Algebra - Exponential Equations

Solve the equation: $\;$ $\sqrt{7^{2x^2 - 5x - 6}} = \left(\sqrt{2}\right)^{3 \log_2 {49}}$

Given equation: $\;\;$ $\sqrt{7^{2x^2 - 5x - 6}} = \left(\sqrt{2}\right)^{3 \log_2 {49}}$

i.e. $\;$ $\left[\left(7\right)^{2x^2 - 5x - 6}\right]^{\frac{1}{2}} = \left[\left(2\right)^{\frac{1}{2}}\right]^{3 \log_2 {7^2}}$

i.e. $\;$ $7^{\frac{2x^2 - 5x - 6}{2}} = 2^{\frac{1}{2} \times 3 \times 2 \log_2 7}$

i.e. $\;$ $7^{\frac{2x^2 - 5x - 6}{2}} = 2^{3 \log_2 7}$

i.e. $\;$ $7^{\frac{2x^2 - 5x - 6}{2}} = 2^{\log_2 {7^3}}$

i.e. $\;$ $7^{\frac{2x^2 - 5x - 6}{2}} = 7^3$

$\implies$ $\dfrac{2x^2 - 5x - 6}{2} = 3$

i.e. $\;$ $2x^2 - 5x - 12 = 0$

i.e. $\;$ $\left(2x + 3\right) \left(x - 4\right) = 0$

$\implies$ $x = \dfrac{-3}{2}$ $\;$ or $\;$ $x = 4$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{-3}{2}, \; 4 \right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $4^{\frac{1}{x} - 2} = \dfrac{\log \sqrt{10}}{2}$

Given equation: $\;\;$ $4^{\frac{1}{x} - 2} = \dfrac{\log \sqrt{10}}{2}$

i.e. $\;$ $4^{\frac{1}{x} - 2} = \dfrac{1}{2} \times \log_{10} {10}^{\frac{1}{2}}$

i.e. $\;$ $4^{\frac{1}{x} - 2} = \dfrac{1}{2} \times \dfrac{1}{2} \times \log_{10} 10$

i.e. $\;$ $4^{\frac{1}{x} - 2} = \dfrac{1}{4} \times 1$

i.e. $\;$ $4^{\frac{1}{x} - 2} \times 4^1 = 1$

i.e. $\;$ $4^{\frac{1}{x} - 1} = 4^0$

$\implies$ $\dfrac{1}{x} - 1 = 0$

i.e. $\;$ $\dfrac{1}{x} = 1$ $\implies$ $x = 1$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{1 \right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $3^{2x+5} = 3^{x+2} + 2$

Given equation: $\;\;$ $3^{2x+5} = 3^{x+2} + 2$

i.e. $\;$ $3^{2x} \times 3^5 - 3^x \times 3^2 - 2 = 0$

i.e. $\;$ $243 \times \left(3^x\right)^2 - 9 \times 3^x - 2 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $3^x = p$ $\;\;\; \cdots \; (2)$

Then equation $(1)$ becomes

$243 p^2 - 9p -2 = 0$

i.e. $\;$ $\left(27p + 2\right) \left(9p - 1\right) = 0$

i.e. $\;$ $p = \dfrac{-2}{27}$ $\;$ or $\;$ $p = \dfrac{1}{9}$

Substituing the value of $p$ in equation $(2)$ gives

when $\;$ $p = \dfrac{-2}{27}$, $\;$ then $\;$ $3^x = \dfrac{-2}{27}$

$\implies$ $x = \log_3 \left(\dfrac{-2}{27}\right)$

But logarithm of a negative number is not defined.

$\therefore \;$ $p = \dfrac{-2}{27}$ $\;$ is not a valid solution.

when $\;$ $p = \dfrac{1}{9}$, $\;$ then $\;$ $3^x = \dfrac{1}{9} = 3^{-2}$ $\implies$ $x = -2$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-2 \right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $4^{x + \sqrt{x^2 - 2}} - 5 \times 2^{x - 1 + \sqrt{x^2 - 2}} = 6$

Given equation: $\;\;$ $4^{x + \sqrt{x^2 - 2}} - 5 \times 2^{x - 1 + \sqrt{x^2 - 2}} = 6$

i.e. $\;$ $\left(2^2\right)^{x + \sqrt{x^2 - 2}} - 5 \times 2^{x + \sqrt{x^2 - 2}} \times 2^{-1} = 6$

i.e. $\;$ $\left(2^{x + \sqrt{x - 2}}\right)^2 - \dfrac{5}{2} \times 2^{x + \sqrt{x^2 - 2}} = 6$ $\;\;\; \cdots \; (1)$

Let $\;$ $2^{x + \sqrt{x - 2}} = p$ $\;\;\; \cdots \; (2)$

Then equation $(1)$ becomes

$p^2 - \dfrac{5}{2} p = 6$

i.e. $\;$ $2p^2 - 5p - 12 = 0$

i.e. $\;$ $\left(p - 4\right) \left(2p + 3\right) = 0$

i.e. $\;$ $p = 4$ $\;$ or $\;$ $p = \dfrac{-3}{2}$

Substituting the values of $p$ in equation $(2)$ gives

when $\;$ $p = 4$, $\;$ $2^{x + \sqrt{x - 2}} = 4$

i.e. $\;$ $2^{x + \sqrt{x - 2}} = 2^2$

$\implies$ $x + \sqrt{x - 2} = 2$

i.e. $\;$ $\sqrt{x - 2} = 2 - x$

i.e. $\;$ $x - 2 = \left(2 - x\right)^2$

i.e. $\;$ $x - 2 = 4 - 4x + x^2$

i.e. $\;$ $x^2 - 5x + 6 = 0$

i.e. $\;$ $\left(x - 3\right) \left(x - 2\right) = 0$

i.e. $\;$ $x = 3$ $\;$ or $\;$ $x = 2$

when $\;$ $p = \dfrac{-3}{2}$, $\;$ $2^{x + \sqrt{x - 2}} = \dfrac{-3}{2}$

i.e. $\;$ $x + \sqrt{x - 2} = \log \left(\dfrac{-3}{2}\right)$

But logarithm of a negative number is not defined.

$\therefore \;$ $p = \dfrac{-3}{2}$ $\;$ is not a valid solution.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2, \; 3 \right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $x^{\left(5 \sin 3x\right) + 2} = \dfrac{1}{\sqrt{x}}$, $\;$ for $\;$ $x \in \left[0, 2 \pi\right]$

Given equation: $\;\;$ $x^{\left(5 \sin 3x\right) + 2} = \dfrac{1}{\sqrt{x}}$, $\;$ for $\;$ $x \in \left[0, 2 \pi\right]$

i.e. $\;$ $x^{\left(5 \sin 3x\right) + 2} = x^{\frac{-1}{2}}$

i.e. $\;$ $x^{\left(5 \sin 3x\right) + 2 + \frac{1}{2}} = 1$

i.e. $\;$ $x^{\left(5 \sin 3x\right) + \frac{5}{2}} = x^0$

$\implies$ $5 \sin 3x + \dfrac{5}{2} = 0$

i.e. $\;$ $\sin 3x = \dfrac{-1}{2}$

i.e. $\;$ $3x = 2 n \pi + \pi + \dfrac{\pi}{6}$; $\;$ $3x = 2n \pi + 2 \pi - \dfrac{\pi}{6}$

i.e. $\;$ $3x = 2 n \pi + \dfrac{7 \pi}{6}$; $\;$ $2 n \pi + \dfrac{11 \pi}{6}$

i.e. $\;$ $x = \dfrac{2n \pi}{3} + \dfrac{7 \pi}{18}$; $\;$ $x = \dfrac{2 n \pi}{3} + \dfrac{11 \pi}{18}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{2n \pi}{3} + \dfrac{7 \pi}{18}, \; \dfrac{2 n \pi}{3} + \dfrac{11 \pi}{18} \right\}$, $\;$ $x \in \left[0, 2\pi\right]$

Algebra - Exponential Equations

Solve the equation: $\;$ $15 \times 2^{x+1} + 15 \times 2^{-x+2} = 135$

Given equation: $\;\;$ $15 \times 2^{x+1} + 15 \times 2^{-x+2} = 135$

i.e. $\;$ $15 \times 2^x \times 2^1 + 15 \times 2^{-x} \times 2^2 = 135$

i.e. $\;$ $30 \times 2^x + \dfrac{60}{2^x} = 135$

i.e. $\;$ $2 \times 2^x + \dfrac{4}{2^x} = 9$ $\;\;\; \cdots \; (1)$

Let $\;$ $2^x = p$ $\;\;\; \cdots \; (2)$

Then equation $(1)$ becomes

$2p + \dfrac{4}{p} = 9$

i.e. $\;$ $2 p^2 - 9p + 4 = 0$

i.e. $\;$ $\left(p - 4\right) \left(2p - 1\right) = 0$

i.e. $\;$ $p = 4$ $\;$ or $\;$ $p = \dfrac{1}{2}$

When $\;$ $p = 4$, $\;$ we have from equation $(2)$,

$2^x = 4 = 2^2$ $\implies$ $x = 2$

When $\;$ $p = \dfrac{1}{2}$, $\;$ we have from equation $(2)$,

$2^x = \dfrac{1}{2} = 2^{-1}$ $\implies$ $x = -1$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-1, \; 2 \right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $7^{\log x} - 5^{\left(\log x\right) + 1} = 3 \times 5^{\left(\log x\right) - 1} - 13 \times 7^{\left(\log x\right) - 1}$

Given equation: $\;\;$ $7^{\log x} - 5^{\left(\log x\right) + 1} = 3 \times 5^{\left(\log x\right) - 1} - 13 \times 7^{\left(\log x\right) - 1}$

i.e. $\;$ $7^{\log x} - 5^{\log x} \times 5^1 = 3 \times 5^{\log x} \times 5^{-1} - 13 \times 7^{\log x} \times 7^{-1}$

i.e. $\;$ $7^{\log x} \left(1 + \dfrac{13}{7}\right) = 5^{\log x} \left(\dfrac{3}{5} + 5\right)$

i.e. $\;$ $7^{\log x} \times \dfrac{20}{7} = 5^{\log x} \times \dfrac{28}{5}$

i.e. $\;$ $\left(\dfrac{7}{5}\right)^{\log x} = \dfrac{28 \times 7}{20 \times 5} = \dfrac{49}{25} = \left(\dfrac{7}{5}\right)^2$

$\implies$ $\log_{10} x = 2$

i.e. $\;$ $x = 10^2 = 100$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{100 \right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $2^{\left(2 \log 4x\right) - 1} - 7^{\log 4x} = 7^{\left(\log 4x\right) - 1} - 3 \times 4^{\log 4x}$

Given equation: $\;\;$ $2^{\left(2 \log 4x\right) - 1} - 7^{\log 4x} = 7^{\left(\log 4x\right) - 1} - 3 \times 4^{\log 4x}$

i.e. $\;$ $\left(2^2\right)^{\log 4x} \times 2^{-1} - 7^{\log 4x} = 7^{\log 4x} \times 7^{-1} - 3 \times 4^{\log 4x}$

i.e. $\;$ $4^{\log 4x} + 3 \times 4^{\log 4x} = \dfrac{7^{\log 4x}}{7} + 7^{\log 4x}$

i.e. $\;$ $4^{\log 4x} \left(\dfrac{1}{2} + 3\right) = 7^{\log 4x} \left(\dfrac{1}{7} + 1\right)$

i.e. $\;$ $4^{\log 4x} \times \dfrac{7}{2} = 7^{\log 4x} \times \dfrac{8}{7}$

i.e. $\;$ $\left(\dfrac{4}{7}\right)^{\log 4x} = \dfrac{8 \times 2}{7 \times 7} = \left(\dfrac{4}{7}\right)^2$

$\implies$ $\log_{10} 4x = 2$

i.e. $\;$ $4x = 10^2 = 100$ $\implies$ $x = 25$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{25 \right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $3^{12x-1} - 9^{6x-1} - 27^{4x-1} + 81^{3x+1} = 2192$

Given equation: $\;\;$ $3^{12x-1} - 9^{6x-1} - 27^{4x-1} + 81^{3x+1} = 2192$

i.e. $\;$ $3^{12x-1} - \left(3^2\right)^{6x-1} - \left(3^3\right)^{4x-1} + \left(3^4\right)^{3x + 1} = 2192$

i.e. $\;$ $3^{12x-1} - 3^{12x-2} - 3^{12x-3} + 3^{12x+4} = 2192$

i.e. $\;$ $3^{12x} \left(3^{-1} - 3^{-2} - 3^{-3} + 3^4\right) = 2192$

i.e. $\;$ $3^{12x} \left(\dfrac{1}{3} - \dfrac{1}{3^2} - \dfrac{1}{3^3} + 3^4\right) = 2192$

i.e. $\;$ $3^{12x} \left(\dfrac{3^2 - 3 - 1}{3^3} + 3^4\right) = 2192$

i.e. $\;$ $3^{12x} \times \dfrac{2192}{27} = 2192$

i.e. $\;$ $3^{12x} = 27 = 3^3$

$\implies$ $12x = 3$

i.e. $\;$ $x = \dfrac{3}{12} = \dfrac{1}{4}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{4} \right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $\left(3^{x^2 - 7.2x + 3.9} - 9 \sqrt{3}\right) \log \left(7 - x\right) = 0$

Given equation: $\;\;$ $\left(3^{x^2 - 7.2x + 3.9} - 9 \sqrt{3}\right) \log \left(7 - x\right) = 0$ $\;\;\; \cdots \; (1)$

$\implies$ $3^{x^2 - 7.2x + 3.9} - 9 \sqrt{3} = 0$ $\;$ or $\;$ $\log \left(7 - x\right) = 0$

i.e. $\;$ $3^{x^2 - 7.2x + 3.9} = 9 \sqrt{3}$ $\;\;\; \cdots \; (2)$ $\;$ or $\;$ $\log \left(7 - x\right) = \log \left(1\right)$ $\;\;\; \cdots \; (3)$

Consider equation $(2)$.

We have, $\;$ $3^{x^2 - 7.2x + 3.9} = 3^2 \times 3^{\frac{1}{2}} = 3^{\frac{5}{2}}$

$\implies$ $x^2 - 7.2x + 3.9 = \dfrac{5}{2}$

i.e. $\;$ $x^2 - 7.2x + 1.4 = 0$

i.e. $\;$ $5x^2 - 36x + 7 = 0$

i.e. $\;$ $\left(x - 7\right) \left(5x - 1\right) = 0$

i.e. $\;$ $x = 7$ $\;$ or $\;$ $x = \dfrac{1}{5}$

When $\;$ $x = 7$, $\;$ equation $(1)$ will have the term $\;$ $\log \left(7 - 7\right) = \log 0$

But $\;$ $\log 0$ $\;$ is not defined.

Therefore, $x = 7$ is not a valid solution.

Consider equation $(3)$.

We have, $\;$ $7 - x = 1$ $\implies$ $x = 6$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{5}, \; 6\right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $9^x - 2^{x + 0.5} = 2^{x + 3.5} - 3^{2x - 1}$

Given equation: $\;\;$ $9^x - 2^{x + 0.5} = 2^{x + 3.5} - 3^{2x - 1}$

i.e. $\;$ $9^x + 3^{2x} \times 3^{-1} = 2^x \times 2^{3.5} + 2^x \times 2^{0.5}$

i.e. $\;$ $9^x + \left(3^2\right)^x \times \dfrac{1}{3} = 2^x \times 2^{\frac{7}{2}} + 2^x \times 2^{\frac{1}{2}}$

i.e. $\;$ $9^x + \dfrac{9^x}{3} = 2^x \times \left(2^7\right)^{\frac{1}{2}} + 2^x \times 2^{\frac{1}{2}}$

i.e. $\;$ $9^x \left(1 + \dfrac{1}{3}\right) = 2^x \times \left(128\right)^{\frac{1}{2}} + 2^x \times 2^{\frac{1}{2}}$

i.e. $\;$ $9^x \times \dfrac{4}{3} = 2^x \times 8 \times 2^{\frac{1}{2}} + 2^x \times 2^{\frac{1}{2}}$

i.e. $\;$ $9^x \times \dfrac{4}{3} = 2^x \times 2^{\frac{1}{2}} \times 9$

i.e. $\;$ $\left(\dfrac{9}{2}\right)^x = 9 \times 9^{\frac{1}{2}} \times \dfrac{1}{2^{2 - \frac{1}{2}}}$

i.e. $\;$ $\left(\dfrac{9}{2}\right)^x = \dfrac{9^{\frac{3}{2}}}{2^{\frac{3}{2}}} = \left(\dfrac{9}{2}\right)^{\frac{3}{2}}$

$\implies$ $x = \dfrac{3}{2}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{3}{2}\right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $2^{2+x} - 2^{2-x} = 15$

Given equation: $\;\;$ $2^{2+x} - 2^{2-x} = 15$

i.e. $\;$ $2^2 \times 2^x - 2^2 \times 2^{-x} = 15$

i.e. $\;$ $4 \times 2^x - \dfrac{4}{2^x} - 15 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $2^x = p$ $\;\;\; \cdots \; (2)$

Then equation $(1)$ becomes

$4p - \dfrac{4}{p} - 15 = 0$

i.e. $\;$ $4p^2 - 15p - 4 = 0$

i.e. $\;$ $\left(p - 4\right) \left(4p + 1\right) = 0$

i.e. $\;$ $p = 4$ $\;$ or $\;$ $p = \dfrac{-1}{4}$

Substituting the value of $p$ in equation $(2)$ gives

when $\;$ $p = 4$, $\;$ $2^x = 4$

i.e. $\;$ $2^x = 2^2$ $\implies$ $x = 2$

when $\;$ $p = \dfrac{-1}{4}$, $\;$ $2^x = \dfrac{-1}{4}$

i.e. $\;$ $2^x = -2^{-2}$ $\implies$ $x = \log_2 \left(-2^2\right)$

But logarithim of a negative number is not defined.

$\therefore \;$ $p = \dfrac{-1}{4}$ $\;$ is not a valid solution.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2\right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $64^{\frac{1}{x}} - 2^{3 + \frac{3}{x}} + 12 = 0$

Given equation: $\;\;$ $64^{\frac{1}{x}} - 2^{3 + \frac{3}{x}} + 12 = 0$

i.e. $\;$ $\left(8^2\right)^{\frac{1}{x}} - 2^3 \times \left(2^3\right)^{\frac{1}{x}} + 12 = 0$

i.e. $\;$ $\left(8^{\frac{1}{x}}\right)^2 - 8 \times 8^{\frac{1}{x}} + 12 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $8^{\frac{1}{x}} = p$ $\;\;\; \cdots \; (2)$

Then equation $(1)$ becomes

$p^2 - 8p + 12 = 0$

i.e. $\;$ $\left(p - 6\right) \left(p - 2\right) = 0$

i.e. $\;$ $p = 6$ $\;\;$ or $\;\;$ $p = 2$

Substituting the value of $p$ in equation $(2)$ gives

when $\;$ $p = 6$, $\;\;$ $8^{\frac{1}{x}} = 6$

i.e. $\;$ $\dfrac{1}{x} = \log_8 6$

i.e. $\;$ $x = \dfrac{1}{\log_8 6} = \log_6 8 = \log_6 2^3 = 3 \log_6 2$

when $\;$ $p = 2$, $\;$ $8^{\frac{1}{x}} = 2$

i.e. $\;$ $2^{\frac{3}{x}} = 2^1$ $\implies$ $\dfrac{3}{x} = 1$ $\implies$ $x = 3$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{3 \log_6 2, \; 3\right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $7 \times 3^{x+1} - 5^{x+2} = 3^{x+4} - 5^{x+3}$

Given equation: $\;\;$ $7 \times 3^{x+1} - 5^{x+2} = 3^{x+4} - 5^{x+3}$

i.e. $\;$ $7 \times 3^{x+1} - 3^{x+1} \times 3^3 = 5^{x+2} - 5^{x+2} \times 5^1$

i.e. $\;$ $3^{x+1} \left(7 - 27\right) = 5^{x+2} \left(1 - 5\right)$

i.e. $\;$ $3^{x+1} \times \left(-20\right) = 5^{x+1} \times 5 \times \left(-4\right)$

i.e. $\;$ $\left(\dfrac{3}{5}\right)^{x+1} = 1$

i.e. $\;$ $\left(\dfrac{3}{5}\right)^{x+1} = \left(\dfrac{3}{5}\right)^0$

$\implies$ $x + 1 = 0$ $\implies$ $x = -1$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = -1$

Algebra - Exponential Equations

Solve the equation: $\;$ $4^x - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$

Given equation: $\;\;$ $4^x - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$

i.e. $\;$ $2^{2x} + 2^{2x-1} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}$

i.e. $\;$ $2^{2x} \left(1 + 2^{-1}\right) = 3^x \left(3^{\frac{1}{2}} + 3^{\frac{-1}{2}}\right)$

i.e. $\;$ $2^{2x} \left(1 + \dfrac{1}{2}\right) = 3^x \left(3^{\frac{1}{2}} + \dfrac{1}{3^{\frac{1}{2}}}\right)$

i.e. $\;$ $2^{2x} \times \dfrac{3}{2} = 3^x \times \dfrac{4}{3^{\frac{1}{2}}}$

i.e. $\;$ $\left(\dfrac{4}{3}\right)^x = \dfrac{4}{3} \times \dfrac{2}{3^{\frac{1}{2}}}$

i.e. $\;$ $\left(\dfrac{4}{3}\right)^{x-1} = \left(\dfrac{4}{3}\right)^{\frac{1}{2}}$

$\implies$ $x - 1 = \dfrac{1}{2}$

i.e. $\;$ $x = \dfrac{3}{2}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \dfrac{3}{2}$

Algebra - Exponential Equations

Solve the equation: $\;$ $2^{2x-3} = 4^{x^2 - 3x -1}$

Given equation: $\;\;$ $2^{2x-3} = 4^{x^2 - 3x -1}$

i.e. $\;$ $2^{2x-3} = \left(2^2\right)^{x^2 - 3x -1}$

i.e. $\;$ $2^{2x - 3} = 2^{2x^2 - 6x - 2}$

$\implies$ $2x - 3 = 2x^2 - 6x - 2$

i.e. $\;$ $2x^2 - 8x + 1 = 0$

i.e. $\;$ $x = \dfrac{8 \pm \sqrt{64 - 8}}{4} = \dfrac{8 \pm \sqrt{56}}{4} = \dfrac{8 \pm 2 \sqrt{14}}{4}$

i.e. $\;$ $x = 2 \pm \dfrac{\sqrt{14}}{2} = 2 \pm \sqrt{\dfrac{7}{2}}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = 2 \pm \sqrt{\dfrac{7}{2}}$

Algebra - Exponential Equations

Solve the equation: $\;$ $2^{2x+1} - 33 \times 2^{x-1} + 4 = 0$

Given equation: $\;\;$ $2^{2x+1} - 33 \times 2^{x-1} + 4 = 0$

i.e. $\;$ $2^{2x} \times 2^1 - 33 \times 2^x \times 2^{-1} + 4 = 0$

i.e. $\;$ $2 \times \left(2^x\right)^2 - \dfrac{33}{2} \times 2^x + 4 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $2^x = p$ $\;\;\; \cdots \; (2)$

Then, in view of equation $(2)$, equation $(1)$ becomes

$2p^2 - \dfrac{33}{2}p + 4 = 0$

i.e. $\;$ $4p^2 - 33p + 8 = 0$

i.e. $\;$ $\left(4p - 1\right) \left(p - 8\right) = 0$

i.e. $\;$ $p = \dfrac{1}{4}$ $\;$ or $\;$ $p = 8$

Substituting the value of $p$ in equation $(2)$ gives

when $\;$ $p = \dfrac{1}{4}$, $\;$ $2^x = \dfrac{1}{4}$

i.e. $\;$ $2^x = 2^{-2}$ $\implies$ $x = -2$

when $\;$ $p = 8$, $\;$ $2^x = 8$

i.e. $\;$ $2^x = 2^3$ $\implies$ $x = 3$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = -2, \; 3$

Algebra - Exponential Equations

Solve the equation: $\;$ $3^{2x+1} + 10 \times 3^x + 3 = 0$

Given equation: $\;\;$ $3^{2x+1} + 10 \times 3^x + 3 = 0$

i.e. $\;$ $3^x \left(3^{x+1} + 10\right) = -3^1$

i.e. $\;$ $3^{x+1} + 10 = -3^{1-x}$

i.e. $\;$ $3^{x+1} + 3^{1-x} = -10$

i.e. $\;$ $3 \left(3^x + 3^{-x}\right) = -10$

i.e. $\;$ $3^x + \dfrac{1}{3^x} = \dfrac{-10}{3}$ $\;\;\; \cdots \; (1)$

Let $\;$ $3^x = p$ $\;\;\; \cdots \; (2)$

Then, equation $(1)$ becomes

$p + \dfrac{1}{p} = \dfrac{-10}{3}$

i.e. $\;$ $3p^2 + 10p + 3 = 0$

i.e. $\;$ $\left(3p + 1\right) \left(p + 3\right) = 0$

i.e. $\;$ $p = \dfrac{-1}{3}$ $\;\;$ or $\;\;$ $p = -3$

Substituting the values of $p$ in equation $(2)$ gives

When $\;$ $p = \dfrac{-1}{3}$ $\implies$ $3^x = \dfrac{-1}{3}$ $\implies$ $x = \log_3 \left(\dfrac{-1}{3}\right)$

But logarithm of a negative number is not defined.

$\implies$ $p = \dfrac{-1}{3}$ $\;$ is not a valid solution.

When $\;$ $p = -3$ $\implies$ $3^x = -3$ $\implies$ $x = \log_3 \left(-3\right)$

But logarithm of a negative number is not defined.

$\implies$ $p = -3$ $\;$ is not a valid solution.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \phi$ $\;$ (null)

Algebra - Exponential Equations

Solve the equation: $\;$ $7^{x+2} - \dfrac{1}{7} \times 7^{x+1} - 14 \times 7^{x-1} + 2 \times 7^x = 48$

Given equation: $\;\;$ $7^{x+2} - \dfrac{1}{7} \times 7^{x+1} - 14 \times 7^{x-1} + 2 \times 7^x = 48$

i.e. $\;$ $7^x \times 7^2 - 7^{-1} \times 7^x \times 7^1 - 2 \times 7 \times 7^x \times 7^{-1} + 2 \times 7^x = 48$

i.e. $\;$ $49 \times 7^x - 7^x - 2 \times 7^x + 2 \times 7^x = 48$

i.e. $\;$ $48 \times 7^x = 48$

i.e. $\;$ $7^x = 1 = 7^0$

$\implies$ $x = 0$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = 0$