Determine all real numbers x > 0 for which log equation is true.

Question

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

Answer 1

$1-2{(\log_x2) }^2=\frac73\cdot \log_x2$

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 $$x=2^{-2/3} or \ \ x=8$$

Answer 2

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 $$ x=2^{-2/3},8. $$

Answer 3

Considering the more general case $$\log_a( x) - \log_x(b) = c - \log_x(d)$$ and transform each logarithm to base $e$. This will give $$\frac{\log (x)}{\log (a)}-\frac{\log (b)}{\log (x)}=c-\frac{\log (d)}{\log (x)}$$ Define $\log(x)=y$ to get $$\frac{y}{\log (a)}-\frac{\log (b)}{y}=c-\frac{\log (d)}{y}$$ This will lead to $$y^2-c \log (a)\,y+\log (a) \log \left(\frac db\right)=0$$ which may have $0,1,2$ solution in the real domain depending on the value of $$\Delta=\log(a)\left(c^2\log(a)-4\log \left(\frac db\right)\right)$$