Question Number 7545 by Tawakalitu. last updated on 03/Sep/16
$${y}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} \:=\:\mathrm{2}^{{x}} \:+\:\mathrm{2}^{{y}} \\ $$$${Find}\:{the}\:{possible}\:{greatest}\:{value}\:{of}\:\mid{x}\:−\:{y}\mid \\ $$
Commented by Yozzia last updated on 03/Sep/16
$$\mathrm{2}^{{x}−{y}} +\mathrm{1}=\mathrm{2}^{−{y}} \left({y}^{\mathrm{2}} +{x}^{\mathrm{2}} \right) \\ $$$$\left({x}−{y}\right){ln}\mathrm{2}={ln}\left(\mathrm{2}^{−{y}} \left({y}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)−\mathrm{1}\right) \\ $$$$\mid{x}−{y}\mid=\frac{\mid{ln}\left(\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}^{{y}} }−\mathrm{1}\right)\mid}{{ln}\mathrm{2}}=\frac{\mid{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}^{{y}} \right)−{yln}\mathrm{2}\mid}{{ln}\mathrm{2}} \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 03/Sep/16
$${Thanks}\:{sir}\:{but}\:{i}\:{think}\:{it}\:{need}\:{to}\:{find}\:{a}\:{numeric}\:{value}\:…..\:{i}\:{mean}\: \\ $$$${a}\:{number} \\ $$
Commented by FilupSmith last updated on 03/Sep/16
$$\mathrm{solve}\:\mathrm{for}\:{x}\:\mathrm{or}\:{y},\:\mathrm{find}\:\mathrm{global}\:\mathrm{maximum}? \\ $$