Dane są liczby:

$a=1+{\log }_{2}3$

$b={\log }_{x}36$

$c=\frac{4}{3}{\log }_{8}6$

gdzie $x\in {\mathbb{R}}_{+}\setminus \left\{0\right\}$.

Aby podane liczby tworzyły ciąg geometryczny, musi zajść:

$ac=b^2$

$\left(1+{\log }_{2}3\right)\cdot \frac{4}{3}{\log }_{8}6={\left({\log }_{x}36\right)}^2$

$\left({\log }_{2}2+{\log }_{2}3\right)\cdot \frac{4}{3}{\log }_{8}6={\left({\log }_{x}36\right)}^2$

${\log }_{2}6\cdot \frac{4}{3}\cdot \frac{{\log }_{2}6}{{\log }_{2}8}={\left({\log }_{x}36\right)}^2$

$\frac{4}{3}\cdot \frac{{\left({\log }_{2}6\right)}^2}{3}={\left({\log }_{x}36\right)}^2$

$\frac{4}{9}{\left({\log }_{2}6\right)}^2={\left({\log }_{x}36\right)}^2$

${\left(\frac{2}{3}{\log }_{2}6\right)}^2={\left({\log }_{x}36\right)}^2$

$\frac{2}{3}{\log }_{2}6={\log }_{x}36\vee \frac{2}{3}{\log }_{2}6=-{\log }_{x}36$

$\frac{2}{3}{\log }_{2}6=\frac{{\log }_{2}36}{{\log }_{2}x}\vee \frac{2}{3}{\log }_{2}6=-\frac{{\log }_{2}36}{{\log }_{2}x}$

$\frac{2}{3}{\log }_{2}6=\frac{2{\log }_{2}6}{{\log }_{2}x}\vee \frac{2}{3}{\log }_{2}6=-\frac{2{\log }_{2}6}{{\log }_{2}x}$

$\frac{1}{3}=\frac{1}{{\log }_{2}x}\vee \frac{1}{3}=-\frac{1}{{\log }_{2}x}$

${\log }_{2}x=3\vee {\log }_{2}x=-3$

$x=8\vee x=\frac{1}{8}$

Zatem podane liczby tworzą ciąg geometryczny, gdy:

$\underline{\underline{x=8\vee x=\frac{1}{8}}}$