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Question Number 207747 by MASANJAJJ last updated on 25/May/24
find the value of y given that y^(y ) =3^(y+9)
$${find}\:{the}\:{value}\:{of}\:{y}\:{given}\:{that}\:{y}^{{y}\:} =\mathrm{3}^{{y}+\mathrm{9}} \\ $$
Commented by A5T last updated on 25/May/24
3^(9+9) =3^(18) =(3^2 )^3^2  =9^9 ⇒y=9
$$\mathrm{3}^{\mathrm{9}+\mathrm{9}} =\mathrm{3}^{\mathrm{18}} =\left(\mathrm{3}^{\mathrm{2}} \right)^{\mathrm{3}^{\mathrm{2}} } =\mathrm{9}^{\mathrm{9}} \Rightarrow{y}=\mathrm{9} \\ $$
Answered by mr W last updated on 25/May/24
y^y =3^y ×3^9   ((y/3))^y =3^9   ((y/3))^(y/3) =3^3   ⇒(y/3)=3 ⇒y=9
$${y}^{{y}} =\mathrm{3}^{{y}} ×\mathrm{3}^{\mathrm{9}} \\ $$$$\left(\frac{{y}}{\mathrm{3}}\right)^{{y}} =\mathrm{3}^{\mathrm{9}} \\ $$$$\left(\frac{{y}}{\mathrm{3}}\right)^{\frac{{y}}{\mathrm{3}}} =\mathrm{3}^{\mathrm{3}} \\ $$$$\Rightarrow\frac{{y}}{\mathrm{3}}=\mathrm{3}\:\Rightarrow{y}=\mathrm{9} \\ $$
Answered by Berbere last updated on 25/May/24
y^y =3^a ⇒a=((yln(y))/(ln(3)))  ⇔3^((yln(y))/(ln(3))) =3^(y+9)   ⇒yln(y)=yln(3)+9ln(3)  yln((y/3))=9ln(3)>0⇒y≥3  y→^f yln((y/3)) increase function over [3,∞[  f′=ln((y/3))+(y/3)≥ln((3/3))+(3/3)=1  f(9)=9ln(3)⇒y=9
$${y}^{{y}} =\mathrm{3}^{{a}} \Rightarrow{a}=\frac{{yln}\left({y}\right)}{{ln}\left(\mathrm{3}\right)} \\ $$$$\Leftrightarrow\mathrm{3}^{\frac{{yln}\left({y}\right)}{{ln}\left(\mathrm{3}\right)}} =\mathrm{3}^{{y}+\mathrm{9}} \\ $$$$\Rightarrow{yln}\left({y}\right)={yln}\left(\mathrm{3}\right)+\mathrm{9}{ln}\left(\mathrm{3}\right) \\ $$$${yln}\left(\frac{{y}}{\mathrm{3}}\right)=\mathrm{9}{ln}\left(\mathrm{3}\right)>\mathrm{0}\Rightarrow{y}\geqslant\mathrm{3} \\ $$$${y}\overset{{f}} {\rightarrow}{yln}\left(\frac{{y}}{\mathrm{3}}\right)\:{increase}\:{function}\:{over}\:\left[\mathrm{3},\infty\left[\right.\right. \\ $$$${f}'={ln}\left(\frac{{y}}{\mathrm{3}}\right)+\frac{{y}}{\mathrm{3}}\geqslant{ln}\left(\frac{\mathrm{3}}{\mathrm{3}}\right)+\frac{\mathrm{3}}{\mathrm{3}}=\mathrm{1} \\ $$$${f}\left(\mathrm{9}\right)=\mathrm{9}{ln}\left(\mathrm{3}\right)\Rightarrow{y}=\mathrm{9} \\ $$$$ \\ $$
Answered by Berbere last updated on 25/May/24
if y∈N^∗   ⇒y=3^a ;a≥1  3^(ay) =3^(y+9)   ay=y+9  y∣ay ⇔y∣y+9⇒y∣(ay−y)=9⇒y∣9  ⇒y∈{1,3,9}⇒y=9 by tcheking
$${if}\:{y}\in\mathbb{N}^{\ast} \\ $$$$\Rightarrow{y}=\mathrm{3}^{{a}} ;{a}\geqslant\mathrm{1} \\ $$$$\mathrm{3}^{{ay}} =\mathrm{3}^{{y}+\mathrm{9}} \\ $$$${ay}={y}+\mathrm{9} \\ $$$${y}\mid{ay}\:\Leftrightarrow{y}\mid{y}+\mathrm{9}\Rightarrow{y}\mid\left({ay}−{y}\right)=\mathrm{9}\Rightarrow{y}\mid\mathrm{9} \\ $$$$\Rightarrow{y}\in\left\{\mathrm{1},\mathrm{3},\mathrm{9}\right\}\Rightarrow{y}=\mathrm{9}\:{by}\:{tcheking} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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