$\log_{3}\left(x+4\right)+\log_{3}\left(x-4\right)=2$

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

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

$2log\left(x\right)-log\left(x+6\right)=0$

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

$\log_2\left(x-2\right)+\log_2\left(x+1\right)=2$

$2\cdot log\left(x\right)-1\cdot log\left(x+6\right)=0$

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