log-4-x-log-y-log-25-x-y-Find-x-y-

Question Number 144563 by mathdanisur last updated on 26/Jun/21

{l o g}_{4} (x) = l o g (y) = {l o g}_{25} (x + y)

F i n d \frac{x}{y} = ?

Answered by imjagoll last updated on 26/Jun/21

let \log_{4} (x) = \log_{10} (y) = \log_{25} (x + y) = k

then {\begin{cases} x = 4^{k} \\ y = 10^{k} \\ x + y = 25^{k} \end{cases}

\Leftrightarrow 4^{k} + 10^{k} = 25^{k}

\Rightarrow 1 + {(\frac{5}{2})}^{k} = {(\frac{5}{2})}^{2 k}

\Rightarrow t^{2} - t - 1 = 0; where t = {(\frac{5}{2})}^{k}

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

\Rightarrow we find \frac{x}{y} = \frac{4^{k}}{10^{k}} = {(\frac{2}{5})}^{k} = \frac{2}{\sqrt{5} + 1}

\frac{x}{y} = \frac{2 (\sqrt{5} - 1)}{4} = \frac{\sqrt{5} - 1}{2}

Commented by mathdanisur last updated on 26/Jun/21

a l o t c o o l t h a n k s S i r, a n s w e r \frac{1}{2} (\sqrt{5} + 1)

Commented by imjagoll last updated on 26/Jun/21

wrong . if \frac{y}{x} = \frac{\sqrt{5} + 1}{2} true

Commented by mathdanisur last updated on 26/Jun/21

a n s w e r : \frac{x}{y} = \frac{\sqrt{5} - 1}{2} a n d \frac{y}{x} = \frac{\sqrt{5} + 1}{2} ?

Commented by liberty last updated on 26/Jun/21

from 4^{k} + 10^{k} = 25^{k} divided by 25^{k}

give {(\frac{2}{5})}^{2 k} + {(\frac{2}{5})}^{k} = 1 where t = {(\frac{2}{5})}^{k}

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

then \frac{x}{y} = \frac{\sqrt{5} - 1}{2} \leftarrow correct

Commented by haladu last updated on 26/Jun/21

N i ce!

Commented by mathdanisur last updated on 27/Jun/21

t h a n k s S i r c o o l

Answered by haladu last updated on 26/Jun/21

\Rightarrow x = 4^{h} = {(2^{h})}^{2}

\Rightarrow y = 10^{h} \Leftrightarrow y^{2} = {(2^{h})}^{2} {(5^{h})}^{2}

\Rightarrow x + y = {(5^{h})}^{2}

y^{2} = x (x + y)

mutiply by \frac{1}{y^{2}}

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

\Rightarrow \frac{x}{y} = \frac{- 1 \pm \sqrt{5}}{2}

Commented by mathdanisur last updated on 27/Jun/21

t h a n k y o u S i r c o o l