Resolver la ecuación logarítmica $\ln\left(x\ln\left(x\right)\right)+x-3=1$

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

$\log_{2}\left(\left(1-x\right)\right)=-2$

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

$\log_{2}\left(\left(x^2-5x-4\right)\right)=1$

$\log_{144}\left(12\right)=x$

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

$3x\cdot \log_{3}\left(4\right)=6$