Solve-the-equation-2-x-3-x-4-x-6-x-9-x-1-

Question Number 176592 by Shrinava last updated on 22/Sep/22

Solve the equation :

2^{x} + 3^{x} - 4^{x} + 6^{x} - 9^{x} = 1

Answered by Frix last updated on 22/Sep/22

obvious solution x = 0

Answered by a.lgnaoui last updated on 23/Sep/22

p o s o n s a = 2^{x} b = 3^{x} a^{2} = 2^{2 x} b^{2} = 3^{2 x} a b = 2^{x} 3^{x} = 6^{x}

(a + b) - (a^{2} + b^{2}) + a b = 1

(a + b) - [{(a + b)}^{2} - 2 a b] + a b = 1

(a + b) - {(a + b)}^{2} + 3 a b = 1 (I)

2^{x} + 3^{x} - {(2^{x} + 3^{x})}^{2} = 1 - 3 \times 6^{x}

(2^{x} + 3^{x}) (1 - 2^{x} - 3^{x}) = 1 - 3 \times 6^{x}

(2^{x} + 3^{x}) (2^{2 x} + 3^{2 x}) = 3 \times 6^{x} - 1

(2^{x} + 3^{x}) [(2^{2 x} + 3^{2 x} + 2 \times 6^{x}) - 2 \times 6^{x}] = 3 \times 6^{x} - 1

{(2^{x} + 3^{x})}^{3} - 2 \times 6^{x} (2^{x} + 3^{x}) = 3 \times 6^{x} - 1

3 a b = 1 + {(a + b)}^{2} - (a + b)

3 \times 6^{x} = 1 + {(2^{x} + 3^{x})}^{2} - (2^{x} + 3^{x})

{(2^{x} + 3^{x})}^{3} - 2 (2^{x} + 3^{x}) \frac{1 + {(2^{x} + 3^{x})}^{2} - (2^{x} + 3^{x})}{3} = {(2^{x} + 3^{x})}^{2} - (2^{x} + 3^{x})

o n p o s e (2^{x} + 3^{x} = z)

z^{3} - \frac{2}{} \frac{[z (1 + 2 z^{2} - z]}{3} = z^{2} - z

z^{2} - \frac{2 (1 + 2 z^{2} - z)}{3} = z - 1

3 z^{2} - 4 z^{2} + 2 z - 2 = 3 z - 3

z^{2} + z + 1 = 0 z = 0

2^{x} = - 3^{x} {(- \frac{2}{3})}^{x} = 1 \Rightarrow x = 0

Δ = 1 - 4 = {[\sqrt{3} i]}^{2}

z 1 = \frac{- 1 \pm \sqrt{3} i}{2}

2^{x} + 3^{x} = \frac{1}{2} + \frac{\sqrt{3}}{2} i

2 (2^{x} + 3^{x}) = 1 + \sqrt{3} i

\dots . s o l u t i o n s v i a c o s h x ?

Commented by Shrinava last updated on 23/Sep/22

Thank you dear professor, yes, gracias

Commented by Frix last updated on 23/Sep/22

(a + b) - (a^{2} + b^{2}) + a b = 1

a = u - v \land b = u + v

2 u - (2 u^{2} + 2 v^{2}) + u^{2} - v^{2} = 1

2 u - u^{2} - 3 v^{2} = 1

v^{2} = - \frac{{(u - 1)}^{2}}{3}

v = \pm \frac{u - 1}{\sqrt{3}} i

a = u \pm \frac{\sqrt{3} (u - 1)}{3} i \land b = u \mp \frac{\sqrt{3} (u - 1)}{3} i

2^{x} = u \pm \frac{\sqrt{3} (u - 1)}{3} i \land 3^{x} = u \mp \frac{\sqrt{3} (u - 1)}{3} i

again obviously x = 0 \land u = 1

x = \frac{\ln (u \pm \frac{\sqrt{3} (u - 1)}{3} i) + 2 n_{1} π i}{\ln 2} \land x = \frac{\ln (u \mp \frac{\sqrt{3} (u - 1)}{3} i) + 2 n_{2} π i}{\ln 3}

I {don}^{'} t think we can find n_{1}, n_{2} \in Z for u \neq 1

p \pm q i = \sqrt{p^{2} + q^{2}} e^{\pm i \tan^{- 1} \frac{q}{p}}

\ln (p \pm q i) = \frac{\ln (p^{2} + q^{2})}{2} + (2 n π \pm \tan^{- 1} \frac{q}{p}) i

we have

\frac{\ln (p - q i)}{\ln 2} = \frac{\ln (p + q i)}{\ln 3}

we need

\frac{2 n_{1} π - \tan^{- 1} t}{\ln 2} = \frac{2 n_{2} π + \tan^{- 1} t}{\ln 3}

\Leftrightarrow

n_{2} = \frac{1 - \frac{\ln 3}{\ln 2}}{2 π} \tan^{- 1} t + \frac{\ln 3}{\ln 2} n_{1} with n_{1}, n_{2} \in Z

Commented by a.lgnaoui last updated on 23/Sep/22

t h a n k y o u

Commented by Tawa11 last updated on 23/Sep/22

Great sirs

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