Find integral solutions of this inequality$$\left (\frac{1}{10}\right )^{\log_{x-3}^{x^2-4x+3}} \ge 1$$

My try :

I took $\log$ on both sides and got $\log_{x-3}^{x-1} \le-1$ but from here I'm unable to move on. What can be the way to solve it from here and also, what are other ways to solve it


You have almost solved it. You just have to find for which values of $x$, the below inequality holds$$\log_{x-3}^{x-1}+1\le0.$$

It's obvious that $x>3$ and $x\neq4$ so we are going to have two intervals$$\left (3,4\right ),\\\left (4,+\infty\right ).$$

The inequality will not hold in the second interval because the base is larger than $1$ so the left hand side of the inequality will remain positive. But in the first interval the base is smaller than $1$ so the logarithm value will be negative.

Now you should solve this equation $\log_{x-3}^{x-1} = -1$ (I leave this part to you). There will be two values for $x$, one positive (name it $x_1$) and one negative (name it $x_2$). The final answer will be $\left (3,x_1\right )$


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