# Solving $\left(\frac{x}{10}\right)^{\log(x)-2}<100$

How to solve the following inequality? $$\left(\frac{x}{10}\right)^{\log(x)-2}<100$$ The solution given is $$x\in(1, 1000)$$

I considered some things in my solving, but I couldn't get the solution to the problem. I would like to know if those assumptions were wrong.

First, I considered $$\log(x)-2 \implies \log(x)-\log(100) \implies \log\left(\frac{x}{100}\right)$$

I did proceed $$\left(\frac{x}{10}\right)^{\log(x)-2}<100 \Longleftrightarrow \left(\frac{x}{10}\right)^{\log\left(\frac{x}{100}\right)}<100 \Longleftrightarrow \frac{x^{\log(\frac{x}{100})}}{\frac{x}{100}}<100 \Longleftrightarrow \frac{100x^{\log(\frac{x}{100})}}{x}<100$$

From $$\log(x), x>0$$, therefore I can multiply both sides by $$x$$ $$x^{\log(\frac{x}{100})}

More directly, one can write the sequence of equivalent inequalities $$10^{(\log(x)-1)(\log(x)-2)} = \left(\frac{x}{10}\right)^{\log(x)-2}<100=10^2 \\ (\log(x)-1)(\log(x)-2) < 2 \\ \log(x)(\log(x)-3) < 0 \\ 0 < \log(x) < 3 \\ 1 < x < 1000$$
As for your solution, it is fine until $$x^{\log(\frac{x}{100})} which is only true for $$x>1$$. If instead $$0, then we'd have $$x^{\log(\frac{x}{100})}1$$
Consider the more general case of $$\left(\frac{x}{10}\right)^{\log(x)-2}=k$$ Let $$x=e^{y+2}$$ which makes th lhs to be $$10^{-y} \exp(y^2+2y)=\exp[y^2-(\log(10)-2)y]$$ So, you face a quadratic equation $$y^2-(\log(10)-2)y-\log(k)=0$$
Solve it for $$y$$ and when done, reuse $$x=e^{y+2}$$.
For $$k=100$$, this should give $$x_{max}=73.89$$.