Question-96636

Question Number 96636 by O Predador last updated on 03/Jun/20

Commented by hknkrc46 last updated on 03/Jun/20

5^{x} - {(9, 8)}^{x} = 7^{x} \Rightarrow 5^{x} - {(\frac{49}{5})}^{x} = 7^{x}

\Rightarrow 5^{x} - \frac{7^{2 x}}{5^{x}} = 7^{x} \Rightarrow 5^{2 x} - 7^{2 x} = 5^{x} \cdot 7^{x}

\Rightarrow \frac{5^{2 x} - 7^{2 x}}{5^{x} \cdot 7^{x}} = 1 \Rightarrow \frac{5^{x}}{7^{x}} - \frac{7^{x}}{5^{x}} = 1

\frac{5^{x}}{7^{x}} = {(\frac{5}{7})}^{x} = t \Rightarrow t - \frac{1}{t} = 1 \Rightarrow t^{2} - t - 1 = 0

\Rightarrow t = \frac{- (- 1) \mp \sqrt{{(- 1)}^{2} - 4 \cdot 1 \cdot (- 1)}}{2 \cdot 1}

\Rightarrow {(\frac{5}{7})}^{x} = \frac{1 \mp \sqrt{5}}{2} \Rightarrow 0 < {(\frac{5}{7})}^{x} = \frac{1 + \sqrt{5}}{2}

\Rightarrow \ln {(\frac{5}{7})}^{x} = \ln \frac{1 + \sqrt{5}}{2}

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

Answered by mr W last updated on 03/Jun/20

{(\frac{1}{1.4})}^{x} - {(1.4)}^{x} = 1

{(1.4)}^{2 x} + {(1.4)}^{x} - 1 = 0

1 . 4^{x} = \frac{\sqrt{5} - 1}{2}

\Rightarrow x = \frac{\ln (\sqrt{5} - 1) - \ln 2}{\ln 7 - \ln 5} \approx - 1.4302

Commented by O Predador last updated on 03/Jun/20

or \log_{\frac{7}{5}} (\frac{- 1 + \sqrt{5}}{2}) ?

Commented by mr W last updated on 03/Jun/20

y e s

Commented by O Predador last updated on 03/Jun/20

Thank you!

Answered by 1549442205 last updated on 03/Jun/20

Equation is equivalent to

5^{x} - {(\frac{49}{5})}^{x} = 7^{x} \Leftrightarrow 5^{2 x} - 7^{2 x} = 7^{x} \times 5^{x} (1)

\Leftrightarrow {(\frac{5}{7})}^{x} - {(\frac{7}{5})}^{x} = 1

et {(\frac{7}{5})}^{x} = y (y > 0)

\Rightarrow \frac{1}{y} - y = 1 \Leftrightarrow y^{2} + y - 1 = 0

\Rightarrow y = \frac{- 1 + \sqrt{5}}{2} \Rightarrow {(\frac{7}{5})}^{t} = \frac{- 1 + \sqrt{5}}{2}

\Rightarrow x = \frac{\ln (\frac{- 1 + \sqrt{5}}{2})}{\ln \frac{7}{5}} \approx - 1.43016