$\log_{8}\left(x\right)=\frac{1}{3}$

$\log_x\left(64\right)=3$

$log\left(x+1\right)=log\left(x-1\right)+3$

$\log\left(x-1\right)+\log\left(x+1\right)=2\log\left(x+2\right)$

$\log_x\left(\frac{1}{8}\right)=-3$

$\log_x\left(100\right)-\log_x\left(25\right)=2$

$\log_2\left(x\right)-\log_2\left(10\right)=\log_2\left(9.3\right)$

$\log\left(\left(y+2\right)\right)=1+\log\left(\left(2y-6\right)\right)$