Question Number 82821 by VBash last updated on 24/Feb/20
$${Log}_{{y}} \:{x}+{Log}_{{x}} \:{y}\:=\mathrm{64} \\ $$$${find}\:{x}\:{and}\:{y} \\ $$
Commented by MJS last updated on 24/Feb/20
$$\mathrm{one}\:\mathrm{equation}\:\mathrm{in}\:\mathrm{two}\:\mathrm{unknown}\:\mathrm{gives}\:\mathrm{a} \\ $$$$\mathrm{parametric}\:\mathrm{solution} \\ $$$$\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:{y}}+\frac{\mathrm{ln}\:{y}}{\mathrm{ln}\:{x}}=\mathrm{64};\:{x}>\mathrm{0}\wedge{y}>\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{ln}\:{x}\right)^{\mathrm{2}} −\left(\mathrm{64ln}\:{y}\right)\mathrm{ln}\:{x}\:+\left(\mathrm{ln}\:{y}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{ln}\:{x}\:=\left(\mathrm{32}\:\pm\sqrt{\mathrm{1023}}\right)\mathrm{ln}\:{y} \\ $$$$\Rightarrow \\ $$$${x}={y}^{\mathrm{32}\pm\sqrt{\mathrm{1023}}} \:\forall\:{y}>\mathrm{0} \\ $$