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Question Number 7545 by Tawakalitu. last updated on 03/Sep/16

y^2  + x^2  = 2^x  + 2^y   Find the possible greatest value of ∣x − y∣

$${y}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} \:=\:\mathrm{2}^{{x}} \:+\:\mathrm{2}^{{y}} \\ $$$${Find}\:{the}\:{possible}\:{greatest}\:{value}\:{of}\:\mid{x}\:−\:{y}\mid \\ $$

Commented by Yozzia last updated on 03/Sep/16

2^(x−y) +1=2^(−y) (y^2 +x^2 )  (x−y)ln2=ln(2^(−y) (y^2 +x^2 )−1)  ∣x−y∣=((∣ln(((x^2 +y^2 )/2^y )−1)∣)/(ln2))=((∣ln(x^2 +y^2 −2^y )−yln2∣)/(ln2))

$$\mathrm{2}^{{x}−{y}} +\mathrm{1}=\mathrm{2}^{−{y}} \left({y}^{\mathrm{2}} +{x}^{\mathrm{2}} \right) \\ $$$$\left({x}−{y}\right){ln}\mathrm{2}={ln}\left(\mathrm{2}^{−{y}} \left({y}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)−\mathrm{1}\right) \\ $$$$\mid{x}−{y}\mid=\frac{\mid{ln}\left(\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}^{{y}} }−\mathrm{1}\right)\mid}{{ln}\mathrm{2}}=\frac{\mid{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}^{{y}} \right)−{yln}\mathrm{2}\mid}{{ln}\mathrm{2}} \\ $$$$ \\ $$

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

$${Thanks}\:{sir}\:{but}\:{i}\:{think}\:{it}\:{need}\:{to}\:{find}\:{a}\:{numeric}\:{value}\:.....\:{i}\:{mean}\: \\ $$$${a}\:{number} \\ $$

Commented by FilupSmith last updated on 03/Sep/16

solve for x or y, find global maximum?

$$\mathrm{solve}\:\mathrm{for}\:{x}\:\mathrm{or}\:{y},\:\mathrm{find}\:\mathrm{global}\:\mathrm{maximum}? \\ $$

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