Question Number 104835 by bemath last updated on 24/Jul/20
$$\begin{cases}{\mathrm{log}\:_{{p}} \left({q}\right)\:=\:{x}^{\mathrm{2}} }\\{\mathrm{log}\:_{{q}} \left({p}^{\mathrm{3}} \right)\:=\:{x}\:}\end{cases} \\ $$$${find}\:{x}\: \\ $$
Answered by john santu last updated on 24/Jul/20
$$\begin{cases}{{q}\:=\:{p}^{{x}^{\mathrm{2}} } \rightarrow{q}^{{x}} \:=\:\left({p}^{{x}^{\mathrm{2}} } \right)^{{x}} }\\{{p}^{\mathrm{3}} \:=\:{q}^{{x}} \rightarrow\:{p}^{\mathrm{3}} \:=\:{p}^{{x}.{x}^{\mathrm{2}} } }\end{cases} \\ $$$$\Rightarrow\:{x}^{\mathrm{3}} \:=\:\mathrm{3}\:;\:{x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:\:\:\left({JS}\:\circledast\right)\: \\ $$
Answered by Dwaipayan Shikari last updated on 24/Jul/20
$${log}_{{p}} \left({q}\right)={x}^{\mathrm{2}} \\ $$$$\mathrm{3}{log}_{{q}} \left({p}\right)={x} \\ $$$$\mathrm{3}{log}_{{q}} {p}.{log}_{{p}} {q}={x}^{\mathrm{3}} \\ $$$$\mathrm{3}={x}^{\mathrm{3}} \\ $$$${x}=\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$