Indices [Exponents] — Exercise 7(B)

$\dfrac{a^{-1}}{a^{-1} + b^{-1}}$ is equal to :

1. $\dfrac{b}{a + b}$

2. $\dfrac{a + b}{a}$

3. $\dfrac{a}{a + b}$

4. a(a + b)

Answer

Simplifying the expression :

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

Hence, Option 1 is the correct option.

If $4^{2x} = \dfrac{1}{32}$, the value of x is :

1. 1.25

2. -1.25

3. 1

4. -1

Answer

Solving, the given expression :

$\Rightarrow 4^{2x} = \dfrac{1}{32} \\[1em] \Rightarrow (2^2)^{2x} = \dfrac{1}{2^5} \\[1em] \Rightarrow 2^{4x} = 2^{-5} \\[1em] \Rightarrow 4x = -5 \\[1em] \Rightarrow x = -\dfrac{5}{4} = -1.25$

Hence, Option 2 is the correct option.

$\dfrac{3x}{y^{-1}} + \dfrac{2y}{x^{-1}}$ is equal to :

1. 6xy

2. 3x² + 2y²

3. 5xy

4. $\dfrac{6}{xy}$

Answer

Simplifying the expression :

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

Hence, Option 3 is the correct option.

$\Big(8^{-\dfrac{4}{3}} ÷ 2^{-2}\Big)$ is equal to :

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

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

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

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

Answer

Simplifying the expression :

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

Hence, Option 1 is the correct option.

If 8^{2x + 5} = 1, value of x is :

1. $-\dfrac{5}{2}$

2. $\dfrac{5}{2}$

3. 2

4. $\dfrac{2}{5}$

Answer

Given,

⇒ 8^{2x + 5} = 1

⇒ 8^{2x + 5} = 8⁰

⇒ 2x + 5 = 0

⇒ 2x = -5

⇒ x = $-\dfrac{5}{2}$.

Hence, Option 1 is the correct option.

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

1. -5

2. 5

3. 0

4. 3

Answer

Given,

⇒ 3^{x + 1} = 9^{x - 2}

⇒ 3^{x + 1} = (3²)^{x - 2}

⇒ 3^{x + 1} = 3^{2x - 4}

⇒ x + 1 = 2x - 4

⇒ 2x - x = 1 + 4

⇒ x = 5.

Hence, Option 2 is the correct option.

If $\sqrt{\dfrac{a}{b}} = \Big(\dfrac{b}{a}\Big)^{1 - 2x}$, the value of x is :

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

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

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

4. $-\dfrac{3}{5}$

Answer

Given,

$\Rightarrow \sqrt{\dfrac{a}{b}} = \Big(\dfrac{b}{a}\Big)^{1 - 2x} \\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{b}{a}\Big)^{1 - 2x} \\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{a}{b}\Big)^{-(1 - 2x)} \\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{a}{b}\Big)^{(2x - 1)} \\[1em] \Rightarrow \dfrac{1}{2} = 2x - 1 \\[1em] \Rightarrow 2x = 1 + \dfrac{1}{2} \\[1em] \Rightarrow 2x = \dfrac{2 + 1}{2} \\[1em] \Rightarrow x = \dfrac{3}{2 \times 2} \\[1em] \Rightarrow x = \dfrac{3}{4}.$

Hence, Option 1 is the correct option.

Solve for x :

2^{2x + 1} = 8

Answer

Given,

⇒ 2^{2x + 1} = 8

⇒ 2^{2x + 1} = 2³

⇒ 2x + 1 = 3

⇒ 2x = 3 - 1

⇒ 2x = 2

⇒ x = $\dfrac{2}{2}$ = 1.

Hence, x = 1.

Solve for x :

2^{5x - 1} = 4 × 2^{3x + 1}

Answer

Given,

⇒ 2^{5x - 1} = 4 × 2^{3x + 1}

⇒ 2^{5x - 1} = 2² × 2^{3x + 1}

⇒ 2^{5x - 1} = 2^{2 + 3x + 1}

⇒ 2^{5x - 1} = 2^{3x + 3}

⇒ 5x - 1 = 3x + 3

⇒ 5x - 3x = 3 + 1

⇒ 2x = 4

⇒ x = $\dfrac{4}{2}$ = 2.

Hence, x = 2.

Solve for x :

3^{4x + 1} = (27)^{x + 1}

Answer

Given,

⇒ 3^{4x + 1} = (27)^{x + 1}

⇒ 3^{4x + 1} = (3³)^{x + 1}

⇒ 3^{4x + 1} = 3^{3(x + 1)}

⇒ 3^{4x + 1} = 3^{3x + 3}

⇒ 4x + 1 = 3x + 3

⇒ 4x - 3x = 3 - 1

⇒ x = 2.

Hence, x = 2.

Solve for x :

(49)^{x + 4} = 7² × (343)^{x + 1}

Answer

Given,

⇒ (49)^{x + 4} = 7² × (343)^{x + 1}

⇒ (7²)^{x + 4} = 7² × (7³)^{x + 1}

⇒ 7^{2(x + 4)} = 7² × 7^{3(x + 1)}

⇒ 7^{2x + 8} = 7² × 7^{3x + 3}

⇒ 7^{2x + 8} = 7^{2 + 3x + 3}

⇒ 2x + 8 = 3x + 5

⇒ 3x - 2x = 8 - 5

⇒ x = 3.

Hence, x = 3.

Find x, if :

4^2x = $\dfrac{1}{32}$

Answer

Given,

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

Hence, x = $-\dfrac{5}{4}$.

Find x, if :

$\sqrt{2^{x + 3}} = 16$

Answer

Given,

⇒ $\sqrt{2^{x + 3}} = 16$

Squaring both sides, we get :

⇒ 2^{x + 3} = 16²

⇒ 2^{x + 3} = (2⁴)²

⇒ 2^{x + 3} = 2⁸

⇒ x + 3 = 8

⇒ x = 8 - 3 = 5.

Hence, x = 5.

Find x, if :

$\Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27}$

Answer

Given,

$\Rightarrow \Big(\sqrt{\dfrac{3}{5}}\Big)^{x + 1} = \dfrac{125}{27} \\[1em] \Rightarrow \Big(\dfrac{3}{5}^{\dfrac{1}{2}}\Big)^{x + 1} = \Big(\dfrac{5}{3}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{3}{5}\Big)^{\dfrac{x + 1}{2}} = \Big(\dfrac{3}{5}\Big)^{-3} \\[1em] \Rightarrow \dfrac{x + 1}{2} = -3 \\[1em] \Rightarrow x + 1 = -6 \\[1em] \Rightarrow x = -6 - 1 = -7.$

Hence, x = -7.

Find x, if :

$\Big(\sqrt[3]{\dfrac{2}{3}}\Big)^{x - 1} = \dfrac{27}{8}$

Answer

Given,

$\Rightarrow \Big(\sqrt[3]{\dfrac{2}{3}}\Big)^{x - 1} = \dfrac{27}{8} \\[1em] \Rightarrow \Big(\dfrac{2}{3}^{\dfrac{1}{3}}\Big)^{x - 1} = \Big(\dfrac{3}{2}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{2}{3}\Big)^{\dfrac{x - 1}{3}} = \Big(\dfrac{2}{3}\Big)^{-3} \\[1em] \Rightarrow \dfrac{x - 1}{3} = -3 \\[1em] \Rightarrow x - 1 = 3 \times -3 \\[1em] \Rightarrow x - 1 = -9 \\[1em] \Rightarrow x = -9 + 1 = -8.$

Hence, x = -8.

Solve :

4^{x - 2} - 2^{x + 1} = 0

Answer

Given,

⇒ 4^{x - 2} - 2^{x + 1} = 0

⇒ (2²)^{x - 2} - 2^{x + 1} = 0

⇒ 2^{2(x - 2)} = 2^{x + 1}

⇒ 2^{2x - 4} = 2^{x + 1}

⇒ 2x - 4 = x + 1

⇒ 2x - x = 1 + 4

⇒ x = 5.

Hence, x = 5.

Solve :

3^x2 : 3^x = 9 : 1

Answer

Given,

⇒ 3^x2 : 3^x = 9 : 1

$\Rightarrow \dfrac{3^{x^2}}{3^x} = \dfrac{9}{1} \\[1em] \Rightarrow \dfrac{3^{x^2}}{3^x} = \dfrac{3^2}{3^0} \\[1em] \Rightarrow 3^{x^2 - x} = 3^{2- 0} \\[1em] \Rightarrow x^2 - x = 2 - 0 \\[1em] \Rightarrow x^2 - x = 2 \\[1em] \Rightarrow x^2 - x - 2 = 0 \\[1em] \Rightarrow x^2 - 2x + x - 2 = 0 \\[1em] \Rightarrow x(x - 2) + 1(x - 2) = 0 \\[1em] \Rightarrow (x + 1)(x - 2) = 0 \\[1em] \Rightarrow x + 1 = 0 \text{ or } x - 2 = 0 \\[1em] \Rightarrow x = -1 \text{ or } x = 2.$

Hence, x = -1 or x = 2.

Solve :

8 × 2^2x + 4 × 2^{x + 1} = 1 + 2^x

Answer

Given,

⇒ 8 × 2^2x + 4 × 2^{x + 1} = 1 + 2^x

⇒ 8 × 2^(x)(2) + 4 × 2^x.2¹ = 1 + 2^x

Substituting 2^x = a, in above equation, we get :

⇒ 8 × a² + 4a × 2 = 1 + a

⇒ 8a² + 8a = 1 + a

⇒ 8a² + 8a - a - 1 = 0

⇒ 8a(a + 1) - 1(a + 1) = 0

⇒ (8a - 1)(a + 1) = 0

⇒ 8a - 1 = 0 or a + 1 = 0

⇒ 8a = 1 or a = -1

a cannot be negative as 2^x, for any value of x is greater than 0.

⇒ 8(2^x) = 1

⇒ 2^x = $\dfrac{1}{8}$

⇒ $2^x = \dfrac{1}{2^3}$

⇒ 2^x = 2^-3

⇒ x = -3.

Hence, x = -3.

Solve :

2^2x + 2^{x + 2} - 4 × 2³ = 0

Answer

Given,

⇒ 2^2x + 2^{x + 2} - 4 × 2³ = 0

⇒ 2^(x)(2) + 2^x.2² - 4 × 8 = 0

Substituting 2^x = a, we get :

⇒ a² + 4a - 32 = 0

⇒ a² + 8a - 4a - 32 = 0

⇒ a(a + 8) - 4(a + 8) = 0

⇒ (a - 4)(a + 8) = 0

⇒ a - 4 = 0 or a + 8 = 0

⇒ a = 4 or a = -8

a cannot be negative as 2^x, for any value of x is greater than 0.

⇒ a = 4

⇒ 2^x = 4

⇒ 2^x = 2²

⇒ x = 2.

Hence, x = 2.

Solve :

$(\sqrt{3})^{x - 3} = (\sqrt[4]{3})^{x + 1}$

Answer

Given,

$\Rightarrow (\sqrt{3})^{x - 3} = (\sqrt[4]{3})^{x + 1} \\[1em] \Rightarrow (3^{\dfrac{1}{2}})^{x - 3} = (3^{\dfrac{1}{4}})^{x + 1} \\[1em] \Rightarrow 3^{\dfrac{x - 3}{2}} = 3^{\dfrac{x + 1}{4}} \\[1em] \Rightarrow \dfrac{x - 3}{2} = \dfrac{x + 1}{4}\\[1em] \Rightarrow 4(x - 3) = 2(x + 1) \\[1em] \Rightarrow 4x - 12 = 2x + 2 \\[1em] \Rightarrow 4x - 2x = 2 + 12 \\[1em] \Rightarrow 2x = 14 \\[1em] \Rightarrow x = \dfrac{14}{2} = 7.$

Hence, x = 7.

Find the values of m and n if :

4^2m = $(\sqrt[3]{16})^{-\dfrac{6}{n}} = (\sqrt{8})^2$

Answer

Given,

4^2m = $(\sqrt[3]{16})^{-\dfrac{6}{n}} = (\sqrt{8})^2$

Considering,

$\Rightarrow 4^{2m} = (\sqrt{8})^2 \\[1em] \Rightarrow (2^2)^{2m} = (\sqrt{8})^2 \\[1em] \Rightarrow 2^{4m} = 8 \\[1em] \Rightarrow 2^{4m} = 2^3 \\[1em] \Rightarrow 4m = 3 \\[1em] \Rightarrow m = \dfrac{3}{4}.$

Considering,

$\Rightarrow (\sqrt[3]{16})^{-\dfrac{6}{n}} = (\sqrt{8})^2 \\[1em] \Rightarrow (16)^{\dfrac{1}{3} \times \Big(-\dfrac{6}{n}\Big)} = 8 \\[1em] \Rightarrow (16)^{-\dfrac{2}{n}} = 8 \\[1em] \Rightarrow (2^4)^{-\dfrac{2}{n}} = 2^3 \\[1em] \Rightarrow (2)^{4 \times -\dfrac{2}{n}} = 2^3 \\[1em] \Rightarrow (2)^{-\dfrac{8}{n}} = 2^3 \\[1em] \Rightarrow -\dfrac{8}{n} = 3 \\[1em] \Rightarrow n = -\dfrac{8}{3}.$

Hence, $m = \dfrac{3}{4} \text{ and } n = -\dfrac{8}{3}$.

Solve for x and y, if :

$(\sqrt{32})^x ÷ 2^{y + 1} = 1 \text{ and } 8^y - 16^{4 - \dfrac{x}{2}} = 0$

Answer

Given,

$\Rightarrow (\sqrt{32})^x ÷ 2^{y + 1} = 1 \\[1em] \Rightarrow \dfrac{(\sqrt{32})^x}{2^{y + 1}} = 1 \\[1em] \Rightarrow \dfrac{(\sqrt{2^5})^x}{2^{y + 1}} = 1 \\[1em] \Rightarrow [(2^5)^{\dfrac{1}{2}}]^x = 2^{y + 1} \\[1em] \Rightarrow 2^{5 \times \dfrac{1}{2} \times x} = 2^{y + 1} \\[1em] \Rightarrow \dfrac{5x}{2} = y + 1 \\[1em] \Rightarrow 5x = 2y + 2 \\[1em] \Rightarrow x = \dfrac{2y + 2}{5} \text{ ......(1)}$

Given,

$\Rightarrow 8^y - 16^{4 - \dfrac{x}{2}} = 0 \\[1em] \Rightarrow 8^y = 16^{4 - \dfrac{x}{2}} \\[1em] \Rightarrow (2^3)^y = (2^4)^{4 - \dfrac{x}{2}} \\[1em] \Rightarrow 2^{3y} = (2)^{4\Big(4 - \dfrac{x}{2}\Big)} \\[1em] \Rightarrow 2^{3y} = 2^{16 - 2x} \\[1em] \Rightarrow 3y = 16 - 2x \\[1em] \Rightarrow 2x = 16 - 3y \text{ ........(2)}$

Substituting value of x from equation (1) in equation (2), we get :

$\Rightarrow 2\Big(\dfrac{2y + 2}{5}\Big) = 16 - 3y \\[1em] \Rightarrow \dfrac{4y + 4}{5} = 16 - 3y \\[1em] \Rightarrow 4y + 4 = 5(16 - 3y) \\[1em] \Rightarrow 4y + 4 = 80 - 15y \\[1em] \Rightarrow 4y + 15y = 80 - 4 \\[1em] \Rightarrow 19y = 76 \\[1em] \Rightarrow y = \dfrac{76}{19} = 4.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{2y + 2}{5} = \dfrac{2 \times 4 + 2}{5} \\[1em] = \dfrac{8 + 2}{5} \\[1em] = \dfrac{10}{5} = 2.$

Hence, x = 2 and y = 4.

Prove that :

$\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a} \Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}$ = 1

Answer

Solving L.H.S. of the given equation :

$\Rightarrow \Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a} \Big(\dfrac{x^c}{x^a}\Big)^{c + a - b} \\[1em] = (x^{a - b})^{a + b - c}.(x^{b - c})^{b + c - a}.(x^{c - a})^{c + a - b} \\[1em] = x^{(a - b)(a + b - c)}.x^{(b - c)(b + c - a)}.x^{(c - a)(c + a - b)} \\[1em] = x^{a^2 + ab - ac - ab - b^2 + bc}.x^{b^2 + bc - ab - bc - c^2 + ac}.x^{c^2 + ac - bc - ac - a^2 + ab} \\[1em] = x^{a^2 - b^2 - ac + bc}.x^{b^2 - c^2 - ab + ac}.x^{c^2 - a^2 - bc + ab} \\[1em] = x^{a^2 - b^2 - ac + bc + b^2 - c^2 - ab + ac + c^2 - a^2 - bc + ab} \\[1em] = x^{a^2 - a^2 - b^2 + b^2 - c^2 + c^2 - ac + ac + bc - bc - ab + ab} \\[1em] = x^0 \\[1em] = 1.$

Since, L.H.S. = R.H.S. = 1.

Hence, proved that $\Big(\dfrac{x^a}{x^b}\Big)^{a + b - c} \Big(\dfrac{x^b}{x^c}\Big)^{b + c - a} \Big(\dfrac{x^c}{x^a}\Big)^{c + a - b}$ = 1.

Prove that :

$\dfrac{x^{a(b - c)}}{x^{b(a - c)}} ÷ \Big(\dfrac{x^b}{x^a}\Big)^c$ = 1

Answer

Solving L.H.S. of the given equation :

$\Rightarrow \dfrac{x^{a(b - c)}}{x^{b(a - c)}} ÷ \Big(\dfrac{x^b}{x^a}\Big)^c = \dfrac{x^{ab - ac}}{x^{ab - bc}} ÷ \dfrac{x^{bc}}{x^{ac}} \\[1em] = x^{ab - ac - (ab - bc)} ÷ x^{bc - ac} \\[1em] = x^{ab - ab - ac + bc} ÷ x^{bc - ac} \\[1em] = x^{bc - ac} ÷ x^{bc - ac} \\[1em] = \dfrac{x^{bc - ac}}{x^{bc - ac}} \\[1em] = 1.$

Since, L.H.S. = R.H.S. = 1.

Hence, proved that $\dfrac{x^{a(b - c)}}{x^{b(a - c)}} ÷ \Big(\dfrac{x^b}{x^a}\Big)^c = 1$.

If a^x = b, b^y = c and c^z = a, prove that :

xyz = 1.

Answer

Given,

⇒ a = c^z .......(1)

⇒ c = b^y .......(2)

⇒ b = a^x ........(3)

Substituting value of c from equation (2) in (1), we get :

⇒ a = (b^y)^z

⇒ a = b^yz

Substituting value of b from equation (3) in above equation, we get :

⇒ a = (a^x)^yz

⇒ a = a^xyz

⇒ xyz = 1.

Hence, proved that xyz = 1.

If a^x = b^y = c^z and b² = ac, prove that :

y = $\dfrac{2xz}{x + z}$.

Answer

Given,

⇒ a^x = b^y = c^z = k (let)

⇒ a^x = k

⇒ a = $k^{\dfrac{1}{x}}$ ........(1)

⇒ b^y = k

⇒ b = $k^{\dfrac{1}{y}}$ ........(2)

⇒ c^z = k

⇒ c = $k^{\dfrac{1}{z}}$ ........(3)

Given,

⇒ b² = ac

Substituting values of a, b and c in above equation, we get :

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

Hence, proved that $y = \dfrac{2xz}{x + z}.$

If 5^-p = 4^-q = 20^r, show that :

$\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} = 0$.

Answer

Given,

⇒ 5^-p = 4^-q = 20^r = k (let)

⇒ 5^-p = k

⇒ 5 = $k^{-\dfrac{1}{p}}$ .......(1)

⇒ 4^-q = k

⇒ 4 = $k^{-\dfrac{1}{q}}$ ......(2)

⇒ 20^r = k

⇒ 20 = $k^{\dfrac{1}{r}}$ ............(3)

We know that,

⇒ 5 × 4 = 20

From equations (1), (2) and (3), we get :

$\Rightarrow k^{-\dfrac{1}{p}} \times k^{-\dfrac{1}{q}} = k^{\dfrac{1}{r}} \\[1em] \Rightarrow k^{-\dfrac{1}{p} + \Big(-\dfrac{1}{q}\Big)} = k^{\dfrac{1}{r}} \\[1em] \Rightarrow k^{-\dfrac{1}{p} - \dfrac{1}{q}} = k^{\dfrac{1}{r}} \\[1em] \Rightarrow -\dfrac{1}{p} - \dfrac{1}{q} = \dfrac{1}{r} \\[1em] \Rightarrow \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} = 0.$

Hence, proved that $\dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r} = 0.$.

If m ≠ n and (m + n)^-1 (m^-1 + n^-1) = m^xn^y;

show that : x + y + 2 = 0

Answer

Given,

$\Rightarrow (m + n)^{-1}(m^{-1} + n^{-1}) = m^xn^y \\[1em] \Rightarrow \dfrac{1}{(m + n)} \times \Big(\dfrac{1}{m} + \dfrac{1}{n}\Big) = m^xn^y \\[1em] \Rightarrow \dfrac{1}{(m + n)} \times \Big(\dfrac{n + m}{mn}\Big) = m^xn^y \\[1em] \Rightarrow \dfrac{1}{mn} = m^xn^y \\[1em] \Rightarrow m^{-1}n^{-1} = m^xn^y \\[1em] \Rightarrow x = -1 \text{ and } y = -1.$

Substituting value in L.H.S. of equation x + y + 2 = 0, we get :

⇒ x + y + 2 = (-1) + (-1) + 2 = -2 + 2 = 0.

Since, L.H.S. = R.H.S. = 0.

Hence, proved that x + y + 2 = 0.

If 5^{x + 1} = 25^{x - 2}; find the value of :

3^{x - 3} × 2^{3 - x}.

Answer

Given,

⇒ 5^{x + 1} = 25^{x - 2}

⇒ 5^{x + 1} = (5²)^{x - 2}

⇒ 5^{x + 1} = 5^{2(x - 2)}

⇒ x + 1 = 2(x - 2)

⇒ x + 1 = 2x - 4

⇒ 2x - x = 1 + 4

⇒ x = 5.

Substituting value of x in 3^{x - 3} × 2^{3 - x}, we get :

⇒ 3^{x - 3} × 2^{3 - x} = 3^{5 - 3} × 2^{3 - 5}

= 3² × 2^(-2)

= $9 \times \dfrac{1}{2^2} = \dfrac{9}{4} = 2\dfrac{1}{4}$.

Hence, 3^{x - 3} × 2^{3 - x} = $2\dfrac{1}{4}$.

If 4^{x + 3} = 112 + 8 × 4^x; find (18x)^3x.

Answer

Given,

⇒ 4^{x + 3} = 112 + 8 × 4^x

⇒ 4^x.4³ = 8(14 + 4^x)

⇒ 64.4^x = 8(14 + 4^x)

⇒ $\dfrac{64.4^x}{8}$ = 14 + 4^x

⇒ 8.4^x = 14 + 4^x

⇒ 8.4^x - 4^x = 14

⇒ 4^x(8 - 1) = 14

⇒ 7.4^x = 14

⇒ 4^x = $\dfrac{14}{7}$

⇒ (2²)^x = 2

⇒ 2^2x = 2¹

⇒ 2x = 1

⇒ x = $\dfrac{1}{2}$.

Substituting value of x in (18x)^3x, we get :

$\Rightarrow (18x)^{3x} = \Big(18 \times \dfrac{1}{2}\Big)^{3 \times \dfrac{1}{2}} \\[1em] = 9^{\dfrac{3}{2}} \\[1em] = (3^2)^{\dfrac{3}{2}} \\[1em] = 3^{2 \times \dfrac{3}{2}} \\[1em] = 3^3 \\[1em] = 27.$

Hence, (18x)^3x = 27.

Solve for x :

$4^{x - 1} \times (0.5)^{3 - 2x} = \Big(\dfrac{1}{8}\Big)^{-x}$

Answer

Given,

$\Rightarrow 4^{x - 1} \times (0.5)^{3 - 2x} = \Big(\dfrac{1}{8}\Big)^{-x} \\[1em] \Rightarrow (2^2)^{x - 1} \times \Big(\dfrac{5}{10}\Big)^{3 - 2x} = \Big(\dfrac{1}{2^3}\Big)^{-x} \\[1em] \Rightarrow 2^{2(x - 1)} \times \Big(\dfrac{1}{2}\Big)^{3 - 2x} = (2^{-3})^{-x} \\[1em] \Rightarrow 2^{2x - 2} \times (2^{-1})^{3 - 2x} = 2^{3x} \\[1em] \Rightarrow 2^{2x - 2} \times 2^{-1(3 - 2x)} = 2^{3x} \\[1em] \Rightarrow 2^{(2x - 2) + [-1(3 - 2x)]} = 2^{3x} \\[1em] \Rightarrow 2^{2x - 2 - 3 + 2x} = 2^{3x} \\[1em] \Rightarrow 2^{4x - 5} = 2^{3x} \\[1em] \Rightarrow 4x - 5 = 3x \\[1em] \Rightarrow 4x - 3x = 5 \\[1em] \Rightarrow x = 5.$

Hence, x = 5.

Solve for x :

(a^{3x + 5})².(a^x)⁴ = a^{8x + 12}

Answer

Given,

⇒ (a^{3x + 5})².(a^x)⁴ = a^{8x + 12}

⇒ a^{2(3x + 5)}.a^4x = a^{8x + 12}

⇒ a^{6x + 10}.a^4x = a^{8x + 12}

⇒ a^{6x + 10 + 4x} = a^{8x + 12}

⇒ a^{10x + 10} = a^{8x + 12}

⇒ 10x + 10 = 8x + 12

⇒ 10x - 8x = 12 - 10

⇒ 2x = 2

⇒ x = $\dfrac{2}{2}$ = 1.

Hence, x = 1.

Solve for x :

$(81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + x\Big(\dfrac{1}{2}\Big)^{-1}.2^0 = 27$

Answer

Given,

$\Rightarrow (81)^{\dfrac{3}{4}} - \Big(\dfrac{1}{32}\Big)^{-\dfrac{2}{5}} + x\Big(\dfrac{1}{2}\Big)^{-1}.2^0 = 27 \\[1em] \Rightarrow (3^4)^{\dfrac{3}{4}} - \Big(\dfrac{1}{2^5}\Big)^{-\dfrac{2}{5}} + (2^{-1})^{-1}x.1 = 27 \\[1em] \Rightarrow 3^{4 \times \dfrac{3}{4}} - (2^{-5})^{-\dfrac{2}{5}} + 2^{-1\times -1}.x = 27 \\[1em] \Rightarrow 3^3 - 2^{-5 \times -\dfrac{2}{5}} + 2x = 27 \\[1em] \Rightarrow 27 - 2^2 + 2x = 27 \\[1em] \Rightarrow 27 - 4 + 2x = 27 \\[1em] \Rightarrow 23 + 2x = 27 \\[1em] \Rightarrow 2x = 27 - 23 \\[1em] \Rightarrow 2x = 4 \\[1em] \Rightarrow x = \dfrac{4}{2} = 2.$

Hence, x = 2.

Solve for x :

2^{3x + 3} = 2^{3x + 1} + 48

Answer

Given,

⇒ 2^{3x + 3} = 2^{3x + 1} + 48

⇒ 2^3x.2³ = 2^3x.2¹ + 48

⇒ 8.2^3x = 2.2^3x + 48

⇒ 8.2^3x - 2.2^3x = 48

⇒ 6.2^3x = 48

⇒ 2^3x = $\dfrac{48}{6}$

⇒ 2^3x = 8

⇒ 2^3x = 2³

⇒ 3x = 3

⇒ x = $\dfrac{3}{3}$ = 1.

Hence, x = 1.

Solve for x :

3(2^x + 1) - 2^{x + 2} + 5 = 0

Answer

Given,

⇒ 3(2^x + 1) - 2^{x + 2} + 5 = 0

⇒ 3.2^x + 3 - 2^x.2² + 5 = 0

⇒ 3.2^x + 3 - 4.2^x + 5 = 0

⇒ -2^x + 8 = 0

⇒ 2^x = 8

⇒ 2^x = 2³

⇒ x = 3.

Hence, x = 3.

Solve for x :

9^{x + 2} = 720 + 9^x

Answer

Given,

⇒ 9^{x + 2} = 720 + 9^x

⇒ 9^x.9² = 720 + 9^x

⇒ 81.9^x - 9^x = 720

⇒ 9^x(81 - 1) = 720

⇒ 9^x.80 = 720

⇒ 9^x = $\dfrac{720}{80}$

⇒ 9^x = 9

⇒ x = 1.

Hence, x = 1.