Question Number 182632 by cortano1 last updated on 12/Dec/22
$$\:\:\mathrm{If}\:\mathrm{log}\:_{\mathrm{11}} \left(\mathrm{3p}\right)=\mathrm{log}\:_{\mathrm{13}} \left(\mathrm{q}+\mathrm{6p}\right)\:=\:\mathrm{log}\:_{\mathrm{143}} \left(\mathrm{q}^{\mathrm{2}} \right) \\ $$$$\:\mathrm{find}\:\frac{\mathrm{p}}{\mathrm{q}}. \\ $$
Answered by manxsol last updated on 12/Dec/22
$$\mathrm{3}{p}=\mathrm{11}^{{k}} \\ $$$${q}+\mathrm{6}{p}=\mathrm{13}^{{k}} \\ $$$${q}^{\mathrm{2}} =\mathrm{11}^{{k}} .\mathrm{13}^{{k}} \\ $$$$................. \\ $$$${q}^{\mathrm{2}} =\mathrm{3}{p}\left({q}+\mathrm{6}{p}\right) \\ $$$${q}^{\mathrm{2}} =\mathrm{3}{pq}+\mathrm{18}{p}^{\mathrm{2}} \\ $$$$\mathrm{1}=\mathrm{3}\left(\frac{{p}}{{q}}\right)+\mathrm{18}\left(\frac{{p}}{{q}}\right)^{\mathrm{2}} \:\:\:\:\:\frac{{p}}{{q}}={x} \\ $$$$\mathrm{18}{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{6}{x}−\mathrm{1}\right)\left(\mathrm{3}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\frac{{p}}{{q}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$