Power Property of Logorithms: Resulting in a quadratic equation

Solve for X:

$\log_3(x+2)\cdot\log_29=4$

$2$

Calculate X:

$2\log(x+4)=1$

$-4+\sqrt{10}$

$2\log(x+1)=\log(2x^2+8x)$

$x=\text{?}$

$-3+\sqrt{10}$

$\frac{1}{2}\log_3(x^4)=\log_3(3x^2+5x+1)$

$x=\text{?}$

$-\frac{5}{4}\pm\frac{\sqrt{17}}{4}$

Find the domain X where the inequality exists

2\log_3x<\log_3(x^2+2x-12)

6 < x

Find the domain of X given the following:

\log_{\frac{1}{7}}(x^2+3x)<2\log_{\frac{1}{7}}(3x+1)

No solution

$\frac{1}{\log_{2x}6}\times\log_236=\frac{\log_5(x+5)}{\log_52}$

$x=\text{?}$

$1.25$

Find X

$\frac{1}{\log_{x^4}2}\times x\log_x16+4x^2=7x+2$

$\frac{-9+\sqrt{113}}{8}$