Logarithm Basics Part II

$\displaystyle log_{0.3} (x-1) < log_{0.09}(x-1)$

$\implies \dfrac{log(x-1)}{log(0.3)} < \dfrac{log(x-1)}{log(0.09)}$ $\left( log_BA=\frac{logA}{logB} \right)$

$\implies \dfrac{log(x-1)}{log(0.3)} < \dfrac{log(x-1)}{2 \cdot log(0.3)}$

$\implies \displaystyle log (x-1)^2> log(x-1)$ (Inequality reversed...think : why ???)

$\implies x^2-2x+1>x-1$

$\implies x^2-3x+2>0$

$\implies x \in (-\infty,1) \cup (2,\infty)$ $\quad \quad \quad .............(1)$

But taking domain of $log$ into consideration :

$x-1>0$

$\implies x>1$ $\quad \quad \quad ............(2)$

So combining $(1) \& (2)$ , the values of $x$ satisfying the above inequality are $\boxed{(2,\infty)}$

enjoy!