Question Number 55969 by Muhammad bello last updated on 07/Mar/19
$$\:\:\:\boldsymbol{\mathrm{solve}}\:\:\:\boldsymbol{\mathrm{for}}\:\:\:\boldsymbol{\mathrm{x}}\:\:\:\&\:\:\boldsymbol{\mathrm{y}} \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{logy}}} \:\:\:=\:\:\mathrm{4} \\ $$$$\:\: \\ $$$$\:\:\:\:\boldsymbol{\mathrm{xy}}\:\:\:\:=\:\:\:\:\mathrm{40}\: \\ $$
Answered by MJS last updated on 07/Mar/19
$${x}^{\mathrm{ln}\:{y}} =\mathrm{4} \\ $$$$\mathrm{ln}\:{x}\:\mathrm{ln}\:{y}\:=\mathrm{ln}\:\mathrm{4}\:\:\left(\mathrm{1}\right) \\ $$$${xy}=\mathrm{40} \\ $$$$\mathrm{ln}\:{x}\:+\mathrm{ln}\:{y}\:=\mathrm{ln}\:\mathrm{40}\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\:{pq}={a}\:\Rightarrow\:{q}=\frac{{a}}{{p}} \\ $$$$\left(\mathrm{2}\right)\:{p}+{q}={b}\:\Rightarrow\:{p}+\frac{{a}}{{p}}={b}\:\Rightarrow\:{p}^{\mathrm{2}} −{bp}+{a}=\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:{p}=\frac{{b}}{\mathrm{2}}\pm\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{a}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{q}=\frac{{b}}{\mathrm{2}}\mp\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{a}}}{\mathrm{2}} \\ $$$$\mathrm{ln}\:{x}\:=\frac{\mathrm{ln}\:\mathrm{40}}{\mathrm{2}}\pm\frac{\sqrt{\left(\mathrm{ln}\:\mathrm{40}\right)^{\mathrm{2}} −\mathrm{4ln}\:\mathrm{4}}}{\mathrm{2}} \\ $$$$\mathrm{ln}\:{y}\:=\frac{\mathrm{ln}\:\mathrm{40}}{\mathrm{2}}\mp\frac{\sqrt{\left(\mathrm{ln}\:\mathrm{40}\right)^{\mathrm{2}} −\mathrm{4ln}\:\mathrm{4}}}{\mathrm{2}} \\ $$$$\Rightarrow\:\left({x}\approx\mathrm{1}.\mathrm{52913}\:\wedge\:{y}\approx\mathrm{26}.\mathrm{1587}\right)\:\vee\:\left({x}\approx\mathrm{26}.\mathrm{1587}\:\wedge\:{y}\approx\mathrm{1}.\mathrm{52913}\right) \\ $$