Finding the number of positive integers $x$ such that $\log_{(x/9)}\left(\frac{x^2}{3}\right)<6+\log_{(3)}\left(\frac{9}{x}\right)$

Question

Find the number of positive integers $x$, such that $$\log_{(x/9)}\left(\frac{x^2}{3}\right) < 6 + \log_{(3)}\left(\frac{9}{x}\right); 1≤x≤100, x≠9$$

Here's what I did:

Using Base Changing Theorem, the inequality can be written as $${\log(x^2/3)\over \log(x/9)} < 6+ {\log(9/x)\over \log(3)}$$ Simplifying it I ended up here: $$ 17(\log (3))²+8\log(x)\log(3)-(\log(x))^2>0$$

After eliminating $log3$, and solving the quadratic in $logx$, I got $$ (logx -4+ \sqrt33)(logx-4- \sqrt33)<0.$$ But this doesn't seem to be the right answer.

What should I do next?

Answer 1

Multiply the equation with -1, then the equation will look like P(log(x)) < 0. Find both the roots and region between the roots will belong to possible values of x which satisfies the given constraints.

Answer 2

I am not quite getting what you got. I shall use the shorthand expression $\log x=l$. $$\frac {2l-\log 3}{l-2\log 3}-\frac {2\log3-l}{\log 3}<6$$ $$\frac {l^2-2l\log 3 +3(\log 3)^2}{\log 3(l-2\log 3)}-6<0$$ $$\frac {l^2-8l\log 3+15(\log 3)^2}{l-2\log 3}<0$$ $$\frac {(l-3\log 3)(l-5\log 3)}{(l-2\log 3)}<0$$ Hence, $l \in (-\infty, 2\log 3)\cup (3\log 3,5\log 3)$ or $x\in (0,9)\cup (27, 243)$.

PS:

Here properties of logarithm have been used, including-

●$\log(\frac ab)=\log a-\log b$

●$\log(x^a)=a\log x$