# Logging out?

Algebra Level 3

$$x > e^{\ln {4}} + \frac{\log _{111} {1024}}{\log _{12321}{1024}} + 2 \times \log _{9} {0.\overline{3}}$$

$$x < \log _{2} {512^{34}} + \frac{\log _{49} {25^{4096}} }{\log _{\sqrt{7}} {5}} - 4086 \times \log _{25}{5}$$

The solution to this inequality is $$x \in (m, n)$$ (of course, $$n > m$$). Enter answer as $$n - m$$

Details

$$0.\overline{3} = 0.33333....$$

×