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 $x$
$$x^{\log(\frac{x}{100})}<x \Longleftrightarrow  \log \left(\frac{x}{100}\right)<1 \Longleftrightarrow  \frac{x}{100}<10 \Longleftrightarrow  \boxed{x<1000}$$

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 $x_{max}=73.89$.
