16-1-x-1-5-4-1-x-1-2-0-find-x-

Question Number 109913 by mohammad17 last updated on 26/Aug/20

\sqrt[x - 1]{16} - 5 \sqrt[x - 1]{4} + 2 = 0 f i n d x ?

Answered by Rasheed.Sindhi last updated on 26/Aug/20

\sqrt[x - 1]{16} - 5 \sqrt[x - 1]{4} + 2 = 0 f i n d x ?

16^{\frac{1}{x - 1}} - 5 {(4)}^{\frac{1}{x - 1}} + 2 = 0

{(4^{\frac{1}{x - 1}})}^{2} - 5 (4^{\frac{1}{x - 1}}) + 2 = 0

y^{2} - 5 y + 2 = 0; [4^{\frac{1}{x - 1}} = y]

y = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2}

4^{\frac{1}{x - 1}} = \frac{5 \pm \sqrt{17}}{2}

(\frac{1}{x - 1}) \log_{2} 4 = \log_{2} (\frac{5 \pm \sqrt{17}}{2})

(\frac{1}{x - 1}) \log_{2} 4 = \log_{2} (\frac{5 \pm \sqrt{17}}{2})

2 (\frac{1}{x - 1}) = \log_{2} (\frac{5 \pm \sqrt{17}}{2})

x = \frac{2}{\log_{2} (\frac{5 \pm \sqrt{17}}{2})} + 1

Commented by mohammad17 last updated on 26/Aug/20

= t h a n k y o u s i r