Change-of-Base Formula for Logarithms: Resulting in a quadratic equation

$\frac{2\log_7(x+1)}{\log_7e}=\ln(3x^2+1)$

$x=\text{?}$

$1,0$

$\frac{\log_4(x^2+8x+1)}{\log_48}=2$

$x=\text{?}$

$-4\pm\sqrt{79}$

Find X

$\frac{\log_84x+\log_8(x+2)}{\log_83}=3$

$-1+\frac{\sqrt{31}}{2}$

Given X>1 find the domain X where it is satisfied:

$\frac{\log_3(x^2+5x+4)}{\log_3x}<\log_x12$

1 < x < -2.5+\frac{\sqrt{57}}{2}

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

$x=\text{?}$

$1.25$