Inverting a Log Function: Using multiple rules

$\frac{1}{\ln4}\cdot\frac{1}{\log_810}=$

$\frac{3}{2}\log e$

$\frac{\log_311}{\log_34}+\frac{1}{\ln3}\cdot2\log3=$

$\log_411+\log e^2$

$\frac{2\log_78}{\log_74}+\frac{1}{\log_43}\times\log_29=$

$7$

$-3(\frac{\ln4}{\ln5}-\log_57+\frac{1}{\log_65})=$

$3\log_5\frac{7}{24}$

$\frac{\log_76-\log_71.5}{3\log_72}\cdot\frac{1}{\log_{\sqrt{8}}2}=$

$1$

$\frac{\log_8x^3}{\log_8x^{1.5}}+\frac{1}{\log_{49}x}\times\log_7x^5=$

$12$

$\frac{1}{\log_x3}\times x^2\log_{\frac{1}{x}}27+4x+6=0$

$x=\text{?}$

$\frac{2}{3}+\frac{\sqrt{22}}{3}$

$\frac{1}{\log_{2x}6}\times\log_236=\frac{\log_5(x+5)}{\log_52}$

$x=\text{?}$

$1.25$

Find X

$\frac{1}{\log_{x^4}2}\times x\log_x16+4x^2=7x+2$

$\frac{-9+\sqrt{113}}{8}$