Algebra - Logarithmic Equations

Solve the equation: $\;$ $\dfrac{1}{6} \log_2 \left(x - 2\right) - \dfrac{1}{3} = \log_{\frac{1}{8}} \sqrt{3x-5}$

Given equation: $\;\;$ $\dfrac{1}{6} \log_2 \left(x - 2\right) - \dfrac{1}{3} = \log_{\frac{1}{8}} \sqrt{3x-5}$

i.e. $\;$ $\dfrac{1}{6} \log_2 \left(x - 2\right) - \log_{\frac{1}{8}} \sqrt{3x-5} = \dfrac{1}{3}$

i.e. $\;$ $\dfrac{1}{6} \log_2 \left(x - 2\right) - \dfrac{\log_2 \sqrt{3x-5}}{\log_2 \dfrac{1}{8}} = \dfrac{1}{3}$

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

i.e. $\;$ $\dfrac{1}{6} \log_2 \left(x - 2\right) - \dfrac{\dfrac{1}{2} \log_2 \left(3x - 5\right)}{-3 \log_2 2} = \dfrac{1}{3}$

i.e. $\;$ $\dfrac{1}{6} \log_2 \left(x-2\right) + \dfrac{1}{6} \log_2 \left(3x-5\right) = \dfrac{1}{3}$

i.e. $\;$ $\dfrac{1}{2} \left[\log_2 \left(x-2\right) + \log_2 \left(3x-5\right)\right] = 1$

i.e. $\;$ $\log_2 \left[\left(x-2\right) \left(3x-5\right)\right] = 2$

i.e. $\;$ $\left(x-2\right) \left(3x-5\right) = 2^2 = 4$

i.e. $\;$ $3x^2 - 11x + 10 = 4$

i.e. $\;$ $3x^2 - 11x + 6 = 0$

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

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

When $\;$ $x = \dfrac{2}{3}$, $\;$ the term $\;$ $\log_2 \left(x-2\right)$ $\;$ in the given problem becomes

$\log_2 \left(\dfrac{2}{3} - 2\right) = \log_2 \left(\dfrac{-4}{3}\right)$

But, logarithim of a negative number is not defined.

$\therefore \;$ $x = \dfrac{2}{3}$ $\;$ is not a valid solution to the given problem.

When $\;$ $x = 3$, $\;$ the given problem becomes

$\dfrac{1}{6} \log_2 \left(3-2\right) - \dfrac{1}{3} = \log_{\frac{1}{8}} \sqrt{9-5}$

i.e. $\;$ $\dfrac{1}{6} \log_2 1 - \log_{\frac{1}{8}} 2 = \dfrac{1}{3}$

i.e. $\;$ $0 + \log_{\frac{1}{8}} 2 = \dfrac{-1}{3}$

i.e. $\;$ $2 = \left(\dfrac{1}{8}\right)^{\frac{-1}{3}}$

i.e. $\;$ $2 = 8^{\frac{1}{3}}$

i.e. $\;$ $2 = 2$ $\;$ which is true.

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log \left(\log x\right) + \log \left(\log x^4 - 3\right) = 0$

Given equation: $\;\;$ $\log \left(\log x\right) + \log \left(\log x^4 - 3\right) = 0$

i.e. $\;$ $\log \left\{\left(\log x\right) \left(\log x^4 - 3\right) \right\} = 0$

i.e. $\;$ $\left(\log x\right) \left(4 \log x - 3\right) = 10^0 = 1$

i.e. $\;$ $4 \left(\log x\right)^2 - 3 \log x - 1 = 0$

i.e. $\;$ $4 \left(\log x\right)^2 - 4 \log x + \log x - 1 = 0$

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

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

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

i.e. $\;$ $x = 10^{\frac{-1}{4}}$ $\;$ or $\;$ $x = 10^1 = 10$

Substituting $\;$ $x = 10^{\frac{-1}{4}}$ $\;$ in the first term of the given question, the term becomes

$\log \left(\log 10^{\frac{-1}{4}}\right) = \log \left(\dfrac{-1}{4} \log 10\right) = \log \left(\dfrac{-1}{4}\right)$

But logarithim of a negative number is not defined.

$\therefore \;$ $x = 10^{\frac{-1}{4}}$ $\;$ is not a valid solution.

Substituing $\;$ $x = 10$ $\;$ in the given equation, we get

$\log \left(\log 10\right) + \log \left(\log 10^4 - 3\right) = 0$

i.e. $\;$ $\log 1 + \log \left(4 \log 10 - 3\right) = 0$

i.e. $\;$ $\log 1 + \log \left(4 - 3\right) = 0$

i.e. $\;$ $\log 1 + \log 1 = 0$

i.e. $\;$ $0 + 0 = 0$ $\;\;\;$ which is true.

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\dfrac{2 \log x}{\log \left(5x - 4\right)} = 1$

Given equation: $\;\;$ $\dfrac{2 \log x}{\log \left(5x - 4\right)} = 1$

i.e. $\;$ $2 \log x = \log \left(5x - 4\right)$

i.e. $\;$ $\log x^2 = \log \left(5x - 4\right)$ $\;\;\; \cdots \; (1)$

Taking antilog on both sides of equation $(1)$ gives

$x^2 = 5x - 4$

i.e. $\;$ $x^2 - 5x + 4 = 0$

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

i.e. $\;$ $x = 4$ $\;$ or $\;$ $x = 1$

Substituting $\;$ $x = 1$ $\;$ in the given equation gives

$\dfrac{2 \log 1}{\log \left(5 - 4\right)} = 1$

i.e. $\;$ $\dfrac{2 \log 1}{\log 1} = 1$

i.e. $\;$ $2 = 1$ $\;$ which is not possible.

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

Substituting $\;$ $x = 1$ $\;$ in the given equation gives

$\dfrac{2 \log 4}{\log \left(20 - 4\right)} = 1$

i.e. $\;$ $\dfrac{\log 4^2}{\log16} = 1$

i.e. $\;$ $1 = 1$ $\;$ which is true.

$\implies$ $x = 4$ $\;$ is a valid solution to the given equation.

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_2 \left(4 \times 3^x - 6\right) - \log_2 \left(9^x - 6\right) = 1$

Given equation: $\;\;$ $\log_2 \left(4 \times 3^x - 6\right) - \log_2 \left(9^x - 6\right) = 1$

i.e. $\;$ $\log_2 \left(\dfrac{4 \times 3^x - 6}{9^x - 6}\right) = 1$

i.e. $\;$ $\dfrac{4 \times 3^x - 6}{9^x - 6} = 2^1 = 2$

i.e. $\;$ $4 \times 3^x - 6 = 2 \times 9^x - 12$

i.e. $\;$ $2 \times 3^{2x} - 4 \times 3^x - 6 = 0$

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

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

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

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

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

i.e. $\;$ $x = \log_3 \left(-1\right)$ $\;$ or $\;$ $x = \log_3 3 = 1$

$\because \;$ logarithim of a negative number is not defined,

$\therefore \;$ $x = \log_3 \left(-1\right)$ $\;$ is not a valid solution.

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

Algebra - Logarithmic Equations

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

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

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

i.e. $\;$ $1 - 8 \left(\log x\right)^2 = \log x - 2 \left(\log x\right)^2$

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

i.e. $\;$ $6 \left(\log x\right)^2 + 3 \log x - 2 \log x - 1 = 0$

i.e. $\;$ $3 \log x \left(2 \log x + 1\right) - 1 \left(2 \log x + 1\right) = 0$

i.e. $\;$ $\left(3 \log x - 1\right) \left(2 \log x + 1\right) = 0$

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

$\implies$ $x = 10^{\frac{1}{3}} = \sqrt[3]{10}$ $\;$ or $\;$ $x = 10^{\frac{-1}{2}} = \dfrac{1}{\sqrt{10}}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{\sqrt{10}}, \; \sqrt[3]{10} \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_2 \left(\dfrac{x}{4}\right) = \dfrac{15}{\log_2 \left(\dfrac{x}{8}\right) - 1}$

Given equation: $\;\;$ $\log_2 \left(\dfrac{x}{4}\right) = \dfrac{15}{\log_2 \left(\dfrac{x}{8}\right) - 1}$

i.e. $\;$ $\log_2 x - \log_2 4 = \dfrac{15}{\log_2 x - \log_2 8 - 1}$ $\;\;\;$ $\left[\because \;\; \log_a \left(\dfrac{m}{n}\right) = \log_a m - \log_a n\right]$

i.e. $\;$ $\log_2 x - \log_2 2^2 = \dfrac{15}{\log_2 x - \log_2 2^3 - 1}$

i.e. $\;$ $\log_2 x - 2 \log_2 2 = \dfrac{15}{\log_2 x - 3 \log_2 2 - 1}$

i.e. $\;$ $\log_2 x - 2 = \dfrac{15}{\log_2 x - 3 - 1}$

i.e. $\;$ $\log_2 x - 2 = \dfrac{15}{\log_2 x - 4}$

i.e. $\;$ $\left(\log_2 x\right)^2 - 2 \log_2 x - 4 \log_2 x + 8 = 15$

i.e. $\;$ $\left(\log_2 x\right)^2 - 6 \log_2 x - 7 = 0$

i.e. $\;$ $\left(\log_2 x\right)^2 - 7 \log_2 x + \log_2 x - 7 = 0$

i.e. $\;$ $\log_2 x \left(\log_2 x - 7\right) + 1 \left(\log_2 x - 7\right) = 0$

i.e. $\;$ $\left(\log_2 x - 7\right) \left(\log_2 x + 1\right) = 0$

i.e. $\;$ $\log_2 x = 7$ $\;$ or $\;$ $\log_2 x = -1$

i.e. $\;$ $x = 2^7 = 128$ $\;$ or $\;$ $x = 2^{-1} = \dfrac{1}{2}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{2}, \; 128 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_4 2^{4x} = 2^{\log_2 4}$

Given equation: $\;\;$ $\log_4 2^{4x} = 2^{\log_2 4}$

i.e. $\;$ $4x \log_4 2 = 4$ $\;\;\;$ $\left[\because \;\; \log_a m^n = n \log_a m; \;\;\; a^{\log_a m} = m\right]$

i.e. $\;$ $x \log_4 4^{\frac{1}{2}} = 1$

i.e. $\;$ $\left(\dfrac{x}{2}\right) \log_4 4 = 1$

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

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_3 \left(4 \times 3^x - 1\right) = 2x + 1$

Given equation: $\;\;$ $\log_3 \left(4 \times 3^x - 1\right) = 2x + 1$

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

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

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

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

Then equation $(1)$ becomes

$3p^2 - 4p + 1 = 0$

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

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

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

when $\;$ $p = 1$, $\;$ then $\;$ $3^x = 1 = 3^0$ $\implies$ $x = 0$

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

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-1, \; 0 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_3 \left(3^x - 6\right) = x - 1$

Given equation: $\;\;$ $\log_3 \left(3^x - 6\right) = x - 1$

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

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

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

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

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

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

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

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_3 \left(\log^2_{\frac{1}{2}} x - 3 \log_{\frac{1}{2}} x+ 5\right) = 2$

Given equation: $\;\;$ $\log_3 \left(\log^2_{\frac{1}{2}} x - 3 \log_{\frac{1}{2}} x + 5\right) = 2$

i.e. $\;$ $\log^2_{\frac{1}{2}} x - 3 \log_{\frac{1}{2}} x + 5 = 3^2 = 9$

i.e. $\;$ $\log^2_{\frac{1}{2}} x - 3 \log_{\frac{1}{2}} x - 4 = 0$

i.e. $\;$ $\log^2_{\frac{1}{2}} x - 4 \log_{\frac{1}{2}} x + \log_{\frac{1}{2}} x - 4 = 0$

i.e. $\;$ $\log_{\frac{1}{2}} x \left(\log_{\frac{1}{2}} x - 4\right) + 1 \left(\log_{\frac{1}{2}} x - 4\right) = 0$

i.e. $\;$ $\left(\log_{\frac{1}{2}}x + 1\right) \left(\log_{\frac{1}{2}}x - 4\right) = 0$

i.e. $\;$ $\log_{\frac{1}{2}}x = -1$ $\;$ or $\;$ $\log_{\frac{1}{2}}x = 4$

i.e. $\;$ $x = \left(\dfrac{1}{2}\right)^{-1} = 2$ $\;$ or $\;$ $x = \left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{16}, \; 2 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_5 \left(\dfrac{2 + x}{10}\right) = \log_5 \left(\dfrac{2}{x + 1}\right)$

Given equation: $\;\;$ $\log_5 \left(\dfrac{2 + x}{10}\right) = \log_5 \left(\dfrac{2}{x + 1}\right)$

i.e. $\;$ $\log_5 \left(\dfrac{2 + x}{10}\right) - \log_5 \left(\dfrac{2}{x + 1}\right) = 0$

i.e. $\;$ $\log_5 \left[\dfrac{\dfrac{2 + x}{10}}{\dfrac{2}{x + 1}}\right] = 0$

i.e. $\;$ $\log_5 \left[\dfrac{\left(2 + x\right) \left(x + 1\right)}{20}\right] = 0$

i.e. $\;$ $\dfrac{\left(2 + x\right) \left(x + 1\right)}{20} = 5^0 = 1$

i.e. $\;$ $x^2 + 3x + 2 = 20$

i.e. $\;$ $x^2 + 3x - 18 = 0$

i.e. $\;$ $\left(x + 6\right) \left(x - 3\right) = 0$

i.e. $\;$ $x = -6$ $\;$ or $\;$ $x = 3$

When $\;$ $x = -6$, $\;$ the terms in the given equation become

$\log_5 \left(\dfrac{2 + x}{10}\right) = \log_5 \left(\dfrac{2 - 6}{10}\right) = \log_5 \left(\dfrac{-4}{10}\right)$

and $\;$ $\log_5 \left(\dfrac{2}{x + 1}\right) = \log_5 \left(\dfrac{2}{-6 + 1}\right) = \log_5 \left(\dfrac{-2}{5}\right)$

But, logarithim of a negative number is not defined.

$\therefore \;$ $x = -6$ is not a valid solution.

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $1 + 2 \log_{\left(x + 2\right)} 5 = \log_5 \left(x + 2\right)$

Given equation: $\;\;$ $1 + 2 \log_{\left(x + 2\right)} 5 = \log_5 \left(x + 2\right)$

i.e. $\;$ $1 + \dfrac{2}{\log_5 \left(x + 2\right)} = \log_5 \left(x + 2\right)$ $\;\;\; \cdots \; (1)$

Let $\;$ $\log_5 \left(x + 2\right) = p$ $\;\;\; \cdots \; (2)$

Then equation $(1)$ becomes

$1 + \dfrac{2}{p} = p$

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

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

i.e. $\;$ $p = 2$ $\;$ or $\;$ $p = -1$

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

when $\;$ $p = 2$, $\;$ then $\;$ $\log_5 \left(x + 2\right) = 2$

i.e. $\;$ $x + 2 = 5^2 = 25$ $\implies$ $x = 23$

when $\;$ $p = -1$, $\;$ then $\;$ $\log_5 \left(x + 2\right) = -1$

i.e. $\;$ $x + 2 = 5^{-1} = \dfrac{1}{5}$ $\implies$ $x = \dfrac{-9}{5}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{-9}{5}, \; 23 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_{0.5x} x^2 - 14 \log_{16x} x^3 + 40 \log_{4x} \sqrt{x} = 0$

Given equation: $\;\;$ $\log_{0.5x} x^2 - 14 \log_{16x} x^3 + 40 \log_{4x} \sqrt{x} = 0$

i.e. $\;$ $2 \log_{0.5x} x - 14 \times 3 \log_{16x} x + 40 \log_{4x} x^{\frac{1}{2}} = 0$

i.e. $\;$ $2 \log_{0.5x} x - 42 \log_{16x} x + 40 \times \dfrac{1}{2} \log_{4x} x = 0$

i.e. $\;$ $\log_{0.5x} x - 21 \log_{16x} x + 10 \log_{4x} x = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $\log_2 x = p$ $\;\;\; \cdots \; (2)$

Now,

$\begin{aligned} \log_{0.5x} x & = \dfrac{\log_2 x}{\log_2 {0.5x}} \\\\ & = \dfrac{\log_2 x}{\log_2 {0.5} + \log_2 x} \\\\ & = \dfrac{\log_2 x}{\log_2 2^{-1} + \log_2 x} \\\\ & = \dfrac{\log_2 x}{-1 \log_2 2 + \log_2 x} \\\\ & = \dfrac{\log_2 x}{\log_2 x - 1} \\\\ & = \dfrac{p}{p - 1} \;\;\; \cdots \; (3) \;\;\; \left[\text{By equation (2)}\right] \end{aligned}$

$\begin{aligned} \log_{16x} x & = \dfrac{\log_2 x}{\log_2 {16x}} \\\\ & = \dfrac{\log_2 x}{\log_2 {16} + \log_2 x} \\\\ & = \dfrac{\log_2 x}{\log_2 2^{4} + \log_2 x} \\\\ & = \dfrac{\log_2 x}{4 \log_2 2 + \log_2 x} \\\\ & = \dfrac{\log_2 x}{\log_2 x + 4} \\\\ & = \dfrac{p}{p + 4} \;\;\; \cdots \; (4) \;\;\; \left[\text{By equation (2)}\right] \end{aligned}$

$\begin{aligned} \log_{4x} x & = \dfrac{\log_2 x}{\log_2 {4x}} \\\\ & = \dfrac{\log_2 x}{\log_2 {4} + \log_2 x} \\\\ & = \dfrac{\log_2 x}{\log_2 2^{2} + \log_2 x} \\\\ & = \dfrac{\log_2 x}{2 \log_2 2 + \log_2 x} \\\\ & = \dfrac{\log_2 x}{\log_2 x + 2} \\\\ & = \dfrac{p}{p + 2} \;\;\; \cdots \; (5) \;\;\; \left[\text{By equation (2)}\right] \end{aligned}$

In view of equations $(3)$, $(4)$ and $(5)$, equation $(1)$ becomes

$\dfrac{p}{p - 1} - \dfrac{21p}{p + 4} + \dfrac{10p}{p + 2} = 0$

i.e. $\;$ $p \left[\dfrac{1}{p - 1} - \dfrac{21}{p + 4} + \dfrac{10}{p + 2}\right] = 0$

$\implies$ $p = 0$ $\;$ or $\;$ $\dfrac{1}{p - 1} - \dfrac{21}{p + 4} + \dfrac{10}{p + 2} = 0$

Case 1: When $\;\;$ $\dfrac{1}{p - 1} - \dfrac{21}{p + 4} + \dfrac{10}{p + 2} = 0$

i.e. $\;$ $\left(p + 4\right) \left(p + 2\right) - 21 \left(p - 1\right) \left(p + 2\right) + 10 \left(p - 1\right) \left(p + 4\right) = 0$

i.e. $\;$ $p^2 + 6p + 8 - 21 p^2 - 21 p + 42 + 10 p^2 + 30 p - 40 = 0$

i.e. $\;$ $10p^2 - 15p - 10 = 0$

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

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

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

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

when $\;\;$ $p = \dfrac{-1}{2}$, $\;$ then $\;$ $\log_2 x = \dfrac{-1}{2}$ $\implies$ $x = 2^{\frac{-1}{2}} = \dfrac{1}{\sqrt{2}}$

when $\;\;$ $p = 2$, $\;$ then $\;$ $\log_2 x = 2$ $\implies$ $x = 2^{2} = 4$

Case 2: When $\;\;$ $p = 0$

$\implies$ $\log_2 x = 0$ $\;\;\;$ [By equation $(2)$]

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

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{\sqrt{2}}, \; 1, \; 4 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log^2_{\frac{1}{2}} \left(4x\right) + \log_2 \left(\dfrac{x^2}{8}\right) = 8$

Given equation: $\;\;$ $\log^2_{\frac{1}{2}} \left(4x\right) + \log_2 \left(\dfrac{x^2}{8}\right) = 8$

i.e. $\;$ $\dfrac{\log^2_2 \left(4x\right)}{\log^2_2 \left(\frac{1}{2}\right)} + \log_2 x^2 - \log_2 8 = 8$

i.e. $\;$ $\left[\dfrac{\log_2 \left(4x\right)}{\log_2 \left(\frac{1}{2}\right)}\right]^2 + 2 \log_2 x - \log_2 2^3 = 8$

i.e. $\;$ $\left[\dfrac{\log_2 4 + \log_2 x}{\log_2 2^{-1}}\right]^2 + 2 \log_2 x - 3 \log_2 2 = 8$

i.e. $\;$ $\left[\dfrac{\log_2 2^2 + \log_2 x}{-1 \times \log_2 2}\right]^2 + 2 \log_2 x - 3 = 8$

i.e. $\;$ $\left[\dfrac{2 \log_2 2 + \log_2 x}{-1}\right]^2 + 2 \log_2 x - 11 = 0$

i.e. $\;$ $\left[2 + \log_2 x\right]^2 + 2 \log_2 x - 11 = 0$

i.e. $\;$ $4 + \log^2_2 x + 4 \log_2 x + 2 \log_2 x - 11 = 0$

i.e. $\;$ $\log^2_2 x + 6 \log_2 x - 7 = 0$

i.e. $\;$ $\log^2_2 x + 7 \log_2 x - \log_2 x - 7 = 0$

i.e. $\;$ $\log_2 x \left(\log_2 x - 1\right) + 7 \left(\log_2 x - 1\right) = 0$

i.e. $\;$ $\left(\log_2 x + 7\right) \left(\log_2 x - 1\right) = 0$

i.e. $\;$ $\log_2 x = -7$ $\;$ or $\;$ $\log_2 x = 1$

i.e. $\;$ $x = 2^{-7}$ $\;$ or $\;$ $x = 2^1 = 2$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2^{-7}, \; 2 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_x \left(9x^2\right) \times \log^2_3 x = 4$

Given equation: $\;\;$ $\log_x \left(9x^2\right) \times \log^2_3 x = 4$

i.e. $\;$ $\left(\log_x 9 + \log_x x^2\right) \times \log^2_3 x = 4$

i.e. $\;$ $\left(\log_x 3^2 + 2 \log_x x\right) \times \log^2_3 x = 4$

i.e. $\;$ $\left(2 \log_x 3 + 2\right) \times \log^2_3 x = 4$

i.e. $\;$ $\left(\log_x 3 + 1\right) \times \log^2_3 x = 2$

i.e. $\;$ $\log_x 3 \times \log^2_3 x + \log^2_3 x = 2$

i.e. $\;$ $\log^2_3 x + \log_x 3 \times \log_3 x \times \log_3 x - 2 = 0$

i.e. $\;$ $\log^2_3 x + \log_x 3 \times \dfrac{1}{\log_x 3} \times \log_3 x - 2 = 0$

i.e. $\;$ $\log^2_3 x + \log_3 x - 2 = 0$

i.e. $\;$ $\log^2_3 x + 2 \log_3 x - \log_3 x - 2 = 0$

i.e. $\;$ $\log_3 x \left(\log_3 x - 1\right) + 2 \left(\log_3 x - 1\right) = 0$

i.e. $\;$ $\left(\log_3 x + 2\right) \left(\log_3 x - 1\right) = 0$

i.e. $\;$ $\log_3 x = -2$ $\;$ or $\;$ $\log_3 x = 1$

i.e. $\;$ $x = 3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}$ $\;$ or $\;$ $x = 3^1 = 3$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{9}, \; 3 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $2 \left(\log_x \sqrt{5}\right)^2 - 3 \log_x \sqrt{5} + 1 = 0$

Given equation: $\;\;$ $2 \left(\log_x \sqrt{5}\right)^2 - 3 \log_x \sqrt{5} + 1 = 0$

i.e. $\;$ $2 \left(\log_x \sqrt{5}\right)^2 - 2 \log_x \sqrt{5} - \log_x \sqrt{5} + 1 = 0$

i.e. $\;$ $2 \log_x \sqrt{5} \left(\log_x \sqrt{5} - 1\right) - 1 \left(\log_x \sqrt{5} - 1\right) = 0$

i.e. $\;$ $\left(2 \log_x \sqrt{5} - 1\right) \left(\log_x \sqrt{5} - 1\right) = 0$

i.e. $\;$ $\log_x \sqrt{5} = \dfrac{1}{2}$, $\;$ or $\;$ $\log_x \sqrt{5} = 1$

i.e. $\;$ $\sqrt{5} = x^{\frac{1}{2}} = \sqrt{x}$ $\;$ or $\;$ $\sqrt{5} = x^1= x$

$\implies$ $x = 5$ $\;$ or $\;$ $x = \sqrt{5}$

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{\log_{\sqrt{x}} \left(x - 2\right)} = 9$

Given equation: $\;\;$ $x^{\log_{\sqrt{x}} \left(x - 2\right)} = 9$ $\;\;\;\; \cdots \; (1)$

Taking log to base $\sqrt{x}$ on either side of equation $(1)$ gives

$\log_{\sqrt{x}} \left[x^{\log_{\sqrt{x}}\left(x - 2\right)}\right] = \log_{\sqrt{x}} 9$

i.e. $\;$ $\log_{\sqrt{x}} \left(x - 2\right) \log_{\sqrt{x}} x = \log_{\sqrt{x}} 9$

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

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

i.e. $\;$ $\log_{\sqrt{x}} \left(x-2\right) = \log_{\sqrt{x}} 3$ $\;\;\; \cdots \; (2)$

Taking anti-log to base $\sqrt{x}$ on either side of equation $(2)$ gives

$x - 2 = 3$ $\implies$ $x = 5$

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $3 + 2 \log_{x+1} 3 = 2 \log_3 \left(x+1\right)$

Given equation: $\;\;$ $3 + 2 \log_{x+1} 3 = 2 \log_3 \left(x+1\right)$

i.e. $\;$ $2 \log_3 \left(x + 1\right) - 2 \log_{x+1} 3 = 3$

i.e. $\;$ $\log_3 \left(x + 1\right) - \log_{x+1} 3 = \dfrac{3}{2}$

i.e. $\;$ $\log_3 \left(x+1\right) - \dfrac{1}{\log_3 \left(x + 1\right)} = \dfrac{3}{2}$ $\;\;\; \cdots \; (1)$

Let $\;$ $\log_3 \left(x + 1\right) = p$ $\;\;\; \cdots \; (2)$

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

$p - \dfrac{1}{p} = \dfrac{3}{2}$

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

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

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

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

when $\;$ $p = \dfrac{-1}{2}$, $\;$ then $\;$ $\log_3 \left(x + 1\right) = \dfrac{-1}{2}$

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

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

when $\;$ $p = 2$, $\;$ then $\;$ $\log_3 \left(x + 1\right) = 2$

i.e. $\;$ $x + 1 = 3^2 = 9$ $\implies$ $x = 8$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{\sqrt{3} - 3}{3}, \; 8 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_4 \left(x + 3\right) - \log_4 \left(x - 1\right) = 2 - \log_4 8$

Given equation: $\;\;$ $\log_4 \left(x + 3\right) - \log_4 \left(x - 1\right) = 2 - \log_4 8$

i.e. $\;$ $\log_4 \left(\dfrac{x + 3}{x - 1}\right) + \log_4 8 = 2$

i.e. $\;$ $\log_4 \left[\dfrac{8 \left(x + 3\right)}{x - 1}\right] = 2$

i.e. $\;$ $\dfrac{8 \left(x + 3\right)}{x - 1} = 4^2 = 16$

i.e. $\;$ $8x + 24 = 16x - 16$

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: $\;$ $\log^2 \left(100 x\right) + \log^2 \left(10 x\right) = 14 + \log \left(\dfrac{1}{x}\right)$

Given equation: $\;\;$ $\log^2 \left(100 x\right) + \log^2 \left(10 x\right) = 14 + \log \left(\dfrac{1}{x}\right)$

i.e. $\;$ $\left(\log 100 + \log x\right)^2 + \left(\log 10 + \log x\right)^2 = 14 + \log 1 - \log x$

i.e. $\;$ $\left(\log 10^2 + \log x\right)^2 + \left(1 + \log x\right)^2 = 14 + 0 - \log x$

i.e. $\;$ $\left(2 \log 10 + \log x\right)^2 + \left(1 + \log x\right)^2 = 14 - \log x$

i.e. $\;$ $\left(2 + \log x\right)^2 + \left(1 + \log x\right)^2 = 14 - \log x$

i.e. $\;$ $4 + \log^2 x + 4 \log x + 1 + \log^2 x + 2 \log x = 14 - \log x$

i.e. $\;$ $2 \log^2 x + 7 \log x - 9 = 0$

i.e. $\;$ $2 \log^2 x + 9 \log x - 2 \log x - 9 = 0$

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

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

i.e. $\;$ $\log x = \dfrac{-9}{2}$ $\;$ or $\;$ $\log x = 1$

$\implies$ $x = 10^{\frac{-9}{2}} = \sqrt{10^{-9}}$ $\;$ or $\;$ $x = 10^1 = 10$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\sqrt{10^{-9}}, \; 10 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log^2_2 x + 2 \log_2 \sqrt{x} - 2 = 0$

Given equation: $\;\;$ $\log^2_2 x + 2 \log_2 \sqrt{x} - 2 = 0$

i.e. $\;$ $\left[\log_2 x\right]^2 + 2 \log_2 \sqrt{x} - 2 = 0$

i.e. $\;$ $\left[\log_2 \left(\sqrt{x}\right)^2\right]^2 + 2 \log_2 \sqrt{x} - 2 = 0$

i.e. $\;$ $\left[2 \log_2 \sqrt{x}\right]^2 + 2 \log_2 \sqrt{x} - 2 = 0$

i.e. $\;$ $4 \log^2_2 \sqrt{x} + 2 \log_2 \sqrt{x} - 2 = 0$

i.e. $\;$ $2 \log^2_2 \sqrt{x} + \log_2 \sqrt{x} - 1 = 0$

i.e. $\;$ $2 \log^2_2 \sqrt{x} + 2 \log_2 \sqrt{x} - \log_2 \sqrt{x} - 1 = 0$

i.e. $\;$ $2 \log_2 \sqrt{x} \left(\log_2 \sqrt{x} + 1\right) - 1 \left(\log_2 \sqrt{x} + 1\right) = 0$

i.e. $\;$ $\left(2 \log_2 \sqrt{x} - 1\right) \left(\log_2 \sqrt{x} + 1\right) = 0$

i.e. $\;$ $\log_2 \sqrt{x} = \dfrac{1}{2}$ $\;$ or $\;$ $\log_2 \sqrt{x} = -1$

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

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

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{4}, \; 2 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log_{\frac{1}{3}} x - 3 \sqrt{\log_{\frac{1}{3}} x} + 2 = 0$

Given equation: $\;\;$ $\log_{\frac{1}{3}} x - 3 \sqrt{\log_{\frac{1}{3}} x} + 2 = 0$

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

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

i.e. $\;$ $\sqrt{\log_{\frac{1}{3}} x} \left(\sqrt{\log_{\frac{1}{3}} x} - 2\right) - 1 \left(\sqrt{\log_{\frac{1}{3}} x} - 2\right) = 0$

i.e. $\;$ $\left(\sqrt{\log_{\frac{1}{3}} x} - 1\right) \left(\sqrt{\log_{\frac{1}{3}} x} - 2\right) = 0$

i.e. $\;$ $\sqrt{\log_{\frac{1}{3}} x} = 1$ $\;$ or $\;$ $\sqrt{\log_{\frac{1}{3}} x} = 2$

i.e. $\;$ $\log_{\frac{1}{3}} x = 1$ $\;$ or $\;$ $\log_{\frac{1}{3}} x = 4$

$\implies$ $x = \left(\dfrac{1}{3}\right)^1 = \dfrac{1}{3}$ $\;$ or $\;$ $x = \left(\dfrac{1}{3}\right)^4 = \dfrac{1}{81}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{3}, \; \dfrac{1}{81} \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\log^2 x - 3 \log x = \log x^2 - 4$

Given equation: $\;\;$ $\log^2 x - 3 \log x = \log x^2 - 4$

i.e. $\;$ $\log^2 x - 3 \log x = 2 \log x - 4$

i.e. $\;$ $\left(\log x\right)^2 - 5 \log x + 4 = 0$

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

i.e. $\;$ $\log x \left(\log x - 4\right) - 1 \left(\log x - 4\right) = 0$

i.e. $\;$ $\left(\log x - 1\right) \left(\log x - 4\right) = 0$

i.e. $\;$ $\log x = 1$ $\;$ or $\;$ $\log x = 4$

$\implies$ $x = 10^1 = 10$ $\;$ or $\;$ $x = 10^4$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10, \; 10^4 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $3 \sqrt{\log_2 x} - \log_2 8x + 1 = 0$

Given equation: $\;\;$ $3 \sqrt{\log_2 x} - \log_2 8x + 1 = 0$

i.e. $\;$ $3 \sqrt{\log_2 x} - \left(\log_2 8 + \log_2 x\right) + 1 = 0$

i.e. $\;$ $3 \sqrt{\log_2 x} - \log_2 2^3 - \log_2 x + 1 = 0$

i.e. $\;$ $3 \sqrt{\log_2 x} - 3 \log_2 2 - \log_2 x + 1 = 0$

i.e. $\;$ $3 \sqrt{\log_2 x} - 3 - \log_2 x + 1 = 0$

i.e. $\;$ $3 \sqrt{\log_2 x} - \log_2 x - 2 = 0$ $\;\;\; \cdots \; (1)$

Let $\;$ $\sqrt{\log_2 x} = p$ $\;\;\; \cdots \; (2)$

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

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

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

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

$\implies$ $p = 1$ $\;$ or $\;$ $p = 2$

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

when $\;$ $p = 1$, $\;$ $\sqrt{\log_2 x} = 1$

i.e. $\;$ $\log_2 x = 1$ $\implies$ $x = 2^1 = 2$

when $\;$ $p = 2$, $\;$ $\sqrt{\log_2 x} = 2$

i.e. $\;$ $\log_2 x = 4$ $\implies$ $x = 2^4 = 16$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2, \; 16 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{\frac{\left(\log x\right) + 7}{4}} = 10^{\left(\log x\right) + 1}$

Given equation: $\;\;$ $x^{\frac{\left(\log x\right) + 7}{4}} = 10^{\left(\log x\right) + 1}$ $\;\;\; \cdots \; (1)$

Taking logarithims on both sides of equation $(1)$ gives

$\log \left(x^{\frac{\log x + 7}{4}}\right) = \log \left(10^{\log x + 1}\right)$

i.e. $\;$ $\left(\dfrac{\log x + 7}{4}\right) \times \log x = \left(\log x + 1\right) \times \log 10$

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

i.e. $\;$ $\left(\log x\right)^2 + 7 \log x = 4 \log x + 4$

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

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

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

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

i.e. $\;$ $\log x = 1$ $\;$ or $\;$ $\log x = -4$

$\implies$ $x = 10^1 = 10$ $\;$ or $x = 10^{-4}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^{-4}, \; 10 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $\left[\sqrt{x}\right]^{\left(\log_5 x\right) - 1} = 5$

Given equation: $\;\;$ $\left[\sqrt{x}\right]^{\left(\log_5 x\right) - 1} = 5$

i.e. $\;$ $\left(\sqrt{x}\right)^{\log_5 x} \times \left(\sqrt{x}\right)^{-1} = 5$

i.e. $\;$ $\dfrac{\left(\sqrt{x}\right)^{\log_5 x}}{\sqrt{x}} = 5$ $\;\;\; \cdots \; (1)$

Taking logarithims to base $5$ on both sides of equation $(1)$ gives

$\log_5 \left[\dfrac{\left(\sqrt{x}\right)^{\log_5 x}}{\sqrt{x}}\right] = \log_5 5$

i.e. $\;$ $\log_5 \left(\sqrt{x}\right)^{\log_5 x} - \log_5 \sqrt{x} = 1$

i.e. $\;$ $\log_5 x \times \log_5 \sqrt{x} - \log_5 \sqrt{x} = 1$

i.e. $\;$ $\log_5 x \times \log_5 x^{\frac{1}{2}} - \log_5 x^{\frac{1}{2}} = 1$

i.e. $\;$ $\dfrac{1}{2} \log_5 x \times \log_5 x - \dfrac{1}{2} \log_5 x = 1$

i.e. $\;$ $\left(\log_5 x\right)^2 - \log_5 x - 2 = 0$

i.e. $\;$ $\left(\log_5 x\right)^2 - 2 \log_5 x + \log_5 x - 2 = 0$

i.e. $\;$ $\log_5 x \left(\log_5 x - 2\right) + 1 \left(\log_5 x - 2\right) = 0$

i.e. $\;$ $\left(\log_5 x + 1\right) \left(\log_5 x - 2\right) = 0$

i.e. $\;$ $\log_5 x = -1$ $\;$ or $\;$ $\log_5 x = 2$

$\implies$ $x = 5^{-1} = \dfrac{1}{5}$, $\;$ or $\;$ $x = 5^2 = 25$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{5}, \; 25 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{\left(\log x\right) + 1} = 10^6$

Given equation: $\;\;$ $x^{\left(\log x\right) + 1} = 10^6$

i.e. $\;$ $x^{\log x} \times x^1 = 10^6$ $\;\;\; \cdots \; (1)$

Taking logarithims on both sides of equation $(1)$ gives

$\log \left(x^{\log x} \times x\right) = \log {10}^6$

i.e. $\;$ $\log \left(x^{\log x}\right) + \log x = 6 \log 10$

i.e. $\;$ $\log x \times \log x + \log x = 6$

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

i.e. $\;$ $\left(\log x\right)^2 + 3 \log x - 2 \log x - 6 = 0$

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

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

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

$\implies$ $x = 10^{-3}$ $\;$ or $x = 10^2$

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{\log_3 x} = 9$

Given equation: $\;\;$ $x^{\log_3 x} = 9$ $\;\;\; \cdots \; (1)$

Taking logarithims to the base $3$ on both sides of equation $(1)$ gives

$\log_3 \left(x^{\log_3 x}\right) = \log_3 9$

i.e. $\;$ $\log_3 x \times \log_3 x = \log_3 3^2$

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

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

i.e. $\;$ $\log_3 x = \pm \sqrt{2}$

i.e. $\;$ $x = 3^{\pm \sqrt{2}}$

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

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{\frac{\log x + 5}{3}} = 10^{5 + \log x}$

Given equation: $\;\;$ $x^{\frac{\log x + 5}{3}} = 10^{5 + \log x}$ $\;\;\; \cdots \; (1)$

Taking logarithims on both sides of equation $(1)$ gives

$\log \left(x^{\frac{\log x + 5}{3}}\right) = \log \left(10^{5 + \log x}\right)$

i.e. $\;$ $\left(\dfrac{\log x + 5}{3}\right) \log x = \left(5 + \log x\right) \log 10$

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

i.e. $\;$ $\dfrac{1}{3} \left(\log x\right)^2 + \dfrac{2}{3} \log x - 5 = 0$

i.e. $\;$ $\left(\log x\right)^2 + 2 \log x - 15 = 0$

i.e. $\;$ $\left(\log x\right)^2 + 5 \log x - 3 \log x - 15 = 0$

i.e. $\;$ $\log x \left(\log x + 5\right) - 3 \left(\log x + 5\right) = 0$

i.e. $\;$ $\left(\log x - 3\right) \left(\log x + 5\right) = 0$

i.e. $\;$ $\log x = 3$ $\;\;$ or $\;\;$ $\log x = -5$

$\implies$ $x = 10^3$ $\;\;$ or $\;\;$ $x = 10^{-5}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10^{-5}, \; 10^3 \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{2 \log x} = 10 x^2$

Given equation: $\;\;$ $x^{2 \log x} = 10 x^2$ $\;\;\; \cdots \; (1)$

Taking logarithims on both sides of equation $(1)$ gives

$\log \left(x^{2 \log x}\right) = \log \left(10 x^2\right)$

i.e. $\;$ $2 \log x \times \log x = \log 10 + \log x^2$

i.e. $\;$ 2 $\left(\log x\right)^2 = 1 + 2 \log x$

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

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

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

i.e. $\;$ $\log x = \dfrac{1 \pm \sqrt{3}}{2}$

$\implies$ $x = 10^{\frac{1 \pm \sqrt{3}}{2}}$

i.e. $\;$ $x = \left(10^{1 \pm \sqrt{3}}\right)^{\frac{1}{2}}$

i.e. $\;$ $x = \sqrt{10^{1 \pm \sqrt{3}}}$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\sqrt{10^{1 + \sqrt{3}}}, \; \sqrt{10^{1 - \sqrt{3}}} \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $9^{\log_3 \left(1 - 2x\right)} = 5x^2 - 5$

Given equation: $\;\;$ $9^{\log_3 \left(1 - 2x\right)} = 5x^2 - 5$

i.e. $\;$ $\left(3^2\right)^{\log_3 \left(1 - 2x\right)} = 5x^2 - 5$

i.e. $\;$ $\left(3\right)^{2 \log_3 \left(1 - 2x\right)} = 5x^2 - 5$

i.e. $\;$ $3^{\log_3 \left(1 - 2x\right)^2} = 5x^2 - 5$

i.e. $\;$ $\left(1 - 2x\right)^2 = 5x^2 - 5$

i.e. $\;$ $1 - 4x + 4x^2 = 5x^2 - 5$

i.e. $\;$ $x^2 + 4x - 6 = 0$

i.e. $\;$ $x = \dfrac{-4 \pm \sqrt{16 + 24}}{2}$

i.e. $\;$ $x = \dfrac{-4 \pm \sqrt{40}}{2}$

i.e. $\;$ $\dfrac{-4 \pm 2 \sqrt{10}}{2}$

i.e. $\;$ $x = -2 \pm \sqrt{10}$

Now, $\;$ $\sqrt{10} \approx 3.16, \;\;\; -2 + \sqrt{10} = 1.16$

and $\;$ $\log_3 \left(1 - 2x\right) = \log_3 \left(1 - 2.32\right) = \log_3 \left(-1.32\right)$

But, logarithim of a negative number is not defined.

$\therefore \;$ $x = -2 + \sqrt{10}$ $\;$ is not a valid solution.

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-2 - \sqrt{10} \right\}$

Algebra - Logarithmic Equations

Solve the equation: $\;$ $x^{1 + \log x} = 10x$

Given equation: $\;\;$ $x^{1 + \log x} = 10x$

i.e. $\;$ $1 + \log x = \log_x {10x}$

i.e. $\;$ $1 + \log x = \log_x {10} + \log_x x$ $\;\;\;$ $\left[\because \; \log_a \left(CD\right) = \log_a C + \log_a D\right]$

i.e. $\;$ $1 + \log x = \dfrac{\log_{10} 10}{\log_{10} x} + 1$ $\;\;\;$ $\left[\because \; \log_b a = \dfrac{\log_m a}{\log_m b}; \;\; \log_a a = 1\right]$

i.e. $\;$ $\log_{10} x = \dfrac{1}{\log_{10} x}$

i.e. $\;$ $\left(\log_{10} x\right)^2 = 1$

i.e. $\;$ $\log_{10} x = \pm 1$

Now, $\;$ $\log_{10} x = 1$ $\implies$ $x = 10^1 = 10$

and $\;$ $\log_{10} x = -1$ $\implies$ $x = 10^{-1} = \dfrac{1}{10} = 0.1$

$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0.1, \; 10 \right\}$