tes

$$\begin{array}{rl}{\log }_b(x)& ={\log }_b\left(c\right){\log }_c\left(x\right)\\ & =\frac{{\log }_c\left(x\right)}{{\log }_c\left(b\right)}\end{array}$$

\begin{aligned}\boxed {\bar x = \frac{1}{n} \sum \limits_{i=1}^{n} {x_i}}

${\displaystyle \overline{x}=\frac{1}{n}\sum _{i=1}^n{x}_i}$ \end{aligned