Question Number 199399 by Calculusboy last updated on 03/Nov/23
$$\boldsymbol{{Solve}}:\:\boldsymbol{{log}}_{\mathrm{2}} \boldsymbol{{r}}+\boldsymbol{{log}}_{\mathrm{3}} \boldsymbol{{p}}=\mathrm{3} \\ $$$$\boldsymbol{{p}}+\boldsymbol{{r}}=\mathrm{11}\:\:\:\boldsymbol{{fund}}\:\boldsymbol{{p}}\:\boldsymbol{{and}}\:\boldsymbol{{r}}. \\ $$
Answered by mr W last updated on 03/Nov/23
$${one}\:{solution}\:\left({p}=\mathrm{9},\:{r}=\mathrm{2}\right)\:{can}\:{be}\:“{seen}'', \\ $$$${other}\:{solution}\:\left({r}\approx\mathrm{10}.\mathrm{3335},\:{p}\approx\mathrm{0}.\mathrm{6665}\right) \\ $$$${can}\:{only}\:{be}\:{approximated}. \\ $$$${generally}\:{there}\:{is}\:{no}\:{exact}\:{solution}. \\ $$