Solve the equation: $\;$ $\log \left(\log x\right) + \log \left(\log x^3 - 2\right) = 0$
Given equation: $\;\;$ $\log \left(\log x\right) + \log \left(\log x^3 - 2\right) = 0$
i.e. $\;$ $\log \left(\log x\right) = - \log \left(3 \log x - 2\right)$
i.e. $\;$ $\log \left(\log x\right) = \log \left(3 \log x - 2\right)^{-1}$ $\;\;\; \cdots \; (1)$
Taking antilog on both sides of equation $(1)$ gives
$\log x = \left(3 \log x - 2\right)^{-1}$
i.e. $\;$ $\log x = \dfrac{1}{3 \log x - 2}$
i.e. $\;$ $3 \left(\log x\right)^2 - 2 \log x - 1 = 0$ $\;\;\; \cdots \; (2)$
Let $\;$ $\log x = p$ $\;\;\; \cdots \; (3)$
Then equation $(2)$ becomes
$3p^2 - 2p -1 = 0$
i.e. $\;$ $\left(3p - 1\right) \left(p - 1\right) = 0$
i.e. $\;$ $p = \dfrac{1}{3}$ $\;$ or $\;$ $p = 1$
Substituting the value of $p$ in equation $(3)$ gives
when $\;$ $p = \dfrac{1}{3}$, $\;$ then $\;$ $\log x = \dfrac{1}{3}$ $\implies$ $x = 10^{\frac{1}{3}}$
when $\;$ $p = 1$, $\;$ then $\;$ $\log x = 1$ $\implies$ $x = 10^1 = 10$
Substituting $\;$ $\log x = \dfrac{1}{3}$ $\;$ in the term $\;$ $\log \left(\log x^3 - 2\right)$ $\;$ i.e. in $\;$ $\log \left(3 \log x - 2\right)$ $\;$ in the given question, the term becomes
$\log \left(3 \times \dfrac{1}{3} - 2\right) = \log \left(-1\right)$
But, logarithim of a negative number is not defined.
$\therefore \;$ $p = \dfrac{1}{3}$ $\;$ is not a valid solution.
Substituting $\;$ $\log x = 1$ $\;$ in the given question gives
$\log 1 + \log \left(3 \times 1 - 2\right) = 0$
i.e. $\;$ $\log 1 + \log 1 = 0$
i.e. $\;$ $0 + 0 = 0$ $\;\;$ which is true.
$\therefore \;$ $\log x = 1$ $\;$ is a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{10 \right\}$