Class 10 Indices Exercise 9 Solutions | Maths Links Readmore Publishers & Distributors

Given,

$\rm 3^{2x + 1} = 9^{2x - 1}$

$\rm or, 3^{2x + 1} = (3^2)^{2x  -1}$

Using the law of indices: $\rm (a^b)^c = a^{bc}$ 

$\rm or, 3^{2x + 1} = 3^{2(2x  -1)}$

The base of the terms in the left-hand and right-hand sides of the equations are the same; hence, we equate their exponents.

$\rm or, 2x + 1 = 2(2x - 1)$

$\rm or, 2x + 1 = 4x - 2$

$\rm or, 2x - 2x + 1 = 4x - 2x - 2$

$\rm or, 1 = 2x - 2$

$\rm or, 1 + 2 = 2x - 2 + 2$

$\rm or, 3 = 2x$

$\rm \therefore x = \frac{3}{2}$

Hence, the required value of x = $\frac{3}{2}$ .

Given,

$\rm 2^{x-4} = 4^{x-6}$

$\rm or, 2^{x-4} = (2^2)^{x-6}$

Using the law of indices: $\rm (a^b)^c = a^{bc}$

$\rm or, 2^{x-4} = 2^{2(x-6)}$

The base of the terms in the left-hand and right-hand sides of the equations are the same; hence, we equate their exponents.

$\rm or, x - 4 = 2(x-6)$

$\rm or, x - 4 = 2x - 12$

$\rm or, x - x -4 = 2x - x - 12$

$\rm or, -4 = x - 12$

$\rm or, -4 + 12 = x - 12 + 12$

$\rm or, 8 = x$

$\rm \therefore x = 8$

Hence, the required value of x = 8.

The required value of x is 1.

$$\rm 4^{2x - 1} = 2^{x+1}$$

By using the law of indices,

$$\rm \left ( 2^2 \right )^{2x - 1} = 2^{x+1}$$

By using the identity that $\rm (a^m)^n = a^{mn}$, we get,

$$\rm 2^{2(2x-1)} = 2^{x+1}$$

Since the bases of both Left Hand and Right Hand sides of the equation are same, we equate their exponents, we get,

$$\rm 2(2x - 1) = x+ 1$$

$$\rm or, 4x - 2 = x + 1$$

$$\rm or, 4x - x = 2 + 1$$

$$\rm or, 3x = 3$$

$$\rm or, \frac{3x}{3} = \frac{3}{3}$$

$$\rm \therefore x = 1$$

Given,

$\rm 3^x + 3^{x+2} = \frac{10}{3}$

By using the law of indices, $\rm a^{m+n} = a^m \cdot a^n$, we get,

$\rm or, 3^x + 3^x \cdot 3^2 = \frac{10}{3}$

$\rm or, 3^x \left ( 1 + 3^2 \right ) = \frac{10}{3}$

$\rm or, 3^x \left ( 1 + 9 \right ) = \frac{10}{3}$

$\rm or, 3^x \cdot 10 = \frac{10}{3}$

Dividing both sides by 10, we get,

$\rm or, 3^x \cdot \frac{10}{10} = \frac{10}{3} \cdot \frac{1}{10}$

$\rm or, 3^x = \frac{1}{3}$

By using the law of indices, $\rm a^{-m} = \frac{1}{a^m}$, we get,

$\rm or, 3^x = 3^{-1}$

The base of the terms in the left-hand and the right-hand sides of the equation are same, so we equate their exponents, we get,

$\rm \therefore x = -1$

Hence, the required value of x = -1. 

Given,

$\rm 2^{x+3} + 2^{x+1} = 80$

By using the law of indices, $\rm a^{m+n} = a^m \cdot a^n$, we get,

$\rm or, 2^x \cdot 2^3 + 2^x \cdot 2^1 = 80$

$\rm or, 2^x \left ( 2^3 + 2^1 \right ) = 80$

$\rm or, 2^x \left ( 8 + 2 \right ) = 80$

$\rm or, 2^x \cdot 10 = 80$

Dividing both sides of the equation by 10, we get,

$\rm or, 2^x \cdot \frac{10}{10} = \frac{80}{10}$

$\rm or, 2^x = 8$

$\rm or, 2^x = 2 \cdot 2 \cdot 2$

$\rm or, 2^x = 2^3$

The base of the terms on both sides of the equation are the same, so we equate their powers, we get,

$\rm \therefore x = 3$

Hence, the required value of x = 3.

Given,

$\rm 2^{x+3} + 2^x = 36$

By using the law of indices, $\rm a^{m+n} = a^m \cdot a^n$, we get,

$\rm or, 2^x \cdot 2^3 + 2^x = 36$

$\rm or, 2^x \cdot ( 2^3 + 1) = 36$

$\rm or, 2^x \cdot (8 + 1) = 36$

$\rm or, 2^x \cdot 9 = 36$

Dividing both sides of the equation by 9, we get,

$\rm or, 2^x \cdot \frac{9}{9} = \frac{36}{9}$

$\rm or, 2^x = 4$

$\rm or, 2^x = 2^2$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 2$

Hence, the required value of x is 2.

Given

$\rm 3^{x+2} + 3^{x+1} = 1 \frac{1}{3}$

By using the law of indices, $\rm a^{m+n} = a^m \cdot a^n$, we get,

$\rm or, 3^x \cdot 3^2 + 3^x \cdot 3^1 = \frac{3 \cdot 1 + 1}{3}$

$\rm or, 3^x ( 3^2 + 3^1) = \frac{4}{3}$

$\rm or, 3^x (9 + 3) = \frac{4}{3}$

$\rm or, 3^x (12) = \frac{4}{3}$

Dividing both sides of the equation by 12, we get,

$\rm or, 3^x \frac{12}{12} = \frac{4}{3} \cdot \frac{1}{12}$

$\rm or, 3^x = \frac{1}{9}$

$\rm or, 3^x = \frac{1}{3^2}$

By the law of indices, $\rm a^{-m} = \frac{1}{a^m}$, we get,

$\rm or 3^x = 3^{-2}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = -2$

Hence, the required value of x is -2.

Given

$\rm 2^x - 2^{x-2} = 6$

By using the law of indices, $\rm a^{m-n} = \frac{a^m}{a^n}$, we get,

$\rm or, 2^x - \frac{2^x}{2^2} = 6$

$\rm or, 2^x - \frac{2^x}{4} = 6$

Multiplying both sides of the equation by 4, we get,

$\rm or, 2^x \cdot 4 - \frac{2^x}{4} \cdot 4 = 6 \cdot 4$

$\rm or, 4 \cdot 2^x - 2^x = 24$

$\rm or,  2^x ( 4 - 1) = 24$

$\rm or, 2^x (3) = 24$

Dividing both sides of the equation by 3, we get,

$\rm or, 2^x \cdot \frac{3}{3} = \frac{24}{3}$

$\rm or, 2^x = 8$

$\rm or, 2^x = 2^3$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 3$

Hence, the required value of x is 3.

Given

$\rm 2^y + 2^{y-2} = 5$

By using the law of indices, $\rm a^{m-n} = \frac{a^m}{a^n}$, we get,

$\rm or, 2^y + \frac{2^y}{2^2} = 5$

$\rm or, 2^y + \frac{2^y}{4} = 5$

Multiplying both sides of the equation by 4, we get,

$\rm or, 2^y \cdot 4 + \frac{2^y}{4} \cdot 4 = 5 \cdot 4$

$\rm or, 4 \cdot 2^y + 2^y = 20$

$\rm or, 2^y ( 4 + 1) = 20$

$\rm or, 2^y \cdot 5 = 20$

Dividing both sides of the equation by 5, we get,

$\rm or, 2^y \cdot \frac{5}{5} = \frac{20}{5}$

$\rm or, 2^y = 4$

$\rm or, 2^y = 2^2$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore y = 2$

Hence, the required value of y = 2.

Given

$\rm 2^{x + 1} - 2^{x} = 8$

By the law of indices, $\rm a^{m+n} = a^{m} \cdot a^{n}$,

$\rm or, 2^{x} \cdot 2^{1} - 2^{x} = 8$

$\rm or, 2^{x} \left ( 2^{1} - 1 \right ) = 8$

$\rm or, 2^{x} (2 - 1) = 8$

$\rm or, 2^{x} \cdot 1 = 8$

$\rm or, 2^{x} = 8$

$\rm or, 2^{x} = 2^3$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 3$

Hence, the required value of x is 3.

Given

$\rm 3^{x + 1} - 3^{x} = 54$

By the law of indices, $\rm a^{m+n} = a^{m} \cdot a^{n}$,

$\rm or, 3^{x} \cdot 3^{1} - 3^{x} = 54$

$\rm or, 3^{x} \left ( 3^{1} - 1 \right ) = 54$

$\rm or, 3^{x} \left ( 3 - 1 \right ) = 54$

$\rm or, 3^{x} \cdot 2 = 54$

Dividing both sides of the equation by 2, we get,

$\rm or, 3^{x} \cdot \frac{2}{2} = \frac{54}{2}$

$\rm or, 3^{x} = 27$

$\rm or, 3^{x} = 3^{3}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 3$

Hence, the required value of x is 3.

Given,

$\rm 2^{x} + 2^{x + 2} = 5$

By the law of indices, $\rm a^{m+n} = a^{m} \cdot a^{n}$,

$\rm or, 2^{x} + 2^{x} \cdot 2^{2} = 5$

$\rm or, 2^{x} \left ( 1 + 2^{2} \right ) = 5$

$\rm or, 2^{x} \cdot (1 + 4) = 5$

$\rm or, 2^{x} \cdot 5 = 5$

Dividing both sides of the equation by 5, we get,

$\rm or, 2^{x} \cdot \frac{5}{5} = \frac{5}{5}$

$\rm or, 2^{x} = 1$

$\rm or, 2^{x} = 2^{0}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 0$

Hence, the required value of x is 0.

Given

$\rm 3^{x+3} + \frac{1}{3^{x}} - 28 = 0$

By using the law of indices, $\rm a^{m+n} = a^{m} \cdot a^{n}$,

$\rm or, 3^{x} \cdot 3^{3} + \frac{1}{3^{x}} - 28 = 0$

$\rm or, 3^{x} \cdot \left ( 3^{x} \cdot 3^{3} \right ) + 3^{x} \cdot \frac{1}{3^{x}} - 3^{x} \cdot 28 = 3^{x} \cdot 0$

$\rm or, 3^{x} \cdot 3^{x} \cdot 27 + 1 - 28 \cdot 3^{x} = 0$

By using the law of indices, $\rm a^{m+n} = a^{m} \cdot a^{n}$,

$\rm or, 3^{x + x} \cdot 27 + 1 - 28 \cdot 3^{x} = 0$

$\rm or,  27 \cdot 3^{2x} - 28 \cdot 3^{x} + 1 = 0$

By using the law of indices, $\rm  \left ( a^{m} \right )^{n} = a^{mn}$,

$\rm or, 27 \cdot \left ( 3^{x} \right) ^{2} - 28 \cdot 3^{x} + 1 = 0$

Let $\rm 3^{x} = a$. Substituting the supposed values in the above equation, we get,

$\rm or, 27 \cdot a^{2} - 28 \cdot a + 1 = 0$

The above equation is quadratic in a. We use the mid-term factorization method to solve the quadratic equation.

$\rm or, 27 \cdot a^{2} - (27 + 1) \cdot a + 1 = 0$

$\rm or, 27 \cdot a^{2} - 27 \cdot a - a + 1 = 0$

$\rm or, 27 a ( a - 1) - 1 (a - 1) = 0$

$\rm or, (27 a - 1) (a - 1) = 0$

Either

$\rm (27 a - 1) = 0$

$\rm or, 27a = 1$

$\rm or, a = \frac{1}{27}$

$\rm or, a = \frac{1}{3^{3}}$

$\rm or, a = 3^{-3}$

$\rm or, 3^{x} = 3^{-3}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = -3$

Or

$\rm (a - 1) = 0$

$\rm or, a = 1$

$\rm or, a = 3^{0}$

$\rm or, 3^{x} = 3^{0}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 0$

Hence, the required values of x are x = {-3, 0}.

Given

$\rm 2^{x+3} + \frac{1}{2^{x}} - 9 = 0$

By using the law of indices, $\rm a^{m+n} = a^{m} \cdot a^{n}$,

$\rm or, 2^{x} \cdot 2^{3} + \frac{1}{2^{x}} - 9 = 0$

$\rm or, 2^{x} \cdot 8 + \frac{1}{2^{x}} - 9 = 0$

Multiplying both sides of the equation by $\rm 2^{x}$, we get,

$\rm or, 2^{x} \cdot 2^{x} \cdot 8 + 2^{x} \cdot \frac{1}{2^{x}} - 2^{x} \cdot 9 = 0$

$\rm or, 2^{x} \cdot 2^{x} \cdot 8 + 1 - 2^{x} \cdot 9 = 0$

By using the law of indices, $\rm a^{m+n} = a^{m} \cdot a^{n}$,

$\rm or, 2^{x + x} \cdot 8 + 1 - 2^{x} \cdot 9 = 0$

$\rm or, 2^{2x} \cdot 8 + 1 - 2^{x} \cdot 9 = 0$

By using the law of indices, $\rm a^{m+n} = a^{m} \cdot a^{n}$,

$\rm or, \left ( 2^{x} \right ) ^{2} \cdot 8 + 1 - 2^{x} \cdot 9 = 0$

Let $\rm 2^{x} = a$. Substituting the supposed values in the above equation, we get,

$\rm or, a^{2} \cdot 8 + 1 - a \cdot 9$

The above equation is quadratic in a. We use the mid-term factorization method to solve the quadratic equation.

$\rm or, 8 a^{2} - 9 a + 1 = 0$

$\rm or, 8 a^{2} - ( 8 + 1) a + 1 = 0$

$\rm or, 8a^{2} - 8a - a + 1 = 0$

$\rm or, 8a ( a - 1) - 1( a- 1) = 0$

$\rm or, (8a - 1)(a - 1) = 0$

Either

$\rm (8a - 1) = 0$

$\rm or, 8a = 1$

$\rm or, a = \frac{1}{8}$

$\rm or, a = \frac{1}{2^{3}}$

By the law of indices, $\rm \frac{1}{a^{m}} = a^{-m}$,

$\rm or, a = 2^{-3}$

$\rm or, 2^{x} = 2^{-3}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = -3$

Or

$\rm (a - 1) = 0$

$\rm or, a = 1$

$\rm or, a = 2^{0}$

$\rm or, 2^{x} = 2^{0}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 0$

Hence, the required values of x are x = {-3, 0}.

Given

$\rm \frac{2^{x+1}}{16} + \frac{16}{2^{x + 1}} = \frac{65}{8}$

$\rm or, \frac{2^{x+1}}{16} + \left ( \frac{2^{x+1}}{16} \right )^{-1} = \frac{65}{8}$ 

Let $\rm a = \frac{2^{x + 1}}{16}$, we get,

$\rm or, a + a^{-1} = \frac{65}{8}$

\[\rm a + \frac{1}{a} = \frac{65}{8}\]

Multiplying both sides by \(\rm a\), we get:

\[\rm a^2 + 1 = \frac{65a}{8}\]

Multiplying both sides by \(\rm 8\), we get:

\[\rm 8a^2 + 8 = 65a\]

Rearranging terms, we have a quadratic equation in \(\rm a\):

\[\rm 8a^2 - 65a + 8 = 0\]

The above equation is quadratic in a. We use the mid-term factorization formula to solve the equation.

$\rm or, 8a^{2} - (64 + 1) a + 8 = 0$

$\rm or, 8a^{2} - 64a - a + 8 = 0$

$\rm or, 8a (a - 8) - 1 (a - 8) = 0$

$\rm or, (8a - 1)(a - 8) = 0$

Either

$\rm (8a - 1) = 0$

$\rm or, 8a = 1$

$\rm or, a = \frac{1}{8}$

Substitute $\rm a = \frac{2^{x + 1}}{16}$ into the equation, we get,

$\rm or, \frac{2^{x + 1}}{16} = \frac{1}{8}$

$\rm or, 2^{x + 1} = 2$

$\rm or, 2^{x + 1} = 2^{1}$

$\rm or, x + 1 = 1$

$\rm or, x = 1 - 1$

$\rm \therefore x = 0$

Or

$\rm (a - 8) = 0$

$\rm or, a = 8$

Substitute $\rm a = \frac{2^{x + 1}}{16}$ into the equation, we get,

$\rm or, \frac{2^{x + 1}}{16} = 8$

$\rm or, 2^{x + 1} = 16 \cdot 8$

$\rm or, 2^{x + 1} = 2^{4} \cdot 2^{3}$

$\rm or, 2^{x + 1} = 2^{4 + 3}$

$\rm or, 2^{x + 1} = 2^{7}$

$\rm or, x + 1 = 7$

$\rm or, x = 7 -1 $

$\rm \therefore x = 6$

Hence, the required values of x are x = {0, 6}.

Given

$\rm 7^x + \frac{1}{7^x} = 49 \frac{1}{49}$

Let us consider $\rm a = 7^x$. Substituting the values of a in the above equations gives us

$\rm a + \frac{1}{a} = 49 \frac{1}{49}$

$\rm or, a + \frac{1}{a} = \frac{49 \cdot 49 + 1}{49}$

$\rm or, a + \frac{1}{a} = \frac{2402}{49}$

Multiplying both sides of the equation by a, we get,

$\rm or, a \cdot a + \frac{1}{a} \cdot a = \frac{2402}{49} \cdot a$

$\rm or, a^2 + 1 = \frac{2402a}{49}$

Multiplying both sides of the equation by 49, we get,

$\rm or, a^2 \cdot 49 + 1 \cdot 49 = \frac{2402a}{49} \cdot 49$

$\rm or, 49 a^2 + 49 = 2402a$

Subtracting 2402a from both sides of the equation, we get,

$\rm or, 49a^2 + 49 - 2402a = 2402a - 2402a$

$\rm or, 49a^2 - 2402a + 49 = 0$

The above equation is quadratic in a. We use the factorization method to solve it.

$\rm or, 49a^2 - (2401 + 1)a + 49 = 0$

$\rm or, 49a^2 - 2401a - a + 49 = 0$

$\rm or, 49a (a - 49) - 1 (a - 49) = 0$

$\rm or, (49a - 1)(a -49) = 0$

Either

$\rm 49a - 1 = 0$

$\rm or, 49a = 1$

$\rm or, a = \frac{1}{49}$

Substituting the value, $\rm a = 7^x$, we get,

$\rm or, 7^x = \frac{1}{7^2}$

$\rm or, 7^x = 7^{-2}$

$\rm \therefore x = -2$

Or

$\rm a - 49 = 0$

$\rm or, a = 49$

Substituting the value, $\rm a = 7^x$, we get,

$\rm or, 7^x = 7^2$

$\rm \therefore x = 2$

Hence, the required values of x = {-2, 2}.

Given

$\rm 2^x + \frac{1}{2^x} = 4 \frac{1}{4}$

Let us consider $\rm a = 2^x$. Substituting the values of a in the above equations gives us

$\rm a + \frac{1}{a} = 4 \frac{1}{4}$

$\rm or, a + \frac{1}{a} = \frac{4 \cdot 4 + 1}{4}$

$\rm or, a + \frac{1}{a} = \frac{17}{4}$

Multiplying both sides of the equation by a, we get,

$\rm or, a \cdot a + \frac{1}{a} \cdot a = 4 \frac{17}{4} \cdot a$

$\rm or, a^2 + 1 = \frac{17a}{4}$

Multiplying both sides of the equation by 4, we get,

$\rm or, a^2 \cdot 4 + 1 \cdot 4 = \frac{17a}{4} \cdot 4$

$\rm or, 4 a^2 + 4 = 17a$

Subtracting 17a from both sides of the equation, we get,

$\rm or, 4a^2 + 4 - 17a = 17a - 17a$

$\rm or, 4a^2 - 17a + 4 = 0$

The above equation is quadratic in a. We use the factorization method to solve it.

$\rm or, 4a^2 - (16 + 1)a + 4 = 0$

$\rm or, 4a^2 - 16a - a + 4 = 0$

$\rm or, 4a (a - 4) - 1 (a - 4) = 0$

$\rm or, (4a - 1)(a -4) = 0$

Either

$\rm 4a - 1 = 0$

$\rm or, 4a = 1$

$\rm or, a = \frac{1}{4}$

Substituting the value, $\rm a = 2^x$, we get,

$\rm or, 2^x = \frac{1}{2^2}$

$\rm or, 2^x = 2^{-2}$

$\rm \therefore x = -2$

Or

$\rm a - 4 = 0$

$\rm or, a = 4$

Substituting the value, $\rm a = 2^x$, we get,

$\rm or, 2^x = 2^2$

$\rm \therefore x = 2$

Hence, the required values of x = {-2, 2}.

Given
\( \rm (3^x + \frac{1}{3^x} = 9 \frac{1}{9})\)

Let \( \rm  a = 3^x\). Substituting the value of \(a\) in the equation gives us:

\(\rm a + \frac{1}{a} = 9 \frac{1}{9}\)

\( \rm a + \frac{1}{a} = \frac{9 \cdot 9 + 1}{9}\)

\( \rm a + \frac{1}{a} = \frac{82}{9}\)

Multiplying both sides of the equation by \(a\), we get:

\( \rm a \cdot a + \frac{1}{a} \cdot a = \frac{82}{9} \cdot a\)

\( \rm a^2 + 1 = \frac{82a}{9}\)

Multiplying both sides of the equation by \(9\), we get:

\( \rm 9a^2 + 9 = 82a\)

Subtracting \(82a\) from both sides of the equation, we get:

\( \rm 9a^2 - 82a + 9 = 82a - 82a\)

\( \rm 9a^2 - 82a + 9 = 0\)

The above equation is quadratic in \(a\). We use the factorization method to solve it:

\(\rm 9a^2 - (81 + 1)a + 9 = 0\)

\(\rm 9a^2 - 81a - a + 9 = 0\)

\(\rm 9a (a - 9) - 1 (a - 9) = 0\)

\(\rm (9a - 1)(a - 9) = 0\)

Either:

\(\rm 9a - 1 = 0\)

\(\rm 9a = 1\)

\(\rm a = \frac{1}{9}\)

Substituting the value, \(a = 3^x\), we get:

\(\rm 3^x = \frac{1}{3^2}\)

\(\rm 3^x = 3^{-2}\)

\(\rm \therefore x = -2\)

Or:

\(\rm a - 9 = 0\)

\(\rm a = 9\)

Substituting the value, \(a = 3^x\), we get:

\(\rm 3^x = 3^2\)

\(\rm \therefore x = 2\)

Hence, the required values of \(x\) are {-2, 2}.

Given

$\rm 5^{x - 2} + 5^{3 - x} = 6$

By using the law of indices, $\rm a^{m-n} = \frac{a^{m}}{a^{n}}$,

$\rm or, \frac{5^{x}}{5^{2}} + \frac{5^{3}}{5^{x}} = 6$

$\rm or, \frac{5^{x}}{25} + \frac{125}{5^{x}} = 6$

Let $\rm 5^{x} = a$

$\rm or, \frac{a}{25} + \frac{125}{a} = 6$

Multiplying both sides of the equation by 25a, we get,

$\rm or, 25a \cdot \frac{a}{25} + 25a \cdot \frac{125}{a} = 25a \cdot 6$

$\rm or, a^{2} + 3125 = 150a$

$\rm or, a^{2} - 150 a + 3125 = 0$

The above equation is quadratic in a. We use the mid-term factorization method to solve it.

$\rm or, a^{2} - (125 + 25) a + 3125 = 0$

$\rm or, a^{2} - 125 a - 25 a + 3125 = 0$

$\rm or, a (a - 125) - 25 (a - 125) = 0$

$\rm or, (a - 25) (a - 125) = 0$

Either

$\rm (a - 25) = 0$

$\rm or, a = 25$

$\rm or, a = 5^{2}$

By supposition, $\rm a = 5^{x}$. We substitute the value for a and we get,

$\rm or, 5^{x} = 5^{2}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 2$

Or

$\rm (a - 125) = 0$

$\rm or, a = 125$

$\rm or, a = 5^{3}$

By supposition, $\rm a = 5^{x}$. We substitute the value for a and we get,

$\rm or, 5^{x} = 5^{3}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 3$

Hence, the required values of x are x = {2,3}.

Given

$\rm 3^{x - 3} + 3^{4 - x} = 4$

By using the law of indices, $\rm a^{m-n} = \frac{a^{m}}{a^{n}}$,

$\rm or, \frac{3^{x}}{3^{3}} + \frac{3^{4}}{3^{x}} = 4$

$\rm or, \frac{3^{x}}{27} + \frac{81}{3^{x}} = 4$

Let $\rm 3^{x} = a$

$\rm or, \frac{a}{27} + \frac{81}{a} = 4$

Multiplying both sides of the equation by 27a, we get,

$\rm or, 27 a \cdot \frac{a}{27} + 27 a \cdot \frac{81}{a} = 27 a \cdot 4$

$\rm or, a^{2} + 2187 = 108 a$

$\rm or, a^{2} - 108 a + 2187 = 0$

The above equation is quadratic in a. We use the mid-term factorization method to solve it.

$\rm or, a^{2} - (27 + 81) a + 2187 = 0$

$\rm or, a^{2} - 27a - 81a + 2187 = 0$

$\rm or, a ( a - 27) - 81 ( a - 27) = 0$

$\rm or, (a - 81)(a - 27) = 0$

Either

$\rm (a - 81) = 0$

$\rm or, a = 81$

$\rm or, a = 3^{4}$

By supposition, $\rm a = 3^{x}$. We substitute the value for a and we get,

$\rm or, 3^{x} = 3^{4}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 4$

Or

$\rm (a - 27) = 0$

$\rm or, a = 27$

$\rm or, a = 3^{3}$

By supposition, $\rm a = 3^{x}$. We substitute the value for a and we get,

$\rm or, 3^{x} = 3^{3}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 3$

Hence, the required values of x are x = {3,4}.

Given

$\rm 5 \cdot 4 ^{x+1} - 16^x = 64$

$\rm or, 5 \cdot 4^{x + 1} - (4^2)^x = 64$

By using the law of indices, $\rm (a^m)^n = (a^n)^m = a^{mn}$, we get,

$\rm or, 5 \cdot 4^{x+1} - (4^x)^2 = 64$

By using the law of indices, $\rm a^{m+n} = a^m \cdot a^n$, we get,

$\rm or, 5 \cdot 4^x \cdot 4^1 - (4^x)^2x = 64$

$\rm or, 20 \cdot 4^x - (4^x)^2 = 4^3$

Let us consider $\rm a = 4^x$, we get,

$\rm or, 20 \cdot a - (a)^2 = 64$

$\rm or, a^2 - 20 a + 64 = 0$

The above equation is quadratic in a. By using the quadratic formula, we get,

$\rm or, a^2 - (16 + 4) a + 64 = 0$

$\rm or, a^2 - 16a - 4a + 64 = 0$

$\rm or, a (a - 16) - 4 (a - 16) = 0$

$\rm or, (a - 4) (a - 16) = 0$

Either

$\rm or, (a - 4) = 0$

$\rm or, a = 4$

By our assumption, $\rm a = 4^x$. We substitute the value back in the equation, we get,

$\rm or, 4^x = 4$

$\rm or, 4^x = 4^1$

The base of the terms on both sides of the equation are the same, so we equate their powers; we get,

$\rm \therefore x = 1$

Or

$\rm or, (a - 16) = 0$

$\rm or, a = 16$

By our assumption, $\rm a = 4^x$. We substitute the value back in the equation, we get,

$\rm or, 4^x = 16$

$\rm or, 4^x = 4^2$

The base of the terms on both sides of the equation are the same, so we equate their powers; we get,

$\rm \therefore x = 2$

Hence, the required values of x = {1, 2}.

Given,

$$\rm 4^x - 5 \cdot 2^x + 4 = 0$$

$$\rm (2^2)^x - 5 \cdot 2^x + 4 = 0$$

$$\rm (2^x)^2 - 5 \cdot (2^x) + 4 = 0$$

Let $\rm a = 2^x$

$$\rm (a)^2 - 5 (a) + 4 = 0$$

Now, we have a quadratic equation in a. We solve the quadratic equation by factorization.

$$\rm a^2 - (1 + 4) a + 4 = 0$$

$$\rm or, a^2 - a - 4a + 4 = 0$$

$$\rm or, a ( a - 1) - 4 ( a - 1) = 0$$

$$\rm or, (a - 4)(a - 1) = 0$$

Either

$$\rm (a - 4) = 0$

$$\rm or, a = 4$$

$$\rm or, 2^x = 4$$

$$\rm or, 2^x = 2^2$$

The base in the left-hand and the right-hand sides of the equations are the same, so we equate their powers.

$$\rm \therefore x = 2$$

Or

$$\rm (a - 1) = 0$$

$$ \rm or, a = 1$$

$$\rm or, a = 2^0$$

$$\rm or, 2^x = 2^0$$

The base in the left-hand and the right-hand sides of the equations are the same, so we equate their powers.

$$\rm \therefore x = 0$$

Hence, the required values of x are x = {0,2}.

Given

$\rm 9^{x} = 4\cdot 3^{x + 1} - 27$

By the law of indices, $\rm a^{m + n} = a^{m} \cdot a^{n}$, we get,

$\rm or, (3^{2})^{x}  = 4 \cdot 3^{x} \cdot 3^{1} - 27$

By the law of indices, $\rm ( a^{m} ) ^{n} = a^{mn}$, we get,

$\rm or, 3^{2x} = 12 \cdot 3^{x} - 27$

By the law of indices, $\rm a^{mn} = ( a^{m} ) ^{n}$, we get,

$\rm or, (3^{x})^{2} = 12 \cdot 3^{x} - 27$

Let $\rm 3^{x} = a$, we get,

$\rm or, (a)^{2} = 12 \cdot a - 27$

The above equation is a quadratic in a. We use the mid-term factorization method to find its roots, we get,

$\rm or, a^{2} - 12 a + 27 = 0$

$\rm or, a^{2} - (3 + 9)a + 27 = 0$

$\rm or, a^{2} - 3a - 9a + 27 = 0$

$\rm or, a(a - 3) - 9(a - 3) = 0$

$\rm or, (a - 9)(a - 3) = 0$

Either

$\rm (a - 9) = 0$

$\rm or, a = 9$

By our supposition earlier, $\rm a = 3^{x}$, we get,

$\rm or, 3^{x} = 3^{2}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x =2$

Or

$\rm (a - 3) = 0$

$\rm or, a = 3$

By our supposition earlier, $\rm a = 3^{x}$, we get,

$\rm or, 3^{x} = 3^{1}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 1$

Hence, the required values of x are x = {1,2}.

Given

$\rm 2^{x} + 2^{x + 1} + 2^{x + 2} + 2^{x + 4} = 62$

By using the law of indices, $\rm a^{m + n} = a^{m} \cdot a^{n}$, we get,

$\rm or, 2^{x} + 2^{x} \cdot 2^{1} + 2^{x} \cdot 2^{2} + 2^{x} \cdot 2^{4} = 62$

$\rm or, 2^{x} \cdot ( 1 + 2 + 2^{2} + 2^{4} ) = 62$

$\rm or, 2^{x} \cdot ( 1 + 2 + 4 + 16 ) = 62$

$\rm or, 2^{x} \cdot 23 = 62$

Dividing both sides of the equation by 23, we get,

$\rm or, 2^{x} \cdot \frac{23}{23} = \frac{62}{23}$

$\rm or, 2^{x} = 2$

$\rm or, 2^{x} = 2^{1}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 1$

Hence, the required value of x is 1.

Given

$\rm 3^{x} + 3^{x + 1} + 3^{x + 2} + 3^{x + 4} = 846$

By the law of indices, $\rm a^{m + n} = a^{m} \cdot a^{n}$, we get,

$\rm or, 3^{x} + 3^{x} \cdot 3^{1} + 3^{x} \cdot 3^{2} + 3^{x} \cdot 3^{4} = 846$

$\rm or, 3^{x} \left ( 1 + 3^{1} + 3^{2} + 3^{4} \right ) = 846$

$\rm or, 3^{x} \left ( 1 + 3 + 9 + 81 \right ) = 846$

$\rm or, 3^{x} \left ( 94 \right ) = 846$

Dividing both sides of the equation by 94, we get,

$\rm or, 3^{x} \frac{94}{94} = \frac{846}{94}$

$\rm or, 3^{x} = 9$

$\rm or, 3^{x} = 3^{2}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 2$

Hence, the required value of x is 2.

Given

$\rm 2^{x - 3} \cdot 2 \cdot a^{1 - x} = 2^{3x - 5} \cdot a^{x - 2}$

By the law of indices, $\rm b^{m - n} = \frac{b^{m}}{b^{n}}$, we get,

$\rm or, \frac{2^{x}}{2^{3}} \cdot 2 \cdot \frac{a^{1}}{a^{x}} = \frac{2^{3x}}{2^{5}} \cdot \frac{a^{x}}{a^{2}}$

$\rm or, \frac{ 2^{x} \cdot 2 \cdot a}{2^{3} \cdot a^{x}} = \frac{2^{3x} \cdot a^{x}}{2^{5} \cdot a^{2}}$

Placing all the terms with variable power on the left-hand side and the remaining terms on the right-hand side of the equation, we get,

$\rm or, \frac{2 ^{x}}{2^{3x} \cdot a^{x} \cdot a^{x}} = \frac{2^{3}}{2 \cdot a\cdot 2^{5} \cdot a^{2}}$

By using the law of indices, $\rm b^{m} \cdot b^{n} = b^{m + n}$, we get,

$\rm or, \frac{2^{x}}{2^{3x} \cdot a^{x + x}} = \frac{2^{3}}{2^{1 + 5} \cdot a^{1 + 2}}$

$\rm or, \frac{2^{x}}{2^{3x} \cdot a^{2x}} = \frac{2^{3}}{2^{6} \cdot a^{3}}$

By using the law of indices, $\rm \frac{b^{m}}{b^{n}} = b^{m - n}$, we get,

$\rm or, \frac{2^{x - 3x}}{a^{2x}} = \frac{2^{3 - 6}}{a^{3}}$

$\rm or, \frac{2^{-2x}}{a^{2x}} = \frac{2^{-3}}{a^{3}}$

By using the law of indices, $\rm a^{-m} = \frac{1}{a^{m}$, we get,

$\rm or, \frac{1}{2^{2x}} \cdot \frac{1}{a^{2x} = \frac{1}{2^{3}} \cdot \frac{1}{a^{3}}$

$\rm or, \frac{1}{2^{2x} \cdot a^{2x}} = \frac{1}{2^{3} \cdot a^{3}}$

By using the law of indices, $\rm b^{m} \cdot c^{m} = (bc)^{m}$, we get,

$\rm or, \frac{1}{ (2a)^{2x}} = \frac{1}{ (2a)^{3}}$

$\rm or, (2a)^{3} = (2a)^{2x}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm or, 3 = 2x$

Dividing both sides of the equation by 2 to free x, we get,

$\rm or, \frac{3}{2} = \frac{2}{2} \cdot x$

$\rm \therefore x = \frac{3}{2}$

Hence, the required value of x is $\rm \frac{3}{2}$.

Given

$\rm 2^{4x + 5} \cdot a^{1 - x} = 4^{x + 3} \cdot 2 \cdot a ^{x -1}$

$\rm or, 2^{4x + 5} \cdot a^{1 - x} = (2^{2})^{x + 3} \cdot 2^{1} \cdot a^{x - 1}$

By the law of indices $\rm (a^{m})^{n} = a^{mn}$, we get,

$\rm or, 2^{4x + 5} \cdot a^{1 - x} = 2^{2 (x + 3)} \cdot 2^{1} \cdot a^{x - 1}$

$\rm or, 2^{4x + 5} \cdot a^{1 - x} = 2^{2x + 6} \cdot 2^{1} \cdot a^{x - 1}$

By the law of indices $\rm a^{m} \cdot a^{n} = a^{m + n}$, we get,

$\rm or, 2^{4x + 5} \cdot a^{1 - x} = { 2^{(2x + 6) + 1} }\cdot a^{x - 1}$

$\rm or, 2^{4x + 5} \cdot a^{1 - x} = 2^{2x + 7} \cdot a^{ x- 1}$

Due to the equality sign, the quantity on the left-hand side of the equation must be equal to the quantity on the right-hand side of the equation. Additionally, each of the two terms on both sides of the equation has the same base, so their exponents must be the same to hold equality. Hence, we equate the exponents of 2 on each side, and we get,

$\rm {4x + 5} = {2x + 7}$

$\rm or, 4x - 2x + 5 = 2x - 2x + 7$

$\rm or, 2x + 5 = 7$

$\rm or, 2x + 5 - 5 = 7 - 5$

$\rm or, 2x = 2$

$\rm \therefore x = 1$

Similarly, when we equate the exponents of a on each side, we get,

$\rm {1 - x} = {x - 1}$

$\rm or, 1 - x = x - 1$

$\rm or, 1 + (1 - x) = 1 + (x - 1)$

$\rm or, 2 -x = x$

$\rm or, 2 - x + x = x + x$

$\rm or, 2  = 2x$

$\rm \therefore x = 1$

Hence, the required value of x is 1.

First, we solve the equation $\rm 3^{x-3} + 3^{4-x} = 4$ to find the possible values of x.

Given

$\rm 3^{x - 3} + 3^{4 - x} = 4$

By using the law of indices, $\rm a^{m-n} = \frac{a^{m}}{a^{n}}$,

$\rm or, \frac{3^{x}}{3^{3}} + \frac{3^{4}}{3^{x}} = 4$

$\rm or, \frac{3^{x}}{27} + \frac{81}{3^{x}} = 4$

Let $\rm 3^{x} = a$

$\rm or, \frac{a}{27} + \frac{81}{a} = 4$

Multiplying both sides of the equation by 27a, we get,

$\rm or, 27 a \cdot \frac{a}{27} + 27 a \cdot \frac{81}{a} = 27 a \cdot 4$

$\rm or, a^{2} + 2187 = 108 a$

$\rm or, a^{2} - 108 a + 2187 = 0$

The above equation is quadratic in a. We use the mid-term factorization method to solve it.

$\rm or, a^{2} - (27 + 81) a + 2187 = 0$

$\rm or, a^{2} - 27a - 81a + 2187 = 0$

$\rm or, a ( a - 27) - 81 ( a - 27) = 0$

$\rm or, (a - 81)(a - 27) = 0$

Either

$\rm (a - 81) = 0$

$\rm or, a = 81$

$\rm or, a = 3^{4}$

By supposition, $\rm a = 3^{x}$. We substitute the value for a and we get,

$\rm or, 3^{x} = 3^{4}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 4$

Or

$\rm (a - 27) = 0$

$\rm or, a = 27$

$\rm or, a = 3^{3}$

By supposition, $\rm a = 3^{x}$. We substitute the value for a and we get,

$\rm or, 3^{x} = 3^{3}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 3$

Hence, the possible values of x are x = {3,4}.

Finally, we substitute x = {3,4} in the second equation $\rm 4^{x-4} + 4^{3 - x} = 1 \frac{1}{4}$ to prove the condition.

Put x = 3

$\rm 4^{3 - 4} + 4^{3 - 3} = 1 \frac{1}{4}$

$\rm or, 4^{-1} + 4^{0} = \frac{4 \cdot 1 + 1}{4}$

By the law of indices $\rm a^{-m} = \frac{1}{a^{m}$ and $\rm a^{0} = 1$

$\rm or, \frac{1}{4} + 1 = \frac{4 + 1}{4}$

$\rm or, \frac{1 + 4}{4} = \frac{5}{4}$

$\rm \therefore \frac{5}{4} = \frac{5}{4}$ which is true.

Put x = 4

$\rm 4^{4 - 4} + 4^{3 - 4} = 1 \frac{1}{4}$

$\rm or, 4^{0} + 4^{-1} = \frac{ 4 \cdot 1 + 1}{4}$

By the law of indices $\rm a^{-m} = \frac{1}{a^{m}$ and $\rm a^{0} = 1$

$\rm or, 1 + \frac{1}{4} = \frac{4 + 1}{4}$

$\rm or,  \frac{ 4 + 1}{4} = \frac{5}{4}$

$\rm \therefore \frac{5}{4} = \frac{5}{4}$ which is true.

Hence, we conclude that the values of x in the equation $\rm 3^{x - 3} + 3^{4-x}$ also satisfy the equation $\rm 4^{x - 4} + 4^{3 - x} = 1 \frac{1}{4}$.