There are some logarithmic inequalities that I cannot solve and would be very happy if you show me the way and explain a bit what should I do! I hope they won't be a problem for you and you would help me!
$\left(x^2+x+1\right)^x<1$
$\frac{lg\left(11\right)-lg\left(-x^2-12x\right)}{lg\left(x+5\right)}>0$
$\log _{\frac{1}{\sqrt{5}}}\left(6^{x+1}-36^x\right)\ge -2$
Thank you in advance!
Hint for the first one
Since $\ln$ is increasing (preserves inequalities) you have
$$\left(x^2+x+1\right)^x<1\iff x\ln \left(x^2+x+1\right)=\ln \left(x^2+x+1\right)^x<\ln 1=0.$$
Now, $x\ln \left(x^2+x+1\right)<0$ implies $x$ and $\ln \left(x^2+x+1\right)$ have different sign. Can you get the sign of $\ln \left(x^2+x+1\right)$ in function of the sign of $x?$
Hint for the third one
$$\log _{\frac{1}{\sqrt{5}}}\left(6^{x+1}-36^x\right)\ge -2\iff 6^{x+1}-36^x\ge 5=(1/\sqrt{5})^{-2}\iff 6^{2x}-6\cdot 6^x+5\le 0.$$