Question

Solve the following inequality:
${\log }_{1/\sqrt{5}}(6^{x+1}-{36}^x)⩾-2$

Answer 1

${\log }_5(6^{x+1}-{36}^x)\le 1$

$6^{x+1}-{36}^x\le 5$

$let{6}^x=z$

$6z-z^2-5\le 0$

$z^2-6z+5\ge 0$

$(z-5)(z-1)\ge 0$

$z\epsilon (-\infty ,1]\cup [5,\infty )$

$and$

$6^{x+1}-{36}^x>0$

$let{6}^x=z$

$6z-z^2\ge 0$

$z^2-6z\le 0$

$z(z-6)\le 0$
$z\epsilon (0,6)$

$6z-z^2\ge 0$

$z^2-6z\le 0$

$z(z-6)\le 0$

$z\epsilon (0,6)$

$So,z\epsilon (0,1]\cup [5,6)$

$hence,$

$x\epsilon (-\infty ,0]\cup [{\log }_{6}5,1)$