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Question Number 140089 by otchereabdullai@gmail.com last updated on 04/May/21
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
$$\mathrm{Given}\:\mathrm{that}\:\mathrm{log}_{\mathrm{4}} \left(\mathrm{y}−\mathrm{1}\right)+\mathrm{log}_{\mathrm{4}} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\mathrm{k} \\ $$$$\mathrm{and}\:\mathrm{log}_{\mathrm{2}} \left(\mathrm{y}+\mathrm{1}\right)−\mathrm{log}_{\mathrm{2}} \mathrm{x}=\mathrm{k}−\mathrm{1} \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{y}^{\mathrm{2}} =\mathrm{1}+\mathrm{8}^{\mathrm{k}} \\ $$$$\mathrm{Hence}\:\mathrm{deduce}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{y}\:\mathrm{and}\:\mathrm{x}\: \\ $$$$\mathrm{when}\:\mathrm{k}=\mathrm{1} \\ $$
Answered by mr W last updated on 04/May/21
log_4  (y−1)((x/y))=k  ⇒ (y−1)((x/y))=4^k =2^(2k)    ...(i)  log_2  ((y+1)/x)=k−1  ⇒ ((y+1)/x)=2^(k−1)    ...(ii)  (i)×(ii):  (y−1)(y+1)(1/y)=2^(3k−1) =(8^k /2)  y^2 −1=(8^k /2)y  ⇒y^2 =1+(8^k /2)y≠1+8^k   something is wrong in question!
$$\mathrm{log}_{\mathrm{4}} \:\left({y}−\mathrm{1}\right)\left(\frac{{x}}{{y}}\right)={k} \\ $$$$\Rightarrow\:\left({y}−\mathrm{1}\right)\left(\frac{{x}}{{y}}\right)=\mathrm{4}^{{k}} =\mathrm{2}^{\mathrm{2}{k}} \:\:\:…\left({i}\right) \\ $$$$\mathrm{log}_{\mathrm{2}} \:\frac{{y}+\mathrm{1}}{{x}}={k}−\mathrm{1} \\ $$$$\Rightarrow\:\frac{{y}+\mathrm{1}}{{x}}=\mathrm{2}^{{k}−\mathrm{1}} \:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\left({y}−\mathrm{1}\right)\left({y}+\mathrm{1}\right)\frac{\mathrm{1}}{{y}}=\mathrm{2}^{\mathrm{3}{k}−\mathrm{1}} =\frac{\mathrm{8}^{{k}} }{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} −\mathrm{1}=\frac{\mathrm{8}^{{k}} }{\mathrm{2}}{y} \\ $$$$\Rightarrow{y}^{\mathrm{2}} =\mathrm{1}+\frac{\mathrm{8}^{{k}} }{\mathrm{2}}{y}\neq\mathrm{1}+\mathrm{8}^{{k}} \\ $$$${something}\:{is}\:{wrong}\:{in}\:{question}! \\ $$

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