<s>[INST] 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 
 [/INST] $1-2{(\log_x2) }^2=\frac73\cdot \log_x2$
Solve the resulting quadratic using $y= \log_x2$
$$1-2y^2=\frac73y$$
$$2y^2 +\frac73y-1=0$$
$$6y^2+7y-3=0$$
$$6y^2 +9y-2y-3=0$$
$$(3y-1)(2y+3)=0$$
Which gives 
$$x=2^{-2/3} or \ \ x=8$$
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