solve-2-x-4-x-8-x-

Question Number 138096 by mr W last updated on 10/Apr/21

s o l v e

2^{x} + 4^{x} = 8^{x}

Commented by MJS_new last updated on 10/Apr/21

the complex solutions are

x = \log_{2} \frac{1 + \sqrt{5}}{2} + \frac{2 n π}{\ln 2} i \lor x = \log_{2} \frac{- 1 + \sqrt{5}}{2} + \frac{(2 n + 1) π}{\ln 2} i; n \in Z

Answered by liberty last updated on 10/Apr/21

\Rightarrow t + t^{2} = t^{3}

\Rightarrow t^{3} - t^{2} - t = 0

\Rightarrow t (t^{2} - t - 1) = 0

\Rightarrow t = 0 r e j e c t e d

\Rightarrow t = \frac{1 + \sqrt{5}}{2}; 2^{x} = \frac{1 + \sqrt{5}}{2}

x = \log_{2} (\frac{1 + \sqrt{5}}{2})

Commented by mr W last updated on 10/Apr/21

n i c e!

Commented by liberty last updated on 10/Apr/21

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Commented by chengulapetrom last updated on 10/Apr/21

a l s o t = \frac{1 - \sqrt{5}}{2} r e j e c t e d