Solve the logarithmic equation $\log \left(4x^2\right)=128$

$\log\left(x\right)=3$

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

$\log_3\left(x+4\right)+\log_3\left(x-4\right)=2$

$\log\left(4X^2\right)=128$

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

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

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