Question Number 170468 by MathsFan last updated on 24/May/22
$$\:\mathrm{G}{iven}\:{that}\:{log}_{\mathrm{4}} \left({y}−\mathrm{1}\right)+{log}_{\mathrm{4}} \left(\frac{{x}}{{y}}\right)={m} \\ $$$$\:{and}\:{log}_{\mathrm{2}} \left({y}+\mathrm{1}\right)−{log}_{\mathrm{2}} {x}={m}−\mathrm{1}, \\ $$$$\:{show}\:{that}\:{y}^{\mathrm{2}} =\mathrm{1}−\mathrm{8}^{{m}} \\ $$
Answered by cortano1 last updated on 24/May/22
$$\:\:\begin{cases}{\mathrm{log}\:_{\mathrm{4}} \left({y}−\mathrm{1}\right)+\mathrm{log}\:_{\mathrm{4}} \left(\frac{{x}}{{y}}\right)={m}}\\{\mathrm{log}\:_{\mathrm{2}} \left({y}+\mathrm{1}\right)−\mathrm{log}\:_{\mathrm{2}} {x}={m}−\mathrm{1}}\end{cases} \\ $$$$\:\begin{cases}{\mathrm{log}\:_{\mathrm{2}} \left(\frac{{y}−\mathrm{1}}{{y}}\right)+\mathrm{log}\:_{\mathrm{2}} {x}\:=\:\mathrm{2}{m}}\\{\mathrm{log}\:_{\mathrm{2}} \left({y}+\mathrm{1}\right)−\mathrm{log}\:_{\mathrm{2}} {x}={m}−\mathrm{1}}\end{cases} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{2}} \left(\frac{{y}^{\mathrm{2}} −\mathrm{1}}{{y}}\right)=\mathrm{3}{m}−\mathrm{1} \\ $$$$\Rightarrow\frac{{y}^{\mathrm{2}} −\mathrm{1}}{{y}}\:=\:\frac{\mathrm{8}^{{m}} }{\mathrm{2}} \\ $$$$ \\ $$
Commented by MathsFan last updated on 24/May/22
$${thanks} \\ $$