Logarithms back on track

First of all, $x > \sqrt[3]{2}$ so that the equation is well defined. We can rewrite the equation as: $\log _{10} \left( {\log _{10} \left( {x^3 - 2} \right)} \right) = - \log _{10} \left( {\log _{10} x} \right) = \log _{10} \left( {\frac{1}{{\log _{10} x}}} \right)$ $\log _{10} \left( {x^3 - 2} \right) = \frac{1}{{\log _{10} x}}$ $\log _{10} x\log _{10} \left( {x^3 - 2} \right) = 1$ $\log _{10} \left( {x^3 - 2} \right)^{\log _{10} x} = 1$ $\left( {x^3 - 2} \right)^{\log _{10} x} = 10$ Now, let's define $f(x) = \left( {x^3 - 2} \right)^{\log _{10} x}$ . In the interval $x \in \left( {\left. {\sqrt[3]{2},\sqrt[3]{3}} \right]} \right., f(x) \le 1$ , because $\left( {x^3 - 2} \right) \le 1$ . It is evident that for $x \in \left[ {\left. {\sqrt[3]{3},\infty } \right)} \right.$ , $f(x)$ is strictly increasing, and as $f\left( {\sqrt[3]{3}} \right) = 1$ , then there exists only one solution to the equation. $\blacksquare$

$f(x)=\log_{10}\left(\log_{10}\left(x^{3}-2\right)\right)$ and $g(x)=\log_{10}\left(\log_{10}\left(x\right)\right)$ both are increasing functions so $f(x)+g(x)$ is also increasing. So the given equation can have atmost one real solution.

In the given equation put $x=2$ in LHS we get $f(x)+g(x)=f(2)+g(2)$ is negative.

Put $x=100$ and $f(x)+g(x)=f(100)+g(100)$ is positive.

Since $f(x)+g(x)$ is continuous in its domain. By bolzano's Intermediate value theorem there is a real number $c$ such that $f(c)+g(c)=0$ which is the only real solution becuase $f(x)+g(x)$ is an increasing function and can cut x-axis at most once.

Moreover by above analysis this $c$ lies somewhere in the interval $(2,100)$

This is actually done by a method known in Real Analysis