Algebra - Logarithmic Equations

Solve the equation: $\;$ $3^{\log_3 {\left(\log \sqrt{x}\right)}} - \log x + \log^2 x - 3 = 0$

Given equation: $\;\;$ $3^{\log_3 {\left(\log \sqrt{x}\right)}} - \log x + \log^2 x - 3 = 0$

i.e. $\;$ $\log \sqrt{x} - \log x + \log x \times \log x - 3 = 0$

i.e. $\;$ $\log x^{\frac{1}{2}} - \log x + \log x \times \log x - 3 = 0$

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

i.e. $\;$ $\log x^{\frac{-1}{2}} + \log x \times \log x - 3 = 0$

i.e. $\;$ $\log x \times \log x - \dfrac{1}{2} \log x - 3 = 0$

i.e. $\;$ $2 \times \log x \times \log x - \log x - 6 = 0$

i.e. $\;$ $\log x \left(2 \log x - 1\right) - 6 = 0$ $\;\;\; \cdots \; (1)$

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

Then equation $(1)$ becomes

$p \left(2p - 1\right) - 6 = 0$

i.e. $\;$ $2p^2 - p - 6 = 0$

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

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

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

when $\;$ $p = \dfrac{-3}{2}$, $\;$ $\log x = \dfrac{-3}{2}$

But logarithim of a number cannot be negative.

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

when $\;$ $p = 2$ $\;$ $\log x = 2$ $\implies$ $x = 10^2 = 100$

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log x - \dfrac{1}{2} \log \left(x - \dfrac{1}{2}\right) = \log \left(x + \dfrac{1}{2}\right) - \dfrac{1}{2} \left(x + \dfrac{1}{8}\right)$

Given equation: $\;\;$ $\log x - \dfrac{1}{2} \log \left(x - \dfrac{1}{2}\right) = \log \left(x + \dfrac{1}{2}\right) - \dfrac{1}{2} \left(x + \dfrac{1}{8}\right)$

i.e. $\;$ $\log x - \log \left(x + \dfrac{1}{2}\right) = \dfrac{1}{2} \log \left(x - \dfrac{1}{2}\right) - \dfrac{1}{2} \log \left(x + \dfrac{1}{8}\right)$

i.e. $\;$ $\log \left(\dfrac{x}{x + \frac{1}{2}}\right) = \dfrac{1}{2} \log \left(\dfrac{x - \frac{1}{2}}{x + \frac{1}{8}}\right)$

i.e. $\;$ $2 \log \left(\dfrac{2x}{2x + 1}\right) = \log \left(\dfrac{\frac{2x - 1}{2}}{\frac{8x + 1}{8}}\right)$

i.e. $\;$ $2 \log \left(\dfrac{2x}{2xx + 1}\right) = \log \left[\dfrac{4 \left(2x - 1\right)}{8x + 1}\right]$

i.e. $\;$ $\log \left(\dfrac{2x}{2x + 1}\right)^2 = \log \left(\dfrac{8x - 4}{8x + 1}\right)$

i.e. $\;$ $\dfrac{4x^2}{4x^2 + 4x + 1} = \dfrac{8x - 4}{8x + 1}$

i.e. $\;$ $32 x^3 + 4x^2 = 32 x^3 + 32 x^2 + 8x - 16 x^2 - 16x - 4$

i.e. $\;$ $12 x^2 - 8x -4 = 0$

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

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

When $\;$ $x = \dfrac{-1}{3}$, $\;$ there are terms like $\;$ $\log \left(\dfrac{-1}{3}\right)$ $\;$ in the given problem.

But logarithim of a negative number is not defined.

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

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $1 - \log 5 = \dfrac{1}{3} \left(\log \dfrac{1}{2} + \log x + \dfrac{1}{3} \log 5\right)$

Given equation: $\;\;$ $1 - \log 5 = \dfrac{1}{3} \left(\log \dfrac{1}{2} + \log x + \dfrac{1}{3} \log 5\right)$

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

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

i.e. $\;$ $1 = \log \left(\dfrac{5^{\frac{1}{9}} \times x^{\frac{1}{3}}}{2^{\frac{1}{3}}}\right) + \log 5$

i.e. $\;$ $1 = \log \left(\dfrac{5^{\frac{1}{9}} \times x^{\frac{1}{3}} \times 5}{2^{\frac{1}{3}}}\right)$

i.e. $\;$ $\dfrac{5^{\frac{10}{9}} \times x^{\frac{1}{3}}}{2^{\frac{1}{3}}} = 10^1 = 10$

i.e. $\;$ $\dfrac{5^{\frac{10}{9}} \times x^{\frac{1}{3}}}{2^{\frac{1}{3}}} = 5 \times 2$

i.e. $\;$ $x^{\frac{1}{3}} = \dfrac{5 \times 2 \times 2^{\frac{1}{3}}}{5^{\frac{10}{9}}}$

i.e. $\;$ $x^{\frac{1}{3}} = 5^{1 - \frac{10}{9}} \times 2^{1 + \frac{1}{3}}$

i.e. $\;$ $x^{\frac{1}{3}} = 5^{\frac{-1}{9}} \times 2^{\frac{4}{3}}$

i.e. $\;$ $x^{\frac{1}{3}} = \left(5^{\frac{-1}{3}} \times 2^4\right)^{\frac{1}{3}}$

i.e. $\;$ $x = 2^4 \times 5^{\frac{-1}{3}}$

i.e. $\;$ $x = \dfrac{16}{\sqrt[3]{5}}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{16}{\sqrt[3]{5}} \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log 5 + \log \left(x + 10\right) - 1 = \log \left(21x - 20\right) - \log \left(2x - 1\right)$

Given equation: $\;\;$ $\log 5 + \log \left(x + 10\right) - 1 = \log \left(21x - 20\right) - \log \left(2x - 1\right)$

i.e. $\;$ $\log_{10} 5 + \log_{10} \left(x + 10\right) - \log_{10} \left(21x - 20\right) + \log_{10} \left(2x - 1\right) = 1$

i.e. $\;$ $\log_{10} \left[\dfrac{5 \left(x + 10\right) \left(2x - 1\right)}{21x - 20}\right] = \log_{10} 10$

i.e. $\;$ $\dfrac{5 \left(2x^2 + 19x - 10\right)}{21x - 20} = 10$

i.e. $\;$ $2x^2 + 19x - 10 = 2 \left(21x - 20\right)$

i.e. $\;$ $2x^2 + 19x - 10 = 42x - 40$

i.e. $\;$ $2x^2 - 23x + 30 = 0$

i.e. $\;$ $\left(x - 10\right) \left(2x - 3\right) = 0$

i.e. $\;$ $x = 10$ $\;$ or $\;$ $x = \dfrac{3}{2}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{3}{2}, \; 10 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_7 \left(2^x - 1\right) + \log_7 \left(2^x - 7\right) = 1$

Given equation: $\;\;$ $\log_7 \left(2^x - 1\right) + \log_7 \left(2^x - 7\right) = 1$

i.e. $\;$ $\log_7 \left[\left(2^x - 1\right) \left(2^x - 7\right)\right] = 1$

$\implies$ By definition, $\;$ $2^{2x} - 8 \times 2^x + 7 = 7^1 = 7$

i.e. $\;$ $2^{2x} - 8 \times 2^x = 0$

i.e. $\;$ $2^x \left(2^x - 8\right) = 0$

i.e. $\;$ $2^x = 0$ $\;$ or $\;$ $2^x = 8$

i.e. $\;$ $x = \log_2 \left(0\right)$ $\;$ or $\;$ $2^x = 2^3$

But $\;$ $\log_2 \left(0\right)$ $\;$ is not defined and is therefore not a valid solution.

Now, $\;$ $2^x = 2^3$ $\implies$ $x = 3$

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_3 \left(x + 1\right) + \log_3 \left(x + 3\right) = 1$

Given equation: $\;\;$ $\log_3 \left(x + 1\right) + \log_3 \left(x + 3\right) = 1$

i.e. $\;$ $\log_3 \left[\left(x + 1\right) \left(x + 3\right)\right] = 1$

$\implies$ By definition, $\;$ $x^2 + 4x + 3 = 3^1 = 3$

i.e. $\;$ $x^2 + 4x = 0$

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

i.e. $\;$ $x = 0$, $\;$ or $\;$ $x = -4$

When $\;$ $x = -4$, $\;$ the Left Hand Side of the given problem becomes

$\log_3 \left(-4 + 1\right) + \log_3 \left(-4 + 3\right) = \log_3 \left(-3\right) + \log_3 \left(-1\right)$

But logarithm of a negative number is not defined.

$\therefore \;$ $x = -4$ is not a valid solution.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_3 \left(\log_9 x + \dfrac{1}{2} + 9^x\right) = 2x$

Given equation: $\;\;$ $\log_3 \left(\log_9 x + \dfrac{1}{2} + 9^x\right) = 2x$

$\implies$ By definition, $\;$ $\log_9 x + \dfrac{1}{2} + 9^x = 3^{2x}$

i.e. $\;$ $\log_9 x + \dfrac{1}{2} + 9^x = \left(3^2\right)^x = 9^x$

i.e. $\;$ $\log_9 x + \dfrac{1}{2} = 0$

i.e. $\;$ $\log_9 x = \dfrac{-1}{2}$

$\implies$ By definition, $\;$ $x = \left(9\right)^{\frac{-1}{2}} = \dfrac{1}{9^{\frac{1}{2}}} = \dfrac{1}{3}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{3} \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_{5-x} \left(x^2 - 2x + 65\right) = 2$

Given equation: $\;\;$ $\log_{5-x} \left(x^2 - 2x + 65\right) = 2$

i.e. $\;$ $x^2 - 2x + 65 = \left(5 - x\right)^2$

i.e. $\;$ $x^2 - 2x + 65 = 25 - 10x + x^2$

i.e. $\;$ $8x = -40$ $\implies$ $x = -5$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-5 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\dfrac{\log_2 \left(9 - 2^x\right)}{3 - x} = 1$

Given equation: $\;\;$ $\dfrac{\log_2 \left(9 - 2^x\right)}{3 - x} = 1$

i.e. $\;$ $\log_2 \left(9 - 2^x\right) = 3 - x$

i.e. $\;$ $9 - 2^x = 2^{3 - x}$

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

i.e. $\;$ $9 \times 2^x - 2^{2x} = 8$

i.e. $\;$ $\left(2^x\right)^2 - 9 \times 2^x + 8 = 0$ $\;\;\; \cdots \; (1)$

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

Then equation $(1)$ becomes

$p^2 - 9p + 8 = 0$

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

i.e. $\;$ $p = 1$ $\;$ or $\;$ $p = 8$

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

when $\;$ $p = 1$,

$2^x = 1$ $\implies$ $x = \log_2 1 = 0$

when $\;$ $p = 8$,

$2^x = 8 = 2^3$ $\implies$ $x = 3$

But when $\;$ $x = 3$, $\;$ the Left Hand Side of the given problem becomes

$\dfrac{\log_2 \left(9 - 2^3\right)}{3 - 3} = \dfrac{\log_2 1}{0} = \dfrac{0}{0}$

$\implies$ $x = 3$ is not a valid solution.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_3 \left(3^x - 8\right) = 2 - x$

Given equation: $\;\;$ $\log_3 \left(3^x - 8\right) = 2 - x$

i.e. $\;$ $3^x - 8 = 3^{2 - x}$

i.e. $\;$ $3^x - 8 = 3^2 \times 3^{-x}$

i.e. $\;$ $3^{2x} - 8 \times 3^x - 9 = 0$ $\;\;\; \cdots \; (1)$

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

Then equation $(1)$ becomes

i.e. $\;$ $p^2 - 8p - 9 = 0$

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

i.e. $\;$ $p = -1$ $\;$ or $\;$ $p = 9$

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

when $\;$ $p = -1$

$3^x = -1$ $\implies$ $x = \log_3 \left(-1\right)$

But logarithim of a negative number is not defined.

$\therefore \;$ $p = -1$ $\;$ is not a valid solution.

when $\;$ $p = 9$

$3^x = 9 = 3^2$ $\implies$ $x = 2$

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_3 \left[1 + \log_3 \left(2^x - 7\right)\right] = 1$

Given equation: $\;\;$ $\log_3 \left[1 + \log_3 \left(2^x - 7\right)\right] = 1$

i.e. $\;$ $1 + \log_3 \left(2^x - 7\right) = 3^1 = 3$

i.e. $\;$ $\log_3 \left(2^x - 7\right) = 2$

i.e. $\;$ $2^x - 7 = 3^2 = 9$

i.e. $\;$ $2^x = 16 = 2^4$

$\implies$ $x = 4$

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_4 \left\{2 \log_3 \left[1 + \log_2 \left(1 + 3 \log_3 x\right)\right] \right\} = \dfrac{1}{2}$

Given equation: $\;\;$ $\log_4 \left\{2 \log_3 \left[1 + \log_2 \left(1 + 3 \log_3 x\right)\right] \right\} = \dfrac{1}{2}$

i.e. $\;$ $2 \log_3 \left[1 + \log_2 \left(1 + 3 \log_3 x\right)\right] = 4^{\frac{1}{2}} = 2$

i.e. $\;$ $\log_3 \left[1 + \log_2 \left(1 + 3 \log_3 x\right)\right] = 1$

i.e. $\;$ $1 + \log_2 \left(1 + 3 \log_3 x\right) = 3^1 = 3$

i.e. $\;$ $\log_2 \left(1 + 3 \log_3 x\right) = 2$

i.e. $\;$ $1 + 3 \log_3 x = 2^2 = 4$

i.e. $\;$ $3 \log_3 x = 3$

i.e. $\;$ $\log_3 x = 1$

i.e. $\;$ $x = 3^1 = 3$

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

Algebra - Logarithmic Equations

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

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

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

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

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

i.e. $\;$ $x = \dfrac{2 \pm \sqrt{4 + 8}}{2} = \dfrac{2 \pm \sqrt{12}}{2} = \dfrac{2 \pm 2 \sqrt{3}}{2}$

i.e. $\;$ $x = 1 \pm \sqrt{3}$

When $\;$ $x = 1 - \sqrt{3}$, $\;$ then we have

$\log_{1 - \sqrt{3} - 1} 3 = 2$ $\implies$ $\log_{-\sqrt{3}} 3 = 2$

But base of a logarithm cannot be negative.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{1 + \sqrt{3} \right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $\left[\left(2^{\sqrt{x} + 3}\right)^{\frac{1}{2 \sqrt{x}}}\right]^{\frac{2}{\sqrt{x} - 1}} = 2$

Given equation: $\;\;$ $\left[\left(2^{\sqrt{x} + 3}\right)^{\frac{1}{2 \sqrt{x}}}\right]^{\frac{2}{\sqrt{x} - 1}} = 2$

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

$\implies$ $\dfrac{\sqrt{x} + 3}{x - \sqrt{x}} = 1$

i.e. $\;$ $\sqrt{x} + 3 = x - \sqrt{x}$

i.e. $\;$ $x - 2 \sqrt{x} - 3 = 0$ $\;\;\; \cdots \; (1)$

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

Then equation $(1)$ becomes

$p^2 - 2p - 3 = 0$

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

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

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

when $\;$ $p = 3$, $\;$ $\sqrt{x} = 3$ $\implies$ $x = 9$

when $\;$ $p = -1$, $\;$ $\sqrt{x} = -1$

But square root of a number cannot be negative.

$\implies$ $p = -1$ $\;$ is not a valid solution.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{9 \right\}$

Algebra - Exponential Equations

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

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

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

i.e. $\;$ $3^{\log \left(\tan x\right)} - 6 \times 3^{\log \left(\tan x\right)^{-1}} = 1$

i.e. $\;$ $3^{\log \left(\tan x\right)} 6 \times 3^{- \log \left(\tan x\right)} = 1$

i.e. $\;$ $3^{\log \left(\tan x\right)} - \dfrac{6}{3^{\log \left(\tan x\right)}} = 1$ $\;\;\; \cdots \; (1)$

Let $\;$ $3^{\log \left(\tan x\right)} = p$ $\;\;\; \cdots \; (2)$

Then equation $(1)$ becomes

$p - \dfrac{6}{p} = 1$

i.e. $\;$ $p^2 - p - 6 = 0$

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

i.e. $\;$ $p = 3$ $\;$ or $\;$ $p = -2$

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

when $\;$ $p = 3$,

$3^{\log \left(\tan x\right)} = 3 = 3^1$

$\implies$ $\log \left(\tan x\right) = 1$

$\implies$ $\tan x = 10^1 = 10$

i.e. $\;$ $x = \tan^{-1} \left(10\right) + n \pi, \;\; n \in Z$

when $\;$ $p = -2$,

$3^{\log \left(\tan x\right)} = -2$

$\implies$ $\log \left(\tan x\right) = \log_3 \left(-2\right)$

But logarithm of a negative number is not defined.

$\implies$ $p = -2$ $\;$ is not a valid solution.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\tan^{-1} \left(10\right) + n \pi, \;\; n \in Z \right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $2 \times 3^{x + 1} - 6 \times 3^{x-1} - 3^x = 9$

Given equation: $\;\;$ $2 \times 3^{x + 1} - 6 \times 3^{x-1} - 3^x = 9$

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

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

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

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

$\implies$ $x + 1 = 2$ $\implies$ $x = 1$

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

Algebra - Exponential Equations

Solve the equation: $\;$ $3^{4x + 8} - 4 \times 3^{2x + 5} + 28 = 2 \log_2 \sqrt{2}$

Given equation: $\;\;$ $3^{4x + 8} - 4 \times 3^{2x + 5} + 28 = 2 \log_2 \sqrt{2}$

i.e. $\;$ $3^{4x} \times 3^8 - 4 \times 3^{2x} \times 3^5 + 28 = 2 \log_2 2^{\frac{1}{2}}$

i.e. $\;$ $\left(3^{2x}\right)^2 \times 6561 - 972 \times 3^{2x} + 28 = 2 \times \dfrac{1}{2} \log_2 2$

i.e. $\;$ $6561 \times \left(3^{2x}\right)^2 - 972 \times 3^{2x} + 28 = 1$

i.e. $\;$ $6561 \times \left(3^{2x}\right)^2 - 972 \times 3^{2x} + 27 = 0$

i.e. $\;$ $243 \times \left(3^{2x}\right)^2 - 36 \times 3^{2x} + 1 = 0$ $\;\;\; \cdots \; (1)$

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

Then equation $(1)$ becomes

$243 p^2 - 36 p + 1 = 0$

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

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

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

when $\;$ $p = \dfrac{1}{27}$, $\;$ $3^{2x} = \dfrac{1}{27} = 3^{-3}$

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

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

$\implies$ $2x = -2$ $\implies$ $x = -1$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-1, \; \dfrac{-3}{2} \right\}$

Algebra - Exponential Equations

Solve the equation: $\;$ $9^{x^2 - 1} - 36 \times 3^{x^2 - 3} + 3 = 0$

Given equation: $\;\;$ $9^{x^2 - 1} - 36 \times 3^{x^2 - 3} + 3 = 0$

i.e. $\;$ $\left[\left(3\right)^2\right]^{x^2 - 1} - 36 \times 3^{x^2} \times 3^{-3} + 3 = 0$

i.e. $\;$ $3^{2x^2 - 2} - \dfrac{36}{27} \times 3^{x^2} + 3 = 0$

i.e. $\;$ $3^{2x^2} \times 3^{-2} - \dfrac{4}{3} \times 3^{x^2} + 3 = 0$

i.e. $\;$ $\dfrac{1}{9} \times \left(3^{x^2}\right)^2 - \dfrac{4}{3} \times 3^{x^2} + 3 = 0$ $\;\;\; \cdots \; (1)$

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

Then equation $(1)$ becomes

$\dfrac{1}{9} p^2 - \dfrac{4}{3} p + 3 = 0$

i.e. $\;$ $p^2 - 12p + 27 = 0$

i.e. $\;$ $\left(p - 3\right) \left(p - 9\right) = 0$

i.e. $\;$ $p = 3$ $\;$ or $\;$ $p = 9$

When $\;$ $p = 3$, $\;$ we have from equation $(2)$,

$3^{x^2} = 3 = 3^1$

$\implies$ $x^2 = 1$ $\implies$ $x = \pm 1$

When $\;$ $p = 9$, $\;$ we have from equation $(2)$,

$3^{x^2} = 9 = 3^2$

$\implies$ $x^2 = 2$ $\implies$ $x = \pm \sqrt{2}$

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

Algebra - Exponential Equations

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

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

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

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

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

i.e. $\;$ $\left(\dfrac{2}{3}\right)^{x^2} = \dfrac{4}{3} \times \dfrac{2}{9} = \left(\dfrac{2}{3}\right)^3$

$\implies$ $x^2 = 3$ $\implies$ $x = \pm \sqrt{3}$

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