solve for y if 2y+2^y^2 +y^3 +4y=9


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Question Number 187963 by Humble last updated on 24/Feb/23

$s o l v e f o r y i f$ $2 y + 2^{y^{2}} + y^{3} + 4 y = 9$
	Answered by a.lgnaoui last updated on 24/Feb/23
	$6y+2^y^2 +y^3 =9 y(y^2 +6)+2^y^2 =9 y^2 +6+(2^y^2 /y)=(9/y) y^2 +6=(1/y)(9−2^y^2 )⇒ (y^2 +6)^2 =(1/y^2 )(9−2^y^2 )^2 z=y^2 (z+6+((9−2^z )/z))(z+6−((9−2^z )/z))=0 ⇒ (z+3)^2 −2^z =0 (1) z^2 +6z−9+2^z =0 (2) (1)+(2)⇒ (z+3)^2 +z^2 +6z−9=0 2z(z+6)=0 { (((z+3)^2 =2^z )),(((z+3)^2 −18+2^z =0)) :} (suite) (z+3)^2 =2^z zlog2=2log(z+3) [(z+3) −3]log2 =log(z+3) z+3=t (t−3)log2=logt tlog2−logt=3log2 log((2^t /t) )=log(2^3 ) (2^t /t)=2^3 (2^(y^2 +3) /(y^2 +3))=2^3 =2^3 (2^y^2 /(y^2 +3)) (2^y^2 /(y^2 +3))=1⇒2^y^2 =y^2 +3 Solution={−1;+1}$
	$6 y + 2^{y^{2}} + y^{3} = 9$ $y (y^{2} + 6) + 2^{y^{2}} = 9$ $y^{2} + 6 + \frac{2^{y^{2}}}{y} = \frac{9}{y}$ $y^{2} + 6 = \frac{1}{y} (9 - 2^{y^{2}}) \Rightarrow$ ${(y^{2} + 6)}^{2} = \frac{1}{y^{2}} {(9 - 2^{y^{2}})}^{2}$ $z = y^{2}$ $(z + 6 + \frac{9 - 2^{z}}{z}) (z + 6 - \frac{9 - 2^{z}}{z}) = 0$ $\Rightarrow$ ${(z + 3)}^{2} - 2^{z} = 0 (1)$ $z^{2} + 6 z - 9 + 2^{z} = 0 (2)$ $(1) + (2) \Rightarrow$ ${(z + 3)}^{2} + z^{2} + 6 z - 9 = 0$ $2 z (z + 6) = 0$ ${\begin{cases} {(z + 3)}^{2} = 2^{z} \\ {(z + 3)}^{2} - 18 + 2^{z} = 0 \end{cases}$ $(s u i t e)$ ${(z + 3)}^{2} = 2^{z}$ $z \log 2 = 2 \log (z + 3)$ $[(z + 3) - 3] \log 2 = \log (z + 3)$ $z + 3 = t$ $(t - 3) \log 2 = logt$ $tlog 2 - logt = 3 \log 2$ $\log (\frac{2^{t}}{t}) = \log (2^{3})$ $\frac{2^{t}}{t} = 2^{3}$ $\frac{2^{y^{2} + 3}}{y^{2} + 3} = 2^{3} = 2^{3} \frac{2^{y^{2}}}{y^{2} + 3}$ $\frac{2^{y^{2}}}{y^{2} + 3} = 1 \Rightarrow 2^{y^{2}} = y^{2} + 3$ $S o l u t i o n = {- 1; + 1}$
	Commented by Humble last updated on 25/Feb/23

	$G r e a t! T h a n k s a l o t$
	Commented by mr W last updated on 25/Feb/23

	$w r o n g!$ $y = - 1 {d o e s n}^{'} t s a t i s f y t h e e q u a t i o n!$ $6 (- 1) + 2^{{(- 1)}^{2}} + {(- 1)}^{3} = - 5 \neq 9$
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