Inverting a Log Function: Using variables

$(\log_7x)^{-1}=$

$\log_x7$

$\frac{\frac{2x}{\log_89}}{\log_98}=$

$2x$

$\frac{4a^2}{\log_79}\colon\log_97=16$

Calculate a.

$\pm2$

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

$12$

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

$x=\text{?}$

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

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

$x=\text{?}$

$1.25$

Find X

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

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