Question Number 155036 by Mr.D.N. last updated on 24/Sep/21
$$\:\mathrm{If}\:\:\boldsymbol{\mathrm{a}}^{\boldsymbol{\mathrm{b}}} =\boldsymbol{\mathrm{b}}^{\boldsymbol{\mathrm{a}}} \:\mathrm{and}\:\boldsymbol{\mathrm{a}}=\mathrm{2}\boldsymbol{\mathrm{b}}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\: \\ $$$$\:\:\mathrm{value}\:\mathrm{of}\:\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} \:? \\ $$
Answered by Rasheed.Sindhi last updated on 24/Sep/21
![(2b)^b =b^(2b) 2b=b^2 b^2 −2b=0 b(b−2)=0 b=0 , 2 a=2b⇒a=0 , 4 [Is (a,b)=(0,0) acceptable ? 0^0 =0^0 ? a^2 +b^2 =0^2 +0^2 =0] a^2 +b^2 =4^2 +2^2 =16+4=20](https://www.tinkutara.com/question/Q155037.png)
$$\left(\mathrm{2}{b}\right)^{{b}} ={b}^{\mathrm{2}{b}} \\ $$$$\mathrm{2}{b}={b}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} −\mathrm{2}{b}=\mathrm{0} \\ $$$${b}\left({b}−\mathrm{2}\right)=\mathrm{0} \\ $$$${b}=\mathrm{0}\:,\:\mathrm{2} \\ $$$${a}=\mathrm{2}{b}\Rightarrow{a}=\mathrm{0}\:,\:\mathrm{4}\:\: \\ $$$$\left[{Is}\:\left({a},{b}\right)=\left(\mathrm{0},\mathrm{0}\right)\:\:{acceptable}\:?\right. \\ $$$$\mathrm{0}^{\mathrm{0}} =\mathrm{0}^{\mathrm{0}} \:? \\ $$$$\left.{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{0}^{\mathrm{2}} +\mathrm{0}^{\mathrm{2}} =\mathrm{0}\right] \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} =\mathrm{16}+\mathrm{4}=\mathrm{20} \\ $$
Commented by Mr.D.N. last updated on 24/Sep/21
$$\:\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir}\:\mathrm{mr}.\mathrm{sindhi}\:\mathrm{its}\:\mathrm{great}\:\mathrm{solution}. \\ $$