1. $3^{x+2} = 81$
Solution: $3^{x+2} = 81$
or, $3^{x+2} = 3^4$
or, $x + 2 = 4$
or, $x = 4 – 2$
or, $x = 2$
Required solution:$x = 2$
2. $5^{3x-7} = 3^{3x-7}$
Solution: $5^{3x-7} = 3^{3x-7}$
or, $\frac{5^{3x-7}}{3^{3x-7}} = 1$
or, $\left(\frac{5}{3}\right)^{3x-7} = \left(\frac{5}{3}\right)^0$
or, $3x – 7 = 0$
or, $3x = 7$
or, $x = \frac{7}{3}$
Required solution:$x = \frac{7}{3}$
3. $2^{x-4} = 4a^{x-6} \quad (a > 0, a \neq 2)$
Solution: $2^{x-4} = 4a^{x-6}$
or, $2^{x-4} = 2^2 \cdot a^{x-6}$
or, $\frac{2^{x-4}}{2^2} = a^{x-6}$
or, $2^{x-4-2} = a^{x-6}$
or, $2^{x-6} = a^{x-6}$
or, $\frac{2^{x-6}}{a^{x-6}} = 1$
or, $\left(\frac{2}{a}\right)^{x-6} = \left(\frac{2}{a}\right)^0$
or, $x – 6 = 0$
or, $x = 6$
Required solution:$x = 6$
4. $(\sqrt{3})^{x+5} = (\sqrt[3]{3})^{2x+5}$
Solution: $(\sqrt{3})^{x+5} = (\sqrt[3]{3})^{2x+5}$
or, $\left(3^{\frac{1}{2}}\right)^{x+5} = \left(3^{\frac{1}{3}}\right)^{2x+5}$
or, $3^{\frac{x+5}{2}} = 3^{\frac{2x+5}{3}}$
or, $\frac{x+5}{2} = \frac{2x+5}{3}$
or, $3(x + 5) = 2(2x + 5)$
or, $3x + 15 = 4x + 10$
or, $4x – 3x = 15 – 10$
or, $x = 5$
Required solution:$x = 5$
5. $(\sqrt[5]{4})^{4x+7} = (\sqrt[11]{64})^{2x+7}$
Solution: $(\sqrt[5]{4})^{4x+7} = (\sqrt[11]{64})^{2x+7}$
or, $\left(4^{\frac{1}{5}}\right)^{4x+7} = \left(4^{\frac{3}{11}}\right)^{2x+7}$
or, $4^{\frac{4x+7}{5}} = 4^{\frac{3(2x+7)}{11}}$
or, $\frac{4x+7}{5} = \frac{6x+21}{11}$
or, $11(4x + 7) = 5(6x + 21)$
or, $44x + 77 = 30x + 105$
or, $44x – 30x = 105 – 77$
or, $14x = 28$
or, $x = \frac{28}{14}$
or, $x = 2$
Required solution:$x = 2$
6. $\frac{3^{3x-4} \cdot a^{2x-5}}{3^{x+1}} = a^{2x-5} \quad (a > 0)$
Solution: $\frac{3^{3x-4} \cdot a^{2x-5}}{3^{x+1}} = a^{2x-5}$
or, $\frac{3^{3x-4}}{3^{x+1}} \cdot \frac{a^{2x-5}}{a^{2x-5}} = 1$
or, $3^{(3x-4) – (x+1)} \cdot 1 = 1$
or, $3^{3x – 4 – x – 1} = 3^0$
or, $3^{2x – 5} = 3^0$
or, $2x – 5 = 0$
or, $2x = 5$
or, $x = \frac{5}{2}$
Required solution:$x = \frac{5}{2}$
7. $\frac{5^{2x} \cdot b^{x-3}}{5^{x+3}} = a^{x-3} \quad (a, b > 0, 5b \neq a)$
Solution: $\frac{5^{2x} \cdot b^{x-3}}{5^{x+3}} = a^{x-3}$
or, $5^{2x – (x+3)} \cdot b^{x-3} = a^{x-3}$
or, $5^{2x – x – 3} \cdot b^{x-3} = a^{x-3}$
or, $5^{x-3} \cdot b^{x-3} = a^{x-3}$
or, $(5b)^{x-3} = a^{x-3}$
or, $\frac{(5b)^{x-3}}{a^{x-3}} = 1$
or, $\left(\frac{5b}{a}\right)^{x-3} = \left(\frac{5b}{a}\right)^0$
or, $x – 3 = 0$
or, $x = 3$
Required solution:$x = 3$
8. $4^{x+2} = 2^{2x+1} + 14$
Solution: $4^{x+2} = 2^{2x+1} + 14$
or, $(2^2)^{x+2} = 2^{2x} \cdot 2^1 + 14$
or, $2^{2x+4} = 2 \cdot 2^{2x} + 14$
or, $2^{2x} \cdot 2^4 = 2 \cdot 2^{2x} + 14$
or, $16 \cdot 2^{2x} – 2 \cdot 2^{2x} = 14$
or, $14 \cdot 2^{2x} = 14$
or, $2^{2x} = \frac{14}{14}$
or, $2^{2x} = 1$
or, $2^{2x} = 2^0$
or, $2x = 0$
or, $x = 0$
Required solution:$x = 0$
9. $5^x + 5^{2-x} = 26$
Solution: $5^x + 5^{2-x} = 26$
or, $5^x + \frac{5^2}{5^x} = 26$
or, $5^x + \frac{25}{5^x} = 26$
or, $a + \frac{25}{a} = 26$ [Let $5^x = a$]
or, $\frac{a^2 + 25}{a} = 26$
or, $a^2 + 25 = 26a$
or, $a^2 – 26a + 25 = 0$
or, $a^2 – 25a – a + 25 = 0$
or, $a(a – 25) – 1(a – 25) = 0$
or, $(a – 25)(a – 1) = 0$
Either,
$a – 25 = 0$
or, $a = 25$
or, $5^x = 5^2$
or, $x = 2$
Or,
$a – 1 = 0$
or, $a = 1$
or, $5^x = 5^0$
or, $x = 0$
Required solution:$x = 0, 2$
10. $3(9^x – 4 \cdot 3^{x-1}) + 1 = 0$
Solution: $3(9^x – 4 \cdot 3^{x-1}) + 1 = 0$
or, $3 \cdot 9^x – 12 \cdot 3^{x-1} + 1 = 0$
or, $3 \cdot (3^2)^x – 12 \cdot \frac{3^x}{3} + 1 = 0$
or, $3 \cdot (3^x)^2 – 4 \cdot 3^x + 1 = 0$
or, $3a^2 – 4a + 1 = 0$ [Let $3^x = a$]
or, $3a^2 – 3a – a + 1 = 0$
or, $3a(a – 1) – 1(a – 1) = 0$
or, $(a – 1)(3a – 1) = 0$
Either,
$a – 1 = 0$
or, $a = 1$
or, $3^x = 3^0$
or, $x = 0$
Or,
$3a – 1 = 0$
or, $3a = 1$
or, $a = \frac{1}{3}$
or, $3^x = 3^{-1}$
or, $x = -1$
Required solution:$x = 0, -1$
11. $4^{1+x} + 4^{1-x} = 10$
Solution: $4^{1+x} + 4^{1-x} = 10$
or, $4 \cdot 4^x + \frac{4}{4^x} = 10$
or, $4a + \frac{4}{a} = 10$ [Let $4^x = a$]
or, $\frac{4a^2 + 4}{a} = 10$
or, $4a^2 + 4 = 10a$
or, $4a^2 – 10a + 4 = 0$
or, $2(2a^2 – 5a + 2) = 0$
or, $2a^2 – 5a + 2 = 0$
or, $2a^2 – 4a – a + 2 = 0$
or, $2a(a – 2) – 1(a – 2) = 0$
or, $(a – 2)(2a – 1) = 0$
Either,
$a – 2 = 0$
or, $a = 2$
or, $(2^2)^x = 2^1$
or, $2^{2x} = 2^1$
or, $2x = 1$
or, $x = \frac{1}{2}$
Or,
$2a – 1 = 0$
or, $2a = 1$
or, $a = \frac{1}{2}$
or, $(2^2)^x = 2^{-1}$
or, $2^{2x} = 2^{-1}$
or, $2x = -1$
or, $x = -\frac{1}{2}$
Required solution:$x = \pm \frac{1}{2}$
12. $2^{2x} – 3 \cdot 2^{x+2} = -32$
Solution: $2^{2x} – 3 \cdot 2^{x+2} = -32$
or, $(2^x)^2 – 3 \cdot 2^x \cdot 2^2 + 32 = 0$
or, $(2^x)^2 – 12 \cdot 2^x + 32 = 0$
or, $a^2 – 12a + 32 = 0$ [Let $2^x = a$]
or, $a^2 – 8a – 4a + 32 = 0$
or, $a(a – 8) – 4(a – 8) = 0$
or, $(a – 8)(a – 4) = 0$
Either,
$a – 8 = 0$
or, $a = 8$
or, $2^x = 2^3$
or, $x = 3$
Or,
$a – 4 = 0$
or, $a = 4$
or, $2^x = 2^2$
or, $x = 2$
Required solution:$x = 2, 3$