Determine all real numbers x > 0 for which log equation is true. 
Determine all real numbers x > 0 for which
  $$\log_4 x - \log_x 16 = \frac{7}{6} - \log_x 8$$

Attemp at solution:
$$\frac{1}{\log_x 4} -\log_x 16 + \log_x 8 = \frac{7}{6}$$
$$\frac{1}{\log_x 4} + \log_x \frac{8}{16} = \frac{7}{6}$$
$$\frac{1+\log_x \frac{1}{2} \cdot \log_x 4}{\log_x 4}  = \frac{7}{6}$$
$$\frac{1+(-2\log_x 2\cdot \log_x 2 )}{\log_x 4}  = \frac{7}{6}$$
$$\frac{1+(-2\log_x 2\cdot \log_x 2 )}{2\log_x 2}  = \frac{7}{6}$$
Not sure how to solve further 

Let
$$
u=\ln x,
$$
then the original equation becomes
$$
\dfrac{u}{2\ln2}-\dfrac{4\ln2}{u}=\dfrac76-\dfrac{3\ln2}{u},
$$
i.e.
$$
3u^2-(7\ln2)u-6(\ln2)^2=0.
$$
Then solution are
$$
u=\dfrac{7\ln2\pm\sqrt{49(\ln2)^2+72(\ln2)^2}}{6}=\dfrac{7\ln2\pm11\ln2}{6}=-\dfrac23\ln2,3\ln2=\ln2^{-2/3},\ln8.
$$
Hence
$$
x=2^{-2/3},8.
$$
