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 $\;\;$ $x = \left\{2 \right\}$