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> How many integers $x$ will satisfy the inequality $$\log\_{20}(x-30)+\log\_{20}(70-x)<2 ?$$
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Using the properties of logarithms we get

$$\log\_{20}((x-30)(70-x))<2$$
$$\Rightarrow \log(-x^2+100x-2100)<2.$$

From here we have

$$-x^2+100x-2100<20^{2}$$
$$\Rightarrow-x^2+100x-2500<0$$
$$\Rightarrow x^2-100x+2500>0$$
$$\Rightarrow (x-50)^2>0.$$

Hence $x \neq 50$. From before we had the restriction that $30 < x <70$ from here I deduced that the answer would be $37$ since we have $31 \leqslant x \leqslant 69$ and $69-31=38$ and then we take off the $50$, hence totalling $37$ integers. The actual answer was $38$, I really cannot see what am I missing here?