Suppose-x-and-y-are-real-such-that-log-4-x-log-6-y-log-9-x-y-find-y-x-

Question Number 167539 by Tawa11 last updated on 19/Mar/22

Suppose x and y are real, such that, \log_{4} x = \log_{6} y = \log_{9} (x + y),

find \frac{y}{x}

Answered by Rasheed.Sindhi last updated on 19/Mar/22

\log_{4} x = \log_{6} y = \log_{9} (x + y) = k (s a y)

x = 4^{k}, y = 6^{k}, x + y = 9^{k}

4^{k} + 6^{k} = 9^{k}

{(\frac{4}{9})}^{k} + {(\frac{6}{9})}^{k} = 1

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

{(\frac{2}{3})}^{k} = u

u^{2} + u - 1 = 0

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

{(\frac{2}{3})}^{k} = \frac{- 1 \pm \sqrt{5}}{2}

\frac{y}{x} = \frac{6^{k}}{4^{k}} = {(\frac{3}{2})}^{k} = {({(\frac{2}{3})}^{k})}^{- 1} = {(\frac{- 1 \pm \sqrt{5}}{2})}^{- 1}

= \frac{2}{- 1 \pm \sqrt{5}} \cdot \frac{- 1 \mp \sqrt{5}}{- 1 \mp \sqrt{5}} = \frac{- 2 (1 \pm \sqrt{5})}{1 - 5}

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

Commented by Tawa11 last updated on 19/Mar/22

God bless you sir .

Answered by Rasheed.Sindhi last updated on 19/Mar/22

\log_{4} x = \log_{6} y = \log_{9} (x + y) = k (s a y)

x = 4^{k}, y = 6^{k}, x + y = 9^{k} \Rightarrow 4^{k} + 6^{k} = 9^{k}

1 + \frac{6^{k}}{4^{k}} = \frac{9^{k}}{4^{k}}

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

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

{(\frac{3}{2})}^{k} = u

u^{2} - u - 1 = 0

u = \frac{1 \pm \sqrt{5}}{2}

{(\frac{3}{2})}^{k} = \frac{1 \pm \sqrt{5}}{2}

\frac{y}{x} = \frac{6^{k}}{4^{k}} = {(\frac{3}{2})}^{k} = \frac{1 \pm \sqrt{5}}{2}