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 $\;\;$ $x = \left\{10 \right\}$