Given that ${{\log }_{8}}(p+2)+{{\log }_{8}}q=r-\tfrac{1}{3}$ and ${{\log }_{2}}(p-2)-lo{{g}_{2}}q=2r+1$ , show that ${{p}^{2}}=4+{{32}^{r}}$.
$\begin{align} & {{\log }_{8}}(p+2)+{{\log }_{8}}q=r-\tfrac{1}{3} \\ & {{\log }_{8}}(p+2)q=r-\tfrac{1}{3} \\ & (p+2)q={{8}^{r-\tfrac{1}{3}}} \\ & (p+2)q={{2}^{3(r-\tfrac{1}{3})}} \\ & q=\frac{{{2}^{3(r-\tfrac{1}{3})}}}{(p+2)}----(i) \\ & {{\log }_{2}}(p-2)-lo{{g}_{2}}q=2r+1 \\ & {{\log }_{2}}\frac{(q-2)}{q}=2r+1 \\ & \frac{p-2}{q}={{2}^{2r+1}} \\ & q=\frac{p-2}{{{2}^{2r+1}}}----(ii) \\ & \text{Equating (i) and (ii)} \\ & \frac{{{2}^{3(r-\tfrac{1}{3})}}}{(p+2)}=\frac{(p-2)}{{{2}^{2r+1}}} \\ & (p-2)(p+2)=({{2}^{3(r-\tfrac{1}{3})}})\cdot ({{2}^{2r+1}}) \\ & {{p}^{2}}-4=({{2}^{3r-1}})({{2}^{2r+1}}) \\ & {{p}^{2}}-4={{2}^{3r-1+2r+1}} \\ & {{p}^{2}}-4={{2}^{5r}} \\ & {{p}^{2}}=4+{{2}^{5r}}=4+{{({{2}^{5}})}^{r}} \\ & {{p}^{2}}=4+{{32}^{r}} \\\end{align}$