$$\frac { 1 } { \log _ { a } n } + \frac { 1 } { \log _ { c } n } = \frac { 2 } { \log _ { b } n } , b ^ { 2 } = a c$$
$a=\frac{b^{2}}{c}$
$n\in \mathrm{C}\setminus 1,0$
$c\in \mathrm{C}$
$b\in \mathrm{C}\setminus 1,0$
$\left\{\begin{matrix}a=\frac{b^{2}}{c}\text{, }n\in \left(0,1\right)\cup \left(1,\infty\right)\text{, }c>1\text{, }b\in \left(0,1\right)\cup \left(1,\infty\right)\text{, }&b\neq 0\text{ and }|b|\leq 1\\a=\frac{b^{2}}{c}\text{, }n\in \left(0,1\right)\cup \left(1,\infty\right)\text{, }c>b^{2}\text{, }b\in \left(0,1\right)\cup \left(1,\infty\right)\text{, }&|b|>1\\a=\frac{b^{2}}{c}\text{, }n\in \left(0,1\right)\cup \left(1,\infty\right)\text{, }c\in \left(1,b^{2}\right)\text{, }b\in \left(0,1\right)\cup \left(1,\infty\right)\text{, }&b<-1\\a=\frac{b^{2}}{c}\text{, }n\in \left(0,1\right)\cup \left(1,\infty\right)\text{, }c\in \left(0,b^{2}\right)\text{, }b\in \left(0,1\right)\text{, }&b>-1\text{ and }b<0\\a=\frac{b^{2}}{c}\text{, }n\in \left(0,1\right)\cup \left(1,\infty\right)\text{, }c\in \left(0,1\right)\text{, }b\in \left(0,1\right)\text{, }&b\leq -1\\a=\frac{b^{2}}{c}\text{, }n\in \left(0,1\right)\cup \left(1,\infty\right)\text{, }c\in \left(b^{2},1\right)\text{, }b\in \left(0,1\right)\text{, }&b\neq 0\text{ and }|b|<1\end{matrix}\right.$