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Question Number 167027 by Tawa11 last updated on 04/Mar/22

Commented by cortano1 last updated on 05/Mar/22

$$\{\begin{array}{l}\log {}_8(\mathrm{p}+2)+\log {}_8\mathrm{q}=\mathrm{r}-\frac{1}{3}\\ \log {}_8(\mathrm{p}-2)-\log {}_8\mathrm{q}=2\mathrm{r}+1\end{array}$$$$\Rightarrow \left(1\right)+\left(2\right)\equiv \log {}_8({\mathrm{p}}^2-4)=3\mathrm{r}+\frac{2}{3}$$$$\Rightarrow {\mathrm{p}}^2-4=8^{3\mathrm{r}+\frac{2}{3}}$$$$\Rightarrow {\mathrm{p}}^2=4+2^{9\mathrm{r}+2}$$$$\Rightarrow {\mathrm{p}}^2=4+4.{512}^{\mathrm{r}}$$$$\Rightarrow {\mathrm{p}}^2=4(1+{512}^{\mathrm{r}})$$$$\Rightarrow \mathrm{p}=2\sqrt{1+{512}^{\mathrm{r}}}$$

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