5-x-7-x-9-8-x-x-

Question Number 126619 by O Predador last updated on 22/Dec/20

5^{x} + 7^{x} = {(9.8)}^{x}

x = ?

Answered by Dwaipayan Shikari last updated on 22/Dec/20

{(\frac{5}{9.8})}^{x} + {(\frac{7}{9.8})}^{x} = 1

\Rightarrow ({(\frac{5}{7})}^{x} a) + a = 1 a = {(\frac{7}{9.8})}^{x} = {(\frac{10}{14})}^{x} = {(\frac{5}{7})}^{x}

\Rightarrow a^{2} + a - 1 = 0 \Rightarrow a = \frac{- 1 \pm \sqrt{5}}{2} = \frac{\sqrt{5} - 1}{2}

{(\frac{5}{7})}^{x} = \frac{\sqrt{5} - 1}{2} \Rightarrow x = \frac{l o g (\frac{\sqrt{5} - 1}{2})}{l o g (\frac{5}{7})} \sim 1.4301679980845866709 \dots

Commented by O Predador last updated on 22/Dec/20

W o n d e r f u l!

Answered by Olaf last updated on 22/Dec/20

5^{x} + 7^{x} = 9 . 8^{x} = {(\frac{7 \times 14}{5 \times 2})}^{x}

5^{x} + 7^{x} = 9 . 8^{x} = 7^{x} {(\frac{7}{5})}^{x}

{(\frac{5}{7})}^{x} + 1 = 9 . 8^{x} = {(\frac{7}{5})}^{x}

Let X = {(\frac{5}{7})}^{x}

X + 1 = \frac{1}{X}

X^{2} + X - 1 = 0

X = \frac{- 1 \pm \sqrt{5}}{2}

\frac{- 1 - \sqrt{5}}{2} impossible because X > 0

X = \frac{- 1 + \sqrt{5}}{2} = {(\frac{5}{7})}^{x}

x \ln \frac{5}{7} = \ln (\frac{\sqrt{5} - 1}{2})

x = \frac{\ln (\frac{\sqrt{5} - 1}{2})}{\ln \frac{5}{7}}

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