Indices [Exponents] — Test Yourself

(20⁰ - 18⁰) x 7⁰ is equal to:

1. 0

2. 1

3. 14

4. none of these

Answer

Given, (20⁰ - 18⁰) x 7⁰

As we know that a⁰ = 1

⇒ (1 - 1) x 1

⇒ 0 x 1

⇒ 0

Hence, option 1 is the correct option.

(-2)^-1 ÷ (-2)^-4 is equal to:

1. 8

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

3. -8

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

Answer

Given,

⇒ (-2)^-1 ÷ (-2)^-4

⇒ (-2)^-1 ÷ $\dfrac{1}{(-2)^4}$

⇒ $\dfrac{1}{(-2)} ÷ \dfrac{1}{(-2)^4}$

⇒ $\dfrac{1}{(-2)} \times (-2)^4$

⇒ (-2)³

⇒ -8.

Hence, option 3 is the correct option.

If $\dfrac{a}{b} = \dfrac{2}{3} ÷ \Big(-\dfrac{2}{3}\Big)^0; \text{then} \Big(\dfrac{a}{b}\Big)^{-2}$ is equal to:

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

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

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

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

Answer

Given, $\dfrac{a}{b} = \dfrac{2}{3} ÷ \Big(-\dfrac{2}{3}\Big)^0$

As we know that a⁰ = 1

$\Rightarrow \dfrac{a}{b} = \dfrac{2}{3} ÷ 1 \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{2}{3}\\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{-2} = \Big(\dfrac{2}{3}\Big)^{-2}\\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{-2} = \Big(\dfrac{3}{2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{-2} = \dfrac{9}{4}.$

Hence, option 3 is the correct option.

If 5^{2x + 3} = 1, the value of x is:

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

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

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

4. $-\dfrac{2}{3}$

Answer

Given, 5^{2x + 3} = 1

As we know that a⁰ = 1,

⇒ 5^{2x + 3} = 5⁰

⇒ 2x + 3 = 0

⇒ 2x = -3

⇒ x = $-\dfrac{3}{2}$

Hence, option 2 is the correct option.

(2 + 3)^-1 x (2^-1 + 3^-1) is equal to :

1. 6

2. -6

3. $\dfrac{1}{6}$

4. $-\dfrac{1}{6}$

Answer

Given,

⇒ (2 + 3)^-1 x (2^-1 + 3^-1)

$\Rightarrow (2 + 3)^{-1} \times (2^{-1} + 3^{-1}) \\[1em] \Rightarrow \Big(\dfrac{1}{5}\Big) \times \Big[\Big(\dfrac{1}{2}\Big) + \Big(\dfrac{1}{3}\Big)\Big]\\[1em] \Rightarrow \dfrac{1}{5} \times \Big[\dfrac{3 + 2}{6}\Big]\\[1em] \Rightarrow \dfrac{1}{5} \times \dfrac{5}{6}\\[1em] \Rightarrow \dfrac{1}{6}.$

Hence, option 3 is the correct option.

Statement 1: $\Big(\dfrac{3}{4}\Big)^{-4} \times \Big(\dfrac{3}{4}\Big)^{-5} = \Big(\dfrac{3}{4}\Big)^{3x}$

⇒ x = -1

Statement 2: $\Big(\dfrac{3}{4}\Big)^{-4 - 5} = \Big(\dfrac{3}{4}\Big)^{3x}$

⇒ 3x = -9

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given,

$\Rightarrow \Big(\dfrac{3}{4}\Big)^{-4} \times \Big(\dfrac{3}{4}\Big)^{-5} = \Big(\dfrac{3}{4}\Big)^{3x}\\[1em] \Rightarrow \Big(\dfrac{3}{4}\Big)^{-4 + (-5)} = \Big(\dfrac{3}{4}\Big)^{3x}\\[1em] \Rightarrow \Big(\dfrac{3}{4}\Big)^{-4 - 5} = \Big(\dfrac{3}{4}\Big)^{3x}\\[1em] \Rightarrow \Big(\dfrac{3}{4}\Big)^{-9} = \Big(\dfrac{3}{4}\Big)^{3x}\\[1em] \Rightarrow 3x = -9\\[1em] \Rightarrow x = -\dfrac{9}{3}\\[1em] \Rightarrow x = -3.$

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Statement 1: $\Big(\dfrac{5}{8}\Big)^{-7} \times \Big(\dfrac{8}{5}\Big)^{-4} = x$

⇒ x = $\Big(\dfrac{5}{8}\Big)^{3}$

Statement 2: $\Big(\dfrac{5}{8}\Big)^{-7} \times \Big(\dfrac{5}{8}\Big)^{4} = x$

⇒ x = $\Big(\dfrac{8}{5}\Big)^{3}$

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given,

$\Rightarrow \Big(\dfrac{5}{8}\Big)^{-7} \times \Big(\dfrac{8}{5}\Big)^{-4} = x\\[1em] \Rightarrow \Big(\dfrac{5}{8}\Big)^{-7} \times \Big(\dfrac{5}{8}\Big)^{4} = x\\[1em] \Rightarrow \Big(\dfrac{5}{8}\Big)^{-7 + 4} = x\\[1em] \Rightarrow \Big(\dfrac{5}{8}\Big)^{-3} = x\\[1em] \Rightarrow x = \Big(\dfrac{8}{5}\Big)^{3}$

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Assertion (A): (3^-7 ÷ 3^-10) x 3^-5 = $\dfrac{1}{9}$.

Reason (R): $\dfrac{1}{3^7} \times 3^{10} \times \dfrac{1}{3^5}$.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given,

(3^-7 ÷ 3^-10) x 3^-5

$\Rightarrow \Big[\Big(\dfrac{1}{3}\Big)^7 ÷ \Big(\dfrac{1}{3}\Big)^{10}\Big] \times \Big(\dfrac{1}{3}\Big)^5 \\[1em] \Rightarrow \Big[\Big(\dfrac{1}{3}\Big)^7 \times 3^{10}\Big] \times \dfrac{1}{3^5} \\[1em] \Rightarrow \dfrac{3^{10}}{3^7 \times 3^5} \\[1em] \Rightarrow \dfrac{3^{10}}{3^{12}} \\[1em] \Rightarrow \dfrac{1}{3^2} \\[1em] \Rightarrow \dfrac{1}{9}.$

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Assertion (A): (1³ + 2³ + 3³)$^\frac{1}{2}$ = x, then x = $\sqrt{1} + \sqrt{8} + \sqrt{27}$.

Reason (R): x = (1 + 8 + 27)$^\frac{1}{2} = 36^\frac{1}{2} = 6$

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given, x = (1³ + 2³ + 3³)$^\frac{1}{2}$

$\Rightarrow (1 + 8 + 27)^\frac{1}{2}\\[1em] \Rightarrow (36)^\frac{1}{2}\\[1em] \Rightarrow \sqrt{36}\\[1em] \Rightarrow 6.$

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Evaluate :

$9^{\dfrac{5}{2}} - 3 \times 8^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}}$

Answer

Simplifying the expression :

$\Rightarrow 9^{\dfrac{5}{2}} - 3 \times 8^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}} = (3^2)^{\dfrac{5}{2}} - 3 \times 1 - \Big(\dfrac{1}{3^4}\Big)^{-\dfrac{1}{2}} \\[1em] = (3)^{2 \times \dfrac{5}{2}} - 3 - (3^{-4})^{-\dfrac{1}{2}} \\[1em] = 3^5 - 3 - 3^{-4 \times -\dfrac{1}{2}} \\[1em] = 243 - 3 - 3^2 \\[1em] = 243 - 3 - 9 \\[1em] = 231.$

Hence, $9^{\dfrac{5}{2}} - 3 \times 8^0 - \Big(\dfrac{1}{81}\Big)^{-\dfrac{1}{2}}$ = 231.

Evaluate :

$(64)^{\dfrac{2}{3}} - \sqrt[3]{125} - \dfrac{1}{2^{-5}} + (27)^{-\dfrac{2}{3}} \times \Big(\dfrac{25}{9}\Big)^{-\dfrac{1}{2}}$

Answer

Simplifying the expression :

$\Rightarrow (64)^{\dfrac{2}{3}} - \sqrt[3]{125} - \dfrac{1}{2^{-5}} + (27)^{-\dfrac{2}{3}} \times \Big(\dfrac{25}{9}\Big)^{-\dfrac{1}{2}} \\[1em] = (2^6)^{\dfrac{2}{3}} - (5^3)^{\dfrac{1}{3}} - 2^5 + (3^3)^{-\dfrac{2}{3}} \times \Big[\Big(\dfrac{5}{3}\Big)^2\Big]^{-\dfrac{1}{2}}\\[1em] = (2)^{6 \times \dfrac{2}{3}} - (5)^{3 \times \dfrac{1}{3}} - 32 + (3)^{3 \times -\dfrac{2}{3}} \times \Big(\dfrac{5}{3}\Big)^{2 \times -\dfrac{1}{2}} \\[1em] = 2^4 - 5^1 - 32 + 3^{-2} \times \Big(\dfrac{5}{3}\Big)^{-1} \\[1em] = 16 - 5 - 32 + \dfrac{1}{3^2} \times \dfrac{3}{5} \\[1em] = -21 + \dfrac{1}{9} \times \dfrac{3}{5} \\[1em] = -21 + \dfrac{3}{45} \\[1em] = -21 + \dfrac{1}{15} \\[1em] = \dfrac{-315 + 1}{15} \\[1em] = \dfrac{-314}{15} \\[1em] = -20\dfrac{14}{15}.$

Hence, $(64)^{\dfrac{2}{3}} - \sqrt[3]{125} - \dfrac{1}{2^{-5}} + (27)^{-\dfrac{2}{3}} \times \Big(\dfrac{25}{9}\Big)^{-\dfrac{1}{2}} = -20\dfrac{14}{15}.$

Evaluate :

$\Big[\Big(-\dfrac{2}{3}\Big)^{-2}\Big]^3 \times \Big(\dfrac{1}{3}\Big)^{-4} \times 3^{-1} \times \dfrac{1}{6}$

Answer

Simplify the expression :

$\Rightarrow \Big[\Big(-\dfrac{2}{3}\Big)^{-2}\Big]^3 \times \Big(\dfrac{1}{3}\Big)^{-4} \times 3^{-1} \times \dfrac{1}{6} \\[1em] = \Big(-\dfrac{2}{3}\Big)^{-6} \times (3^{-1})^{-4} \times \dfrac{1}{3} \times \dfrac{1}{6} \\[1em] = \Big(\dfrac{3}{2}\Big)^6 \times 3^4 \times \dfrac{1}{18} \\[1em] = \dfrac{3^6 \times 3^4}{2^6 \times 18} \\[1em] = \dfrac{3^{6 + 4}}{2^6 \times (2 \times 3 \times 3)} \\[1em] = \dfrac{3^{10}}{2^{6 + 1} \times 3^2} \\[1em] = \dfrac{3^{10 - 2}}{2^7} \\[1em] = \dfrac{3^8}{2^7} \\[1em] = 3^8 ÷ 2^7.$

Hence, $\Big[\Big(-\dfrac{2}{3}\Big)^{-2}\Big]^3 \times \Big(\dfrac{1}{3}\Big)^{-4} \times 3^{-1} \times \dfrac{1}{6} = 3^8 ÷ 2^7.$

Simplify :

$\dfrac{3 \times 9^{n + 1} - 9 \times 3^{2n}}{3 \times 3^{2n + 3} - 9^{n + 1}}$

Answer

Simplify the expression :

$\Rightarrow \dfrac{3 \times 9^{n + 1} - 9 \times 3^{2n}}{3 \times 3^{2n + 3} - 9^{n + 1}} = \dfrac{3 \times (3^2)^{n + 1} - 9 \times 3^{2n}}{3 \times 3^{2n}.3^3 - (3^2)^{n + 1}} \\[1em] = \dfrac{3 \times 3^{2(n + 1)} - 9 \times 3^{2n}}{81.3^{2n} - 3^{2(n + 1)}} \\[1em] = \dfrac{3 \times 3^{2n + 2} - 9 \times 3^{2n}}{81.3^{2n} - 3^{2n + 2}} \\[1em] = \dfrac{3 \times 3^{2n} \times 3^2 - 9 \times 3^{2n}}{81.3^{2n} - 3^{2n}.3^2} \\[1em] = \dfrac{3^{2n}(3 \times 3^2 - 9)}{3^{2n}(81 - 3^2)} \\[1em] = \dfrac{27 - 9}{81 - 9} \\[1em] = \dfrac{18}{72} \\[1em] = \dfrac{1}{4}.$

Hence, $\dfrac{3 \times 9^{n + 1} - 3 \times 3^{2n}}{3 \times 3^{2n + 3} - 9^{n + 1}} = \dfrac{1}{4}$.

Solve :

3^{x - 1} × 5^{2y - 3} = 225

Answer

Solving the expression :

$\Rightarrow 3^{x - 1} \times 5^{2y - 3} = 225 \\[1em] \Rightarrow 3^x.3^{-1} \times 5^{2y}.5^{-3} = 3^2 \times 5^2 \\[1em] \Rightarrow \dfrac{3^x}{3^1} \times \dfrac{5^{2y}}{5^3} = 3^2 \times 5^2 \\[1em] \Rightarrow 3^x \times 5^{2y} = 3^2 \times 5^2 \times 3^1 \times 5^3 \\[1em] \Rightarrow 3^x \times 5^{2y} = 3^{2 + 1} \times 5^{2 + 3} \\[1em] \Rightarrow 3^x \times 5^{2y} = 3^3 \times 5^5 \\[1em] \Rightarrow x = 3 \text{ and } 2y = 5 \\[1em] \Rightarrow x = 3 \text{ and } y = \dfrac{5}{2} = 2\dfrac{1}{2}.$

Hence, x = 3 and y = $2\dfrac{1}{2}$.

If $\Big(\dfrac{a^{-1}b^2}{a^2b^{-4}}\Big)^7 ÷ \Big(\dfrac{a^3b^{-5}}{a^{-2}b^3}\Big)^{-5} = a^x.b^y$, find x + y.

Answer

Given,

$\Rightarrow \Big(\dfrac{a^{-1}b^2}{a^2b^{-4}}\Big)^7 ÷ \Big(\dfrac{a^3b^{-5}}{a^{-2}b^3}\Big)^{-5} = a^x.b^y \\[1em] \Rightarrow (a^{-1 - 2}b^{2 - (-4)})^7 ÷ (a^{3 - (-2)}b^{-5 - 3})^{-5} = a^x.b^y \\[1em] \Rightarrow (a^{-3}b^6)^7 ÷ (a^5b^{-8})^{-5} = a^x.b^y \\[1em] \Rightarrow (a^{-3 \times 7}.b^{6 \times 7}) ÷ (a^{5 \times -5}.b^{-8 \times -5}) = a^x.b^y \\[1em] \Rightarrow (a^{-21}.b^{42}) ÷ (a^{-25}.b^{40}) = a^x.b^y \\[1em] \Rightarrow \dfrac{a^{-21}.b^{42}}{a^{-25}.b^{40}} = a^xb^y \\[1em] \Rightarrow a^{-21 - (-25)}.b^{42 - 40} = a^xb^y \\[1em] \Rightarrow a^{-21 + 25}.b^{2} = a^x.b^y \\[1em] \Rightarrow a^4.b^2 = a^x.b^y \\[1em] \Rightarrow x = 4 \text{ and } y = 2.$

x + y = 4 + 2 = 6.

Hence, x + y = 6.

If 3^{x + 1} = 9^{x - 3}, find the value of 2^{1 + x}.

Answer

Given,

⇒ 3^{x + 1} = 9^{x - 3}

⇒ 3^{x + 1} = (3²)^{x - 3}

⇒ 3^{x + 1} = 3^{2(x - 3)}

⇒ 3^{x + 1} = 3^{2x - 6}

⇒ x + 1 = 2x - 6

⇒ 2x - x = 1 + 6

⇒ x = 7.

Substituting value of x in 2^{1 + x}, we get :

⇒ 2^{1 + 7} = 2⁸ = 256.

Hence, 2^{1 + x} = 256.

If 2^x = 4^y = 8^z and $\dfrac{1}{2x} + \dfrac{1}{4y} + \dfrac{1}{8z}$ = 4, find the value of x.

Answer

Given,

⇒ 2^x = 4^y = 8^z

⇒ 2^x = (2²)^y = (2³)^z

⇒ 2^x = 2^2y = 2^3z

⇒ x = 2y = 3z

⇒ x = 2y and x = 3z

$\Rightarrow y = \dfrac{x}{2} \text{ and } z = \dfrac{x}{3}$.

Substituting value of y and z in $\dfrac{1}{2x} + \dfrac{1}{4y} + \dfrac{1}{8z} = 4$, we get :

$\Rightarrow \dfrac{1}{2x} + \dfrac{1}{4 \times \dfrac{x}{2}} + \dfrac{1}{8 \times \dfrac{x}{3}} = 4 \\[1em] \Rightarrow \dfrac{1}{2x} + \dfrac{1}{2x} + \dfrac{3}{8x} = 4 \\[1em] \Rightarrow \dfrac{2}{2x} + \dfrac{3}{8x} = 4 \\[1em] \Rightarrow \dfrac{8 + 3}{8x} = 4 \\[1em] \Rightarrow \dfrac{11}{8x} = 4 \\[1em] \Rightarrow x = \dfrac{11}{8 \times 4} = \dfrac{11}{32}.$

Hence, x = $\dfrac{11}{32}$.

If $\dfrac{9^n.3^2.3^n - (27)^n}{(3^m.2)^3} = 3^{-3}$.

Show that : m - n = 1.

Answer

Given,

$\Rightarrow \dfrac{9^n.3^2.3^n - (27)^n}{(3^m.2)^3} = 3^{-3} \\[1em] \Rightarrow \dfrac{(3^2)^n.3^2.3^n - (3^3)^n}{(3^m)^3.(2)^3} = 3^{-3} \\[1em] \Rightarrow 3^{2n}.3^2.3^n - 3^{3n} = 3^{-3}.(3^m)^3.(2)^3 \\[1em] \Rightarrow 9.3^{2n + n} - 3^{3n} = 3^{-3}.3^{3m}.8 \\[1em] \Rightarrow 9.3^{3n} - 3^{3n} = 8.3^{3m - 3} \\[1em] \Rightarrow 3^{3n}(9 - 1) = 8.3^{3m - 3} \\[1em] \Rightarrow 8.3^{3n} = 8.3^{3(m - 1)} \\[1em] \Rightarrow 3^{3n} = 3^{3(m - 1)} \\[1em] \Rightarrow 3n = 3(m - 1) \\[1em] \Rightarrow n = m - 1 \\[1em] \Rightarrow m - n = 1.$

Hence, proved that m - n = 1.

Solve for x : $(13)^{\sqrt{x}} = 4^4 - 3^4 - 6.$

Answer

Given,

$\Rightarrow (13)^{\sqrt{x}} = 4^4 - 3^4 - 6 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 256 - 81 - 6 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 169 \\[1em] \Rightarrow (13)^{\sqrt{x}} = 13^2 \\[1em] \Rightarrow \sqrt{x} = 2$

Squaring both sides of the above equation, we get :

$\Rightarrow (\sqrt{x})^2 = 2^2 \\[1em] \Rightarrow x = 4.$

Hence, x = 4.

If 3^4x = (81)^-1 and $(10)^{\dfrac{1}{y}} = 0.0001$, find the value of 2^-x × 16^y.

Answer

Given,

$\Rightarrow 3^{4x} = (3^4)^{-1} \\[1em] \Rightarrow 3^{4x} = 3^{-4} \\[1em] \Rightarrow 4x = -4 \\[1em] \Rightarrow x = -\dfrac{4}{4} \\[1em] \Rightarrow x = -1.$

Given,

$\Rightarrow (10)^{\dfrac{1}{y}} = 0.0001 \\[1em] \Rightarrow (10)^{\dfrac{1}{y}} = \dfrac{1}{10^4} \\[1em] \Rightarrow (10)^{\dfrac{1}{y}} = 10^{-4} \\[1em] \Rightarrow \dfrac{1}{y} = -4 \\[1em] \Rightarrow y = -\dfrac{1}{4}.$

Substituting value of x and y in 2^-x × 16^y, we get :

$\Rightarrow 2^{-x} \times 16^y = 2^{-(-1)} \times 16^{-\dfrac{1}{4}} \\[1em] = 2^1 \times (2^4)^{-\dfrac{1}{4}} \\[1em] = 2 \times 2^{4 \times -\dfrac{1}{4}} \\[1em] = 2 \times 2^{-1} \\[1em] = 2 \times \dfrac{1}{2} \\[1em] = 1.$

Hence, 2^-x × 16^y = 1.

If (a^m)ⁿ = a^m.aⁿ, find the value of :

m(n - 1) - (n - 1)

Answer

Given,

⇒ (a^m)ⁿ = a^m.aⁿ

⇒ a^mn = a^{m + n}

⇒ mn = m + n

⇒ mn - m = n

⇒ m(n - 1) = n

⇒ m = $\dfrac{n}{n - 1}$ .......(1)

Substituting value of m from equation (1) in m(n - 1) - (n - 1), we get :

$\Rightarrow m(n - 1) - (n - 1) = \dfrac{n}{n - 1} \times (n - 1) - (n - 1) \\[1em] = n - (n - 1) \\[1em] = n - n + 1 \\[1em] = 1.$

Hence, m(n - 1) - (n - 1) = 1.

If m = $\sqrt[3]{15} \text{ and } n = \sqrt[3]{14}$, find the value of m - n - $\dfrac{1}{m^2 + mn + n^2}$.

Answer

Given,

$\Rightarrow m = \sqrt[3]{15} \text{ and } n = \sqrt[3]{14} \\[1em] \Rightarrow m = (15)^{\dfrac{1}{3}} \text{ and } n = (14)^{\dfrac{1}{3}}$

Cubing both sides, we get :

$\Rightarrow m^3 = [(15)^{\dfrac{1}{3}}]^3 \text{ and } n^3 = [(14)^{\dfrac{1}{3}}]^3 \\[1em] \Rightarrow m^3 = (15)^{\dfrac{1}{3} \times 3} \text{ and } n^3 = (14)^{\dfrac{1}{3} \times 3} \\[1em] \Rightarrow m^3 = 15 \text{ and } n^3 = 14.$

Simplifying the expression $m - n - \dfrac{1}{m^2 + mn + n^2}$, we get :

$\Rightarrow \dfrac{m(m^2 + mn + n^2) - n(m^2 + mn + n^2) - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \dfrac{m^3 + m^2n + mn^2 - nm^2 - mn^2 - n^3 - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \dfrac{m^3 - n^3 - 1}{m^2 + mn + n^2} \\[1em]$

Substituting value of m³ and n³ in above equation, we get :

$\Rightarrow \dfrac{15 - 14 - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \dfrac{1 - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \dfrac{0}{m^2 + mn + n^2} \\[1em] \Rightarrow 0.$

Hence, $m - n - \dfrac{1}{m^2 + mn + n^2} = 0$

Evaluate :

$\Big(\dfrac{x^q}{x^r}\Big)^{\dfrac{1}{qr}} \times \Big(\dfrac{x^r}{x^p}\Big)^{\dfrac{1}{rp}} \times \Big(\dfrac{x^p}{x^q}\Big)^{\dfrac{1}{pq}}$

Answer

Simplifying the expression :

$\Rightarrow \Big(\dfrac{x^q}{x^r}\Big)^{\dfrac{1}{qr}} \times \Big(\dfrac{x^r}{x^p}\Big)^{\dfrac{1}{rp}} \times \Big(\dfrac{x^p}{x^q}\Big)^{\dfrac{1}{pq}} \\[1em] = (x^{q - r})^{\dfrac{1}{qr}} \times (x^{r - p})^{\dfrac{1}{rp}} \times (x^{p - q})^{\dfrac{1}{pq}} \\[1em] = (x)^{\dfrac{q - r}{qr}} \times (x)^{\dfrac{r - p}{rp}} \times (x)^{\dfrac{p - q}{pq}} \\[1em] = x^{\Big(\dfrac{q - r}{qr} + \dfrac{r - p}{rp} + \dfrac{p - q}{pq}\Big)} \\[1em] = x^{\Big(\dfrac{p(q - r) + q(r - p) + r(p - q)}{pqr}\Big)} \\[1em] = x^{\Big(\dfrac{pq - pr + qr - qp + rp - rq}{pqr}\Big)} \\[1em] = x^{\Big(\dfrac{0}{pqr}\Big)} \\[1em] = x^0 \\[1em] = 1.$

Hence, $\Big(\dfrac{x^q}{x^r}\Big)^{\dfrac{1}{qr}} \times \Big(\dfrac{x^r}{x^p}\Big)^{\dfrac{1}{rp}} \times \Big(\dfrac{x^p}{x^q}\Big)^{\dfrac{1}{pq}} = 1$.

Prove that :

$\dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} - b^{-1}} = \dfrac{2b^2}{b^2 - a^2}$

Answer

To prove:

$\dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} - b^{-1}} = \dfrac{2b^2}{b^2 - a^2}$

Solving L.H.S. of the above equation, we get :

$\Rightarrow \dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} - b^{-1}} = \dfrac{\dfrac{1}{a}}{\dfrac{1}{a} + \dfrac{1}{b}} + \dfrac{\dfrac{1}{a}}{\dfrac{1}{a} - \dfrac{1}{b}} \\[1em] = \dfrac{\dfrac{1}{a}}{\dfrac{b + a}{ab}} + \dfrac{\dfrac{1}{a}}{\dfrac{b - a}{ab}} \\[1em] = \dfrac{ab}{a(b + a)} + \dfrac{ab}{a(b - a)} \\[1em] = \dfrac{b}{b + a} + \dfrac{b}{b - a} \\[1em] = \dfrac{b(b - a) + b(b + a)}{(b + a)(b - a)} \\[1em] = \dfrac{b^2 - ba + b^2 + ab}{b^2 - a^2} \\[1em] = \dfrac{2b^2}{b^2 - a^2}.$

Since, L.H.S. = R.H.S. = $\dfrac{2b^2}{b^2 - a^2}$

Hence, proved that $\dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} - b^{-1}} = \dfrac{2b^2}{b^2 - a^2}$.

Prove that :

$\dfrac{a + b + c}{a^{-1} b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}} = abc$

Answer

To prove:

$\dfrac{a + b + c}{a^{-1} b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}} = abc$

Solving L.H.S. of the above equation, we get :

$\Rightarrow \dfrac{a + b + c}{a^{-1} b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}} = \dfrac{a + b + c}{\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca}} \\[1em] = \dfrac{a + b + c}{\dfrac{c + a + b}{abc}} \\[1em] = \dfrac{abc(a + b + c)}{a + b + c} \\[1em] = abc.$

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

Hence, proved that $\dfrac{a + b + c}{a^{-1} b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}} = abc$.

Find the value of x:

(3 + 4) (3² + 4²) (3⁴ + 4⁴) (3⁸ + 4⁸) (3¹⁶ + 4¹⁶) (3³² + 4³²) = (4^x - 3^x)

Answer

We know the algebraic identity:

(a - b) (a + b) = a² - b²

The given expression can also be written as,

(4 + 3) (4² + 3²) (4⁴ + 3⁴) (4⁸ + 3⁸) (4¹⁶ + 3¹⁶) (4³² + 3³²) = (4^x - 3^x)

To use our identity, we need a (4 - 3) term at the beginning. Since 4 - 3 = 1, multiplying the left side by (4 - 3) does not change its value:

∴ Given expression can be re-written as:

(4 - 3)(4 + 3) (4² + 3²) (4⁴ + 3⁴) (4⁸ + 3⁸) (4¹⁶ + 3¹⁶) (4³² + 3³²) = (4^x - 3^x)

From the algebraic identity,

(4 - 3) (4 + 3) = (4² - 3²)

(4² - 3²) (4² + 3²) = (4⁴ - 3⁴)

(4⁴ - 3⁴) (4⁴ + 3⁴) = (4⁸ - 3⁸)

(4⁸ - 3⁸) (4⁸ + 3⁸) = (4¹⁶ - 3¹⁶)

(4¹⁶ - 3¹⁶)(4¹⁶ + 3¹⁶) = (4³² - 3³²)

(4³² - 3³²)(4³² + 3³²) = (4⁶⁴ - 3⁶⁴)

4⁶⁴ - 3⁶⁴ = 4^x - 3^x

∴ x = 64

Hence, x = 64.