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 $\;\;$ $x = \left\{4 \right\}$