Rules of Logarithms Combined: Resulting in a quadratic equation

$\log_7x^4-\log_72x^2=3$

?=x

$\log_ax-\log_ay=\log_a\frac{x}{y}$

$\log_7x^4-\log_72x^2=$

$\log_7\frac{x^4}{2x^2}=3$

$7^3=\frac{x^2}{2}$

We multiply by: $2$

$2\cdot7^3=x^2$

Extract the root

$x=\sqrt{680}=7\sqrt{14}$

$x=-\sqrt{680}=-7\sqrt{14}$

$-7\sqrt{14\text{ }}\text{ , }7\sqrt{14}$

$\log7x+\log(x+1)-\log7=\log2x-\log x$

$?=x$

Defined domain

x>0

x+1>0

x>-1

$\log7x+\log\left(x+1\right)-\log7=\log2x-\log x$

$\log\frac{7x\cdot\left(x+1\right)}{7}=\log\frac{2x}{x}$

We reduce by: $7$ and by $X$

$x\left(x+1\right)=2$

$x^2+x-2=0$

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

$x+2=0$

$x=-2$

Undefined domain x>0

$x-1=0$

$x=1$

Defined domain

$1$

Solve for X:

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

$2$

$\log_2x+\log_2\frac{x}{2}=5$

?=x

$8$

$\log_4x^2\cdot\log_716=2\log_78$

?=x

$\pm\sqrt{8}$

$\log_2(x^2+3x+3)\cdot\log_3\frac{1}{4}=-2\log_3(\frac{4x+2}{-2})$

?=x

$-1,-4$

Find X

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

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

$\log3x+\log(x-1)=3$

$?=x$

$18.8$

$\log_4x+\log_4(x+2)=2$

$-1+\sqrt{17}$

$\log_27\cdot\log_48\cdot\log_3x^2=\log_24\cdot\log_47\cdot\log_38$

?=x

$-2,2$

$\log7\times\ln x=\ln7\cdot\log(x^2+8x-8)$

?=x

$1$

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

$x=\text{?}$

$-3+\sqrt{10}$

Calculate X:

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

$-4+\sqrt{10}$

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

$x=\text{?}$

$-4\pm\sqrt{79}$

$\ln(4x+3)-\ln(x^2-8)=2$

?=x

$3.18$

?=a

$\ln(a+5)+\ln(a+7)=0$

$-6+\sqrt{2}$

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

$x=\text{?}$

$1,0$

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

$x=\text{?}$

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

$\log_4(3x^2+8x-10)-\log_4(-x^2-x+12.5)=0$

?=x

$-3.75,1.5$

$x=\text{?}$

$\ln(x+5)+\ln x≤\ln4+\ln2x$

0 < X \le 3