Tricky Quadratic Logarithm

$\begin{aligned} \log_4 x^2 + \log_x 16 - 5 & < 0 \\ \log_4 x^2 + \frac{\log_4 16}{\log_4 x} - 5 & < 0 \\ 2 \log_4 x + \frac{2}{\log_4 x} - 5 & < 0 \\ 2 \log_4^2 x - 5 \log_4 x + 2 & < 0 \\ \left(2\log_4 x - 1 \right) \left(\log_4 x - 2 \right) & < 0 \\ \Rightarrow (x - 2)(x -16) & < 0 \end{aligned}$

Therefore, the integer solutions are $3,4,5,...15$ , a total of $\boxed{13}$ solutions.

$\log_{4} x^2\ + \log_{x} 16\ - 5 < 0$

$2\log_{4} x\ + 2\log_{x} 4\ - 4 - 1 < 0$

$2\log_{4} x\ + 2\log_{x} 4\ - 4 - (\log_{4} x)(\log_{x} 4) < 0$

Substitute $A$ = $\log_{4} x$ and $B$ = $\log_{x} 4$ :

$2A + 2B - 4 - AB < 0$

$0 < AB - 2A - 2B + 4$

$0 < (A - 2)(B - 2)$

$0 < (\log_{4} x\ - 2)(\log_{x} 4\ - 2)$

$A$ = $\log_{4} x = 2$ when $x = 16$ , and $B$ = $\log_{x} 4 = 2$ when $x = 2$ .

Now if $x > 16$ , then $A - 2 > 0$ but $B - 2 < 0$ .

If $2$ < $x$ < $16$ , then $A - 2 < 0$ and $B - 2 < 0$ .

If $x < 2$ , then $A - 2 < 0$ but $B - 2 > 0$ .

Hence, the positive integer solutions will be in the domain (2 , 16) or $x$ $\epsilon$ {3 , 4 , 5 , 6 ,..., 14 , 15} .

As a result, there are $\boxed{13}$ positive integer solutions to this logarithm inequality.

Graphing, the curve crosses y=0 at x=2 and x=16 , remaining y<0 in this range. $\therefore$ x=3, 4, . . . 15. Quant 13 integer x.