solve-for-y-if-2y-2-y-2-y-3-4y-9-

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

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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