Given-that-log-4-y-1-log-4-x-y-k-and-log-2-y-1-log-2-x-k-1-Show-that-y-2-1-8-k-Hence-deduce-the-value-of-y-and-x-when-k-1-

Question Number 140089 by otchereabdullai@gmail.com last updated on 04/May/21

Given that \log_{4} (y - 1) + \log_{4} (\frac{x}{y}) = k

and \log_{2} (y + 1) - \log_{2} x = k - 1

Show that y^{2} = 1 + 8^{k}

Hence deduce the value of y and x

when k = 1

Answered by mr W last updated on 04/May/21

\log_{4} (y - 1) (\frac{x}{y}) = k

\Rightarrow (y - 1) (\frac{x}{y}) = 4^{k} = 2^{2 k} \dots (i)

\log_{2} \frac{y + 1}{x} = k - 1

\Rightarrow \frac{y + 1}{x} = 2^{k - 1} \dots (i i)

(i) \times (i i) :

(y - 1) (y + 1) \frac{1}{y} = 2^{3 k - 1} = \frac{8^{k}}{2}

y^{2} - 1 = \frac{8^{k}}{2} y

\Rightarrow y^{2} = 1 + \frac{8^{k}}{2} y \neq 1 + 8^{k}

s o m e t h i n g i s w r o n g i n q u e s t i o n!

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