the value of $n$ is

1. $\dfrac{1}{2}$

2. 1

3. 2

4. $\dfrac{7}{2}$

Answer

We know that,

For any non-zero rational number ($a$)

$\Big(\dfrac{p}{q}\Big)^{-m} = \Big(\dfrac{q}{p}\Big)^{m}$

Solving R.H.S.,

$\Rightarrow \Big(\dfrac{y}{x}\Big)^{n-3}\\[1em] \Rightarrow \Big(\dfrac{x}{y}\Big)^{-(n-3)}\\[1em] \Rightarrow \Big(\dfrac{x}{y}\Big)^{3-n}$

So,

$\Rightarrow \Big(\dfrac{x}{y}\Big)^{n-1} = \Big(\dfrac{x}{y}\Big)^{3-n}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow n - 1 = 3 - n\\[1em] \Rightarrow n + n = 3 + 1\\[1em] \Rightarrow 2n = 4\\[1em] \Rightarrow n = 2$

Hence, option 3 is the correct option.

If $(25)^x = (125)^y$, then $x : y$ equals

1. 1 : 1

2. 1 : 3

3. 2 : 3

4. 3 : 2

Answer

We know that,

For any non-zero rational number ($a$)

$(a^m)^n = a^{mn}$

We know that 25 = 5² and 125 = 5³

Solving,

$\Rightarrow (25)^x = (125)^y\\[1em] \Rightarrow (5^2)^x = (5^3)^y\\[1em] \Rightarrow 5^{2x} = 5^{3y}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 2x = 3y\\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{3}{2}\\[1em] \Rightarrow x : y = 3 : 2$

Hence, option 4 is the correct option.

The simplified value of $\dfrac{2 \cdot 3^{n + 1} + 7 \cdot 3^{n - 1}}{3^{n + 2} - 2\left(\dfrac{1}{3}\right)^{1 - n}}$ is

1. -1

2. 0

3. 1

4. 3

Answer

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m + n}$ and $\Big(\dfrac{1}{a}\Big)^{m} = a^{-m}$

Solving the numerator,

$\Rightarrow 2 \cdot 3^{n+1} + 7 \cdot 3^{n-1}\\[1em] \Rightarrow 2 \cdot 3^{n-1} \cdot 3^2 + 7 \cdot 3^{n-1}\\[1em] \Rightarrow 3^{n-1}(2 \times 9 + 7)\\[1em] \Rightarrow 3^{n-1}(18 + 7)\\[1em] \Rightarrow 25 \cdot 3^{n-1}$

Solving the denominator,

$\Rightarrow 3^{n+2} - 2\Big(\dfrac{1}{3}\Big)^{1-n}\\[1em] \Rightarrow 3^{n+2} - 2 \cdot 3^{-(1-n)}\\[1em] \Rightarrow 3^{n+2} - 2 \cdot 3^{n-1}\\[1em] \Rightarrow 3^{n-1} \cdot 3^3 - 2 \cdot 3^{n-1}\\[1em] \Rightarrow 3^{n-1}(27 - 2)\\[1em] \Rightarrow 25 \cdot 3^{n-1}$

Substituting the values of numerator and denominator in the given expression,

$\Rightarrow \dfrac{2 \cdot 3^{n + 1} + 7 \cdot 3^{n - 1}}{3^{n + 2} - 2\Big(\dfrac{1}{3}\Big)^{1 - n}}\\[1em] \Rightarrow \dfrac{25 \cdot 3^{n - 1}}{25 \cdot 3^{n - 1}}\\[1em] \Rightarrow 1$

Hence, option 3 is the correct option.

Fill in the blanks:

(i) The value of (-5)^-3 is .........

(ii) $\dfrac{1}{512}$ expressed as a power with base 8 is .........

(iii) The value of $\left(\dfrac{1}{2}\right)^{-2} + \left(\dfrac{1}{3}\right)^{-2} + \left(\dfrac{1}{4}\right)^{-2}$ is .........

(iv) (-3)⁴ ÷ 3⁴ is equal to .........

(v) If $3 \cdot 3^n = 3$, then the value of $n$ is .........

(vi) The multiplicative inverse of 10¹⁰ is .........

(vii) (5⁰ + 6⁰ + 7⁰) = .........

Answer

(i) We know that,

For any non-zero rational number ($a$)

$a^{-n} = \dfrac{1}{a^n}$

$\Rightarrow (-5)^{-3}\\[1em] \Rightarrow \dfrac{1}{(-5)^3}\\[1em] \Rightarrow \dfrac{1}{(-5) \times (-5) \times (-5)}\\[1em] \Rightarrow \dfrac{1}{-125}\\[1em] \Rightarrow -\dfrac{1}{125}$

Hence, the value of (-5)^-3 is $\mathbf{-\dfrac{1}{125}}$.

(ii) We know that 512 = 8 × 8 × 8 = 8³

$\Rightarrow \dfrac{1}{512}\\[1em] \Rightarrow \dfrac{1}{8^3}\\[1em] \Rightarrow 8^{-3}$

Hence, $\mathbf{\dfrac{1}{512}}$ expressed as a power with base 8 is $\mathbf{8^{-3}}$.

(iii) We know that,

For any non-zero rational number ($a$)

$\Big(\dfrac{p}{q}\Big)^{-m} = \Big(\dfrac{q}{p}\Big)^{m}$

$\Rightarrow \Big(\dfrac{1}{2}\Big)^{-2} + \Big(\dfrac{1}{3}\Big)^{-2} + \Big(\dfrac{1}{4}\Big)^{-2}\\[1em] \Rightarrow \Big(\dfrac{2}{1}\Big)^{2} + \Big(\dfrac{3}{1}\Big)^{2} + \Big(\dfrac{4}{1}\Big)^{2}\\[1em] \Rightarrow 4 + 9 + 16\\[1em] \Rightarrow 29$

Hence, the value is 29.

(iv) Solving,

$\Rightarrow (-3)^4 \div 3^4\\[1em] \Rightarrow \dfrac{(-3) \times (-3) \times (-3) \times (-3)}{3 \times 3 \times 3 \times 3}\\[1em] \Rightarrow \dfrac{81}{81}\\[1em] \Rightarrow 1$

Hence, (-3)⁴ ÷ 3⁴ is equal to 1.

(v) We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$

$\Rightarrow 3 \cdot 3^n = 3\\[1em] \Rightarrow 3^{1+n} = 3^1$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 1 + n = 1\\[1em] \Rightarrow n = 0$

Hence, the value of $\bm{n}$ is 0.

(vi) As the multiplicative inverse of $a^m = a^{-m}$

$\Rightarrow 10^{-10}\\[1em] \Rightarrow \dfrac{1}{10^{10}}$

Hence, the multiplicative inverse of 10¹⁰ is $\mathbf{10^{-10}}$.

(vii) We know that $a^0 = 1$

⇒ 5⁰ + 6⁰ + 7⁰

⇒ 1 + 1 + 1

⇒ 3

Hence, (5⁰ + 6⁰ + 7⁰) = 3.

Write true (T) or false (F):

(i) $a^m \times b^n = (ab)^{mn}$

(ii) (10)^-100 = (100)^-10

(iii) The reciprocal of $\left(\dfrac{3}{7}\right)^{-2}$ is $\left(\dfrac{7}{3}\right)^{2}$.

(iv) $(-a)^{-m} = \dfrac{1}{a^m}$

(v) 10⁵ × 10⁵ = 10¹⁰

(vi) $p^{\dfrac{-2}{5}} \times p^{\dfrac{2}{5}} = 1$

(vii) (-8)^-3 × (-8)^-4 = (-8)^-12

Answer

(i) The law $a^m \times b^n$ can be combined into a single power only when the bases are the same or the exponents are the same. The correct law is $a^m \times b^m = (ab)^m$.

Taking an example with $a = 2, b = 3, m = 2$ and $n = 3$,

$\Rightarrow a^m \times b^n = 2^2 \times 3^3 = 4 \times 27 = 108\\[1em] \Rightarrow (ab)^{mn} = (2 \times 3)^{2 \times 3} = 6^6 = 46656$

Since, L.H.S. ≠ R.H.S.

Hence, the given statement is False.

(ii) We know that 100 = 10²

Solving R.H.S.,

$\Rightarrow (100)^{-10}\\[1em] \Rightarrow (10^2)^{-10}\\[1em] \Rightarrow 10^{-20}$

Since (10)^-100 ≠ 10^-20,

Hence, the given statement is False.

(iii) As the reciprocal of $a^{-m} = a^m$

Solving,

$\Rightarrow \Big(\dfrac{3}{7}\Big)^{-2}\\[1em] \Rightarrow \Big(\dfrac{7}{3}\Big)^{2}$

The reciprocal of $\Big(\dfrac{7}{3}\Big)^{2}$ is $\Big(\dfrac{3}{7}\Big)^{2}$ and not $\Big(\dfrac{7}{3}\Big)^{2}$.

Hence, the given statement is False.

(iv) Solving L.H.S.,

$\Rightarrow (-a)^{-m}\\[1em] \Rightarrow \dfrac{1}{(-a)^m}$

When $m$ is odd, $(-a)^m = -a^m$ and so $(-a)^{-m} = -\dfrac{1}{a^m}$, which is not equal to $\dfrac{1}{a^m}$.

For example, taking $a = 2$ and $m = 3$,

$\Rightarrow (-2)^{-3} = \dfrac{1}{(-2)^3} = \dfrac{1}{-8} = -\dfrac{1}{8}\\[1em] \Rightarrow \dfrac{1}{2^3} = \dfrac{1}{8}$

Since, L.H.S. ≠ R.H.S.

Hence, the given statement is False.

(v) We know that,

$a^m \times a^n = a^{m+n}$

$\Rightarrow 10^5 \times 10^5\\[1em] \Rightarrow 10^{5+5}\\[1em] \Rightarrow 10^{10}$

Since, L.H.S. = R.H.S.

Hence, the given statement is True.

(vi) We know that,

$a^m \times a^n = a^{m+n}$ and $a^0 = 1$

$\Rightarrow p^{\dfrac{-2}{5}} \times p^{\dfrac{2}{5}}\\[1em] \Rightarrow p^{\dfrac{-2}{5}+\dfrac{2}{5}}\\[1em] \Rightarrow p^{0}\\[1em] \Rightarrow 1$

Since, L.H.S. = R.H.S.

Hence, the given statement is True.

(vii) We know that,

$a^m \times a^n = a^{m+n}$

$\Rightarrow (-8)^{-3} \times (-8)^{-4}\\[1em] \Rightarrow (-8)^{-3+(-4)}\\[1em] \Rightarrow (-8)^{-7}$

Since (-8)^-7 ≠ (-8)^-12,

Hence, the given statement is False.

Assertion (A): $\dfrac{2^3}{5^3}$ can also be expressed as $\left(\dfrac{5}{2}\right)^{-3}$.

Reason (R): For any non-zero real number $x$ and for any positive integers $m$ and $n$,

$\left(x^m\right)^n = x^{m + n}$

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

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

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

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

Answer

We know that,

$\dfrac{a^m}{b^m} = \Big(\dfrac{a}{b}\Big)^m$ and $\Big(\dfrac{p}{q}\Big)^{-m} = \Big(\dfrac{q}{p}\Big)^{m}$

Solving,

$\Rightarrow \dfrac{2^3}{5^3}\\[1em] \Rightarrow \Big(\dfrac{2}{5}\Big)^{3}\\[1em] \Rightarrow \Big(\dfrac{5}{2}\Big)^{-3}$

So, Assertion (A) is true.

The correct law of exponents is $\left(x^m\right)^n = x^{mn}$ and not $x^{m + n}$.

So, Reason (R) is false.

Hence, option 3 is the correct option.

Assertion (A): The reciprocal of $\left(\dfrac{4}{5}\right)^{3}$ is $\left(\dfrac{4}{5}\right)^{-3}$.

Reason (R): If $p = 0$, then the reciprocal of $\dfrac{p}{q}$ does not exist.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

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

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

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

Answer

The reciprocal of $a^m = a^{-m}$

Solving,

$\Rightarrow \Big(\dfrac{4}{5}\Big)^{3} \times \Big(\dfrac{4}{5}\Big)^{-3}\\[1em] \Rightarrow \Big(\dfrac{4}{5}\Big)^{3 + (-3)}\\[1em] \Rightarrow \Big(\dfrac{4}{5}\Big)^{0}\\[1em] \Rightarrow 1$

So, Assertion (A) is true.

If $p = 0$, then $\dfrac{p}{q} = 0$. Since the product of 0 with any number is 0 and never 1, the reciprocal of 0 does not exist.

So, Reason (R) is also true.

However, Assertion (A) follows from the law of reciprocals of powers and not from the fact that 0 has no reciprocal.

So, Reason (R) is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

When simplified, $(x^{-1} + y^{-1})^{-1}$ is equal to:

1. $x + y$

2. $\dfrac{xy}{x+y}$

3. $xy$

4. $\dfrac{1}{xy}$

Answer

We know that,

For any non-zero rational number ($a$)

$a^{-n} = \dfrac{1}{a^n}$

Solving,

$\Rightarrow (x^{-1} + y^{-1})^{-1}\\[1em] \Rightarrow \Big(\dfrac{1}{x} + \dfrac{1}{y}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{y + x}{xy}\Big)^{-1}\\[1em] \Rightarrow \dfrac{xy}{x + y}$

Hence, option 2 is the correct option.

If $a^x = b^y = c^z$ and $b^2 = ac$, then $y$ equals:

1. $\dfrac{2xz}{x+z}$

2. $\dfrac{xz}{2(x-z)}$

3. $\dfrac{xz}{2(z-x)}$

4. $\dfrac{2xz}{x-z}$

Answer

Let $a^x = b^y = c^z = k$. So,

$\Rightarrow a = k^{\dfrac{1}{x}}, \quad b = k^{\dfrac{1}{y}}, \quad c = k^{\dfrac{1}{z}}$

Given, $b^2 = ac$

Substituting the values,

$\Rightarrow \Big(k^{\dfrac{1}{y}}\Big)^2 = k^{\dfrac{1}{x}} \times k^{\dfrac{1}{z}}\\[1em] \Rightarrow k^{\dfrac{2}{y}} = k^{\dfrac{1}{x}+\dfrac{1}{z}}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow \dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z}\\[1em] \Rightarrow \dfrac{2}{y} = \dfrac{z + x}{xz}\\[1em] \Rightarrow y = \dfrac{2xz}{x + z}$

Hence, option 1 is the correct option.

If $3^a \cdot 4^5 = (18)^{10}$, the value of $a$ is:

1. 10

2. 12

3. 20

4. 25

Answer

We know that,

For any non-zero rational numbers ($a$ and $b$)

$(ab)^m = a^mb^m$ and $(a^m)^n = a^{mn}$

We know that 4 = 2² and 18 = 2 × 3²

Solving R.H.S.,

$\Rightarrow (18)^{10}\\[1em] \Rightarrow (2 \times 3^2)^{10}\\[1em] \Rightarrow 2^{10} \times 3^{20}$

Solving L.H.S.,

$\Rightarrow 3^a \cdot 4^5\\[1em] \Rightarrow 3^a \cdot (2^2)^5\\[1em] \Rightarrow 3^a \cdot 2^{10}$

So,

$\Rightarrow 3^a \cdot 2^{10} = 2^{10} \times 3^{20}\\[1em] \Rightarrow 3^a = 3^{20}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow a = 20$

Hence, option 3 is the correct option.

If $2^{12} - 2^{10} = 3 \times 2^x$, then the value of $x$ is:

1. 10

2. 12

3. 20

4. 21

Answer

We know that,

For any non-zero rational number ($a$)

$a^m \times a^n = a^{m+n}$

Solving L.H.S.,

$\Rightarrow 2^{12} - 2^{10}\\[1em] \Rightarrow 2^{10} \times 2^2 - 2^{10}\\[1em] \Rightarrow 2^{10}(4 - 1)\\[1em] \Rightarrow 3 \times 2^{10}$

So,

$\Rightarrow 3 \times 2^{10} = 3 \times 2^x\\[1em] \Rightarrow 2^{10} = 2^x$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x = 10$

Hence, option 1 is the correct option.

The value of (7^-1 - 8^-1)^-1 - (3^-1 - 4^-1)^-1 is:

1. 56

2. 44

3. 20

4. 1

Answer

We know that,

For any non-zero rational number ($a$)

$a^{-n} = \dfrac{1}{a^n}$

Solving the first bracket,

$\Rightarrow (7^{-1} - 8^{-1})^{-1}\\[1em] \Rightarrow \Big(\dfrac{1}{7} - \dfrac{1}{8}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{8 - 7}{56}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{1}{56}\Big)^{-1}\\[1em] \Rightarrow 56$

Solving the second bracket,

$\Rightarrow (3^{-1} - 4^{-1})^{-1}\\[1em] \Rightarrow \Big(\dfrac{1}{3} - \dfrac{1}{4}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{4 - 3}{12}\Big)^{-1}\\[1em] \Rightarrow \Big(\dfrac{1}{12}\Big)^{-1}\\[1em] \Rightarrow 12$

So,

$\Rightarrow 56 - 12\\[1em] \Rightarrow 44$

Hence, option 2 is the correct option.

The cells of bacteria doubles itself after every one hour. How many cells will there be after 8 hours?

1. (2 × 8) times the original number

2. 8² times the original number

3. 2⁸ times the original number

4. 8⁸ times the original number

Answer

Let the original number of cells be $N$.

Since the cells double after every one hour,

$\Rightarrow \text{After 1 hour} = N \times 2 = 2^1N\\[1em] \Rightarrow \text{After 2 hours} = 2^1N \times 2 = 2^2N\\[1em] \Rightarrow \text{After 3 hours} = 2^2N \times 2 = 2^3N$

Continuing this pattern, after 8 hours the number of cells

$\Rightarrow 2^8N$

So, there will be 2⁸ times the original number of cells.

Hence, option 3 is the correct option.

The value of 2¹ - 2^-1 + 2¹ - 2^-1 + 2¹ - 2^-1 + ........ (up to 100 terms) is equal to:

1. 150

2. 100

3. 80

4. 75

Answer

We know that,

For any non-zero rational number ($a$)

$a^{-n} = \dfrac{1}{a^n}$

Solving one pair of terms,

$\Rightarrow 2^1 - 2^{-1}\\[1em] \Rightarrow 2 - \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{4 - 1}{2}\\[1em] \Rightarrow \dfrac{3}{2}$

Since there are 100 terms in all, the number of such pairs

$\Rightarrow \dfrac{100}{2}\\[1em] \Rightarrow 50$

So, the value of the given expression

$\Rightarrow 50 \times \dfrac{3}{2}\\[1em] \Rightarrow \dfrac{150}{2}\\[1em] \Rightarrow 75$

Hence, option 4 is the correct option.