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Question

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

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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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