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x-y-y-x-3-1-x-y-3-2-solve-the-equation-




Question Number 14483 by chux last updated on 01/Jun/17
x^y +y^x =3.....(1)  x+y=3.....(2)    solve the equation
$$\mathrm{x}^{\mathrm{y}} +\mathrm{y}^{\mathrm{x}} =\mathrm{3}…..\left(\mathrm{1}\right) \\ $$$$\mathrm{x}+\mathrm{y}=\mathrm{3}…..\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17
(x,y)=(1,2)
$$\left({x},{y}\right)=\left(\mathrm{1},\mathrm{2}\right) \\ $$
Commented by chux last updated on 01/Jun/17
please show workings
$$\mathrm{please}\:\mathrm{show}\:\mathrm{workings} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17
y^x =3−x^y ⇒x^y <3⇒ylogx<log3  x^y =3−y^x ⇒y^x <3⇒xlogy<log3  ylogx−xlogy<0⇒x^y <y^x ⇒x<y  x+y=3,x<y⇒x=1,y=2 .
$${y}^{{x}} =\mathrm{3}−{x}^{{y}} \Rightarrow{x}^{{y}} <\mathrm{3}\Rightarrow{ylogx}<{log}\mathrm{3} \\ $$$${x}^{{y}} =\mathrm{3}−{y}^{{x}} \Rightarrow{y}^{{x}} <\mathrm{3}\Rightarrow{xlogy}<{log}\mathrm{3} \\ $$$${ylogx}−{xlogy}<\mathrm{0}\Rightarrow{x}^{{y}} <{y}^{{x}} \Rightarrow{x}<{y} \\ $$$${x}+{y}=\mathrm{3},{x}<{y}\Rightarrow{x}=\mathrm{1},{y}=\mathrm{2}\:. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Jun/17
Commented by chux last updated on 01/Jun/17
thanks.... please can it be solved  by any numerical method.
$$\mathrm{thanks}….\:\mathrm{please}\:\mathrm{can}\:\mathrm{it}\:\mathrm{be}\:\mathrm{solved} \\ $$$$\mathrm{by}\:\mathrm{any}\:\mathrm{numerical}\:\mathrm{method}.\: \\ $$$$ \\ $$
Commented by chux last updated on 01/Jun/17
please help me sirs
$$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sirs} \\ $$
Commented by mrW1 last updated on 01/Jun/17
For equations like x^y +y^x =a there is  no general analytical solution. For  special values as in your case you  can find the right solution  through “try and error”, “guess” or  intuition. But for any other values  you can only get numerical approximation  of the solution. An usual way is  using graphics.  For example  x^y +y^x =3.5  x+y=5.5  you can get 2 solutions using graphic  x≈0.7544 or 4.7456  y≈4.7456 or 0.7544  Here you can not guess the solution.
$${For}\:{equations}\:{like}\:{x}^{{y}} +{y}^{{x}} ={a}\:{there}\:{is} \\ $$$${no}\:{general}\:{analytical}\:{solution}.\:{For} \\ $$$${special}\:{values}\:{as}\:{in}\:{your}\:{case}\:{you} \\ $$$${can}\:{find}\:{the}\:{right}\:{solution} \\ $$$${through}\:“{try}\:{and}\:{error}'',\:“{guess}''\:{or} \\ $$$${intuition}.\:{But}\:{for}\:{any}\:{other}\:{values} \\ $$$${you}\:{can}\:{only}\:{get}\:{numerical}\:{approximation} \\ $$$${of}\:{the}\:{solution}.\:{An}\:{usual}\:{way}\:{is} \\ $$$${using}\:{graphics}. \\ $$$${For}\:{example} \\ $$$${x}^{{y}} +{y}^{{x}} =\mathrm{3}.\mathrm{5} \\ $$$${x}+{y}=\mathrm{5}.\mathrm{5} \\ $$$${you}\:{can}\:{get}\:\mathrm{2}\:{solutions}\:{using}\:{graphic} \\ $$$${x}\approx\mathrm{0}.\mathrm{7544}\:{or}\:\mathrm{4}.\mathrm{7456} \\ $$$${y}\approx\mathrm{4}.\mathrm{7456}\:{or}\:\mathrm{0}.\mathrm{7544} \\ $$$${Here}\:{you}\:{can}\:{not}\:{guess}\:{the}\:{solution}. \\ $$
Commented by chux last updated on 01/Jun/17
Thanks ..... I really appreciate this.
$$\mathrm{Thanks}\:…..\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{this}. \\ $$

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