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\log x=5\log a+2\log b-0,5\log c-\log8

\log x=2-\log a+3\log c-2\log\left(a-c\right)

\ln x=-3\ln a-\ln b+\frac{1}{3}\ln c

\ln x=1+\ln10-2\ln7+4\ln2

\log2-\log x+2\log3=0

\log(x+4)-\log(2x-3)=\log2

\log_3(x+1)-\log_3(x-1)=2

2\log_2x-3\log_2^2x+1=0

\log^2x-8,6\log x+4,8=0

$\left\{\begin{array}{l}x+2y=5\\ 2x-3y=3\end{array}\right.$

$\left\{\begin{array}{l}y+3=2x\\ 4x-2y=3\end{array}\right.$

$\left\{\begin{array}{l}5\left(x+y\right)-3\left(x-y\right)=4\\ \frac{x}{2}-\frac{y - 1}{3}=1\end{array}\right.$

4x^2-3x+2>0

x\left(5-x\right)\ge6

6x\le x^2+5

\left(2x+3\right)^2>9

x^2<8

3x^2>9x

x^4-3x^2=4

x^4=3x^2-2

x^3-3x^2=x-3

x^3+5x^2-x=5

(2x-1)^3+4x-2=0

\frac{1}{x^2-x-6}+\frac{2}{x^2+x-6}=\frac{4}{x^2-9}-\frac{2}{x^2-4}

\frac{20}{x^2+3x+2}-\frac{2}{x^2-3x+2}=\frac{8}{x^2-1}-\frac{5}{x^2-4}

\frac{1}{x^2+2x+4}-\frac{1}{x^2+2x+5}=\frac{1}{12}

\frac{1}{x^2-3x+3}+\frac{2}{x^2-3x+4}=\frac{6}{x^2-3x+5}

\left|x-1\right|-2x=4

\left|3x+1\right|-2x=0

2\left|x-2\right|=\left|x\right|

\left|x\right|+\left|x-5\right|=5

\left|x-2\right|-\left|5+x\right|=3

\left|x^2-1\right|+x=5

7^{x-1}>7^{2x+4}

0,99^x<0,99^8

e^{0,1}<e^{10x+7}

0,5^{x^2}>0,5^{5x}

0,0625<0,5^{5-x^2}

729>9^{x^2-1}

\log_x^29+\log_x9-6=0

\log_x^28+\log_x4+\log_x2=0

5^{\log x}=7

x^{\log0,25x}=1

e^{\log x}=10

$\begin{array}{l}\left\{\begin{array}{l}\frac{3}{x}+\frac{2}{y}=1\\ \frac{1}{x}-\frac{3}{y}=2\end{array}\right.\end{array}$

$\left\{\begin{array}{l}\frac{x + y}{x - y}=\frac{3}{7}\\ xy-2y=-14\end{array}\right.$

$\left\{\begin{array}{l}{x}^2=21+y^2\\ x+y=7\end{array}\right.$

$\left\{\begin{array}{l}{x}^2-y=14\\ 3x+y=4\end{array}\right.$

$\left\{\begin{array}{l}{y}^2-xy=12\\ x^2-xy=-3\end{array}\right.$

$\left\{\begin{array}{l}{x}^2+4xy+4y^2=100\\ x-2y=2\end{array}\right.$

$\left\{\begin{array}{l}6x+2y-z=2\\ 4x-y+3z=-3\\ 3x+2y-2z=3\end{array}\right.$

$\left\{\begin{array}{l}x+2y=1\\ x+3y+2z=1\\ y-2z=1\end{array}\right.$

$\left\{\begin{array}{l}x-y+z=2\\ 2x-3y-z+1=0\\ 2y+z=x\end{array}\right.$

\left(5-x\right)\left(1+x\right)\left(2x+6\right)<0

\left(3-x\right)^2\left(4-x\right)>0

\left(x^2+x+13\right)\left(x-2\right)\le0

\left(2x+1\right)\left(x^2-3x-4\right)\left(x-6\right)\le0

x^3\le4x\left(x-1\right)

x^3<\frac{1}{x}

\left|x-1\right|\ge4