find the value of y given that y^(y ) =3^(y+9)


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Question Number 207747 by MASANJAJJ last updated on 25/May/24

$f i n d t h e v a l u e o f y g i v e n t h a t y^{y} = 3^{y + 9}$
	Answered by mr W last updated on 25/May/24

	$y^{y} = 3^{y} \times 3^{9}$ ${(\frac{y}{3})}^{y} = 3^{9}$ ${(\frac{y}{3})}^{\frac{y}{3}} = 3^{3}$ $\Rightarrow \frac{y}{3} = 3 \Rightarrow y = 9$
	Answered by Berbere last updated on 25/May/24

	$y^{y} = 3^{a} \Rightarrow a = \frac{y l n (y)}{l n (3)}$ $\Leftrightarrow 3^{\frac{y l n (y)}{l n (3)}} = 3^{y + 9}$ $\Rightarrow y l n (y) = y l n (3) + 9 l n (3)$ $y l n (\frac{y}{3}) = 9 l n (3) > 0 \Rightarrow y ⩾ 3$ $y \overset{f}{\to} y l n (\frac{y}{3}) i n c r e a s e f u n c t i o n o v e r [3, \infty [$ $f^{'} = l n (\frac{y}{3}) + \frac{y}{3} ⩾ l n (\frac{3}{3}) + \frac{3}{3} = 1$ $f (9) = 9 l n (3) \Rightarrow y = 9$
	Answered by Berbere last updated on 25/May/24
	$if y∈N^∗ ⇒y=3^a ;a≥1 3^(ay) =3^(y+9) ay=y+9 y∣ay ⇔y∣y+9⇒y∣(ay−y)=9⇒y∣9 ⇒y∈{1,3,9}⇒y=9 by tcheking$
	$i f y \in N^{*}$ $\Rightarrow y = 3^{a}; a ⩾ 1$ $3^{a y} = 3^{y + 9}$ $a y = y + 9$ $y ∣ a y \Leftrightarrow y ∣ y + 9 \Rightarrow y ∣ (a y - y) = 9 \Rightarrow y ∣ 9$ $\Rightarrow y \in {1, 3, 9} \Rightarrow y = 9 b y t c h e k i n g$
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