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Question Number 188284 by Humble last updated on 27/Feb/23
a friend shared this challenging problem to me. 2^y ×y^2 +(2y)^(2y) =272 no inspection approach! thank you all.
$${a}\:{friend}\:{shared}\:{this}\:{challenging}\: \\ $$$${problem}\:{to}\:{me}. \\ $$$$\mathrm{2}^{\boldsymbol{{y}}} ×\boldsymbol{{y}}^{\mathrm{2}} +\left(\mathrm{2}\boldsymbol{{y}}\right)^{\mathrm{2}\boldsymbol{{y}}} \:=\mathrm{272} \\ $$$$\boldsymbol{{no}}\:\boldsymbol{{inspection}}\:\boldsymbol{{approach}}! \\ $$$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{all}}. \\ $$
Commented by Frix last updated on 27/Feb/23
I don′t think there′s a method for an exact solution. You can approximate or use your brains. I showed that 2^y ×y^2 +(2y)^(2y) =16+256 ⇒ y=2 If you don′t like this approach, find another one!
$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{method}\:\mathrm{for}\:\mathrm{an}\:\mathrm{exact} \\ $$$$\mathrm{solution}.\:\mathrm{You}\:\mathrm{can}\:\mathrm{approximate}\:\mathrm{or}\:\mathrm{use} \\ $$$$\mathrm{your}\:\mathrm{brains}.\:\mathrm{I}\:\mathrm{showed}\:\mathrm{that} \\ $$$$\mathrm{2}^{{y}} ×{y}^{\mathrm{2}} +\left(\mathrm{2}{y}\right)^{\mathrm{2}{y}} =\mathrm{16}+\mathrm{256} \\ $$$$\Rightarrow\:{y}=\mathrm{2} \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{like}\:\mathrm{this}\:\mathrm{approach},\:\mathrm{find}\:\mathrm{another} \\ $$$$\mathrm{one}! \\ $$
Answered by a.lgnaoui last updated on 28/Feb/23
Authere methode Remarque: 272=2^4 +2^8 2^y ×y^2 +2^(2y) ×y^(2y) =272 (2^y )×y^2 +(2^y )^2 ×(y^2 )^y =2^4 +2^8 =2^4 +(2^4 )^2 { (((2^y )×y^2 = a)),(((2^y )^2 ×(y^y )^2 =b)) :} (b/a)=2^y ×y^(2y−2) =2^y ×y^(2(y−1)) a+b=2^y y^2 +2^y ×y^2 (2^y ×y^(2(y−1)) ) =y^2 2^y +2^(2y) y^(2y) =2^y y^2 (1+2^y y^(2(y−1)) ) =2^y y^2 (1+(2^(2y) /4))=2^y y^2 (1+(((2^y )^2 )/4)) a+b=272=2^4 +2^8 a+b= 2^4 (1+(1/(16)) ) { ((y^2 2^y =2^4 (1))),((((2^(2y) /4))=(2^4 /4^2 ) (2))) :} y=2
$${Authere}\:{methode} \\ $$$${Remarque}:\:\:\:\:\mathrm{272}=\mathrm{2}^{\mathrm{4}} +\mathrm{2}^{\mathrm{8}} \\ $$$$\mathrm{2}^{{y}} ×{y}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}{y}} ×{y}^{\mathrm{2}{y}} =\mathrm{272} \\ $$$$\left(\mathrm{2}^{{y}} \right)×{y}^{\mathrm{2}} +\left(\mathrm{2}^{{y}} \right)^{\mathrm{2}} ×\left({y}^{\mathrm{2}} \right)^{{y}} =\mathrm{2}^{\mathrm{4}} +\mathrm{2}^{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}^{\mathrm{4}} +\left(\mathrm{2}^{\mathrm{4}} \right)^{\mathrm{2}} \\ $$$$\begin{cases}{\left(\mathrm{2}^{{y}} \right)×{y}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:=\:{a}}\\{\left(\mathrm{2}^{{y}} \right)^{\mathrm{2}} ×\left({y}^{{y}} \right)^{\mathrm{2}} ={b}}\end{cases} \\ $$$$\frac{{b}}{{a}}=\mathrm{2}^{{y}} ×{y}^{\mathrm{2}{y}−\mathrm{2}} =\mathrm{2}^{{y}} ×{y}^{\mathrm{2}\left({y}−\mathrm{1}\right)} \\ $$$${a}+{b}=\mathrm{2}^{{y}} {y}^{\mathrm{2}} +\mathrm{2}^{{y}} ×{y}^{\mathrm{2}} \left(\mathrm{2}^{{y}} ×{y}^{\mathrm{2}\left({y}−\mathrm{1}\right)} \right) \\ $$$$={y}^{\mathrm{2}} \mathrm{2}^{{y}} +\mathrm{2}^{\mathrm{2}{y}} {y}^{\mathrm{2}{y}} =\mathrm{2}^{{y}} {y}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}^{{y}} {y}^{\mathrm{2}\left({y}−\mathrm{1}\right)} \right) \\ $$$$=\mathrm{2}^{{y}} {y}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{2}^{\mathrm{2}{y}} }{\mathrm{4}}\right)=\mathrm{2}^{{y}} {y}^{\mathrm{2}} \left(\mathrm{1}+\frac{\left(\mathrm{2}^{{y}} \right)^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:{a}+{b}=\mathrm{272}=\mathrm{2}^{\mathrm{4}} +\mathrm{2}^{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:{a}+{b}=\:\mathrm{2}^{\mathrm{4}} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{16}}\:\right) \\ $$$$\:\:\:\:\begin{cases}{{y}^{\mathrm{2}} \mathrm{2}^{{y}} =\mathrm{2}^{\mathrm{4}} \:\:\:\:\:\:\:\left(\mathrm{1}\right)}\\{\left(\frac{\mathrm{2}^{\mathrm{2}{y}} }{\mathrm{4}}\right)=\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{4}^{\mathrm{2}} }\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{y}}=\mathrm{2} \\ $$$$\: \\ $$

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