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

$$\mathrm{Given}\mathrm{that}{\log }_4(\mathrm{y}-1)+{\log }_4\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\mathrm{k}$$$$\mathrm{and}{\log }_2(\mathrm{y}+1)-{\log }_2\mathrm{x}=\mathrm{k}-1$$$$\mathrm{Show}\mathrm{that}{\mathrm{y}}^2=1+8^{\mathrm{k}}$$$$\mathrm{Hence}\mathrm{deduce}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{y}\mathrm{and}\mathrm{x}$$$$\mathrm{when}\mathrm{k}=1$$

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

$${\log }_4(y-1)\left(\frac{x}{y}\right)=k$$$$\Rightarrow (y-1)\left(\frac{x}{y}\right)=4^k=2^{2k}...\left(i\right)$$$${\log }_2\frac{y+1}{x}=k-1$$$$\Rightarrow \frac{y+1}{x}=2^{k-1}...\left(ii\right)$$$$\left(i\right)\times \left(ii\right):$$$$(y-1)(y+1)\frac{1}{y}=2^{3k-1}=\frac{8^k}{2}$$$$y^2-1=\frac{8^k}{2}y$$$$\Rightarrow y^2=1+\frac{8^k}{2}y\ne 1+8^k$$$$somethingiswronginquestion!$$

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