Question Number 148573 by mathdanisur last updated on 29/Jul/21
$${log}_{\sqrt{\boldsymbol{{x}}}} \:\frac{{x}}{{y}}\:=\:{A}\:\:\Rightarrow\:\:{log}_{\sqrt{\boldsymbol{{y}}}} \:\frac{{y}}{{x}}\:=\:? \\ $$
Answered by Ar Brandon last updated on 29/Jul/21
$${A}=\mathrm{log}_{\sqrt{{x}}} \frac{{x}}{{y}}=\mathrm{log}_{\sqrt{{x}}} {x}−\mathrm{log}_{\sqrt{{x}}} {y} \\ $$$$\:\:\:\:=\mathrm{log}_{\sqrt{{x}}} \sqrt{{x}}\overset{\mathrm{2}} {\:}−\frac{\mathrm{log}_{\sqrt{{y}}} {y}}{\mathrm{log}_{\sqrt{{y}}} \sqrt{{x}}}=\mathrm{2}−\frac{\mathrm{2}}{\mathrm{log}_{\sqrt{{y}}} \sqrt{{x}}} \\ $$$$\Rightarrow\mathrm{log}_{\sqrt{{y}}} \sqrt{{x}}=\frac{\mathrm{2}}{\mathrm{2}−{A}} \\ $$$$\Rightarrow\mathrm{log}_{\sqrt{{y}}} \frac{{y}}{{x}}=\mathrm{2}−\mathrm{log}_{\sqrt{{y}}} {x}=\mathrm{2}−\frac{\mathrm{4}}{\mathrm{2}−{A}} \\ $$
Commented by puissant last updated on 29/Jul/21
$$\mathrm{log}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\mathrm{log}\left(\mathrm{x}\right)−\mathrm{log}\left(\mathrm{y}\right).. \\ $$
Commented by Ar Brandon last updated on 29/Jul/21
$$\mathrm{Merci}\:\mathrm{pour}\:\mathrm{la}\:\mathrm{remarque}. \\ $$
Commented by mathdanisur last updated on 29/Jul/21
$${Thank}\:{you}\:{Sir} \\ $$
Commented by Ar Brandon last updated on 29/Jul/21
$$\mathrm{You}'\mathrm{re}\:\mathrm{welcome}\:! \\ $$
Answered by EDWIN88 last updated on 29/Jul/21
$$\mathrm{log}\:_{\sqrt{{x}}} \left(\frac{{x}}{{y}}\right)={A}\:\Rightarrow\mathrm{2log}\:_{{x}} \left(\frac{{x}}{{y}}\right)={A} \\ $$$$\mathrm{log}\:_{{x}} \left(\frac{{y}}{{x}}\right)=−\frac{{A}}{\mathrm{2}}\Rightarrow\frac{{y}}{{x}}={x}^{−\frac{{A}}{\mathrm{2}}} \\ $$$$\Rightarrow{y}={x}^{\mathrm{1}−\frac{{A}}{\mathrm{2}}} \\ $$$$\mathrm{log}\:_{\sqrt{{y}}} \left(\frac{{y}}{{x}}\right)=\mathrm{2log}\:_{{y}} \left(\frac{{y}}{{x}}\right) \\ $$$$=\:\mathrm{2log}_{{y}} \:\left({x}^{−\frac{{A}}{\mathrm{2}}} \right) \\ $$$$=−{A}\mathrm{log}\:_{{y}} \left({x}\right)\:=−{A}\mathrm{log}\:_{\left({x}^{\mathrm{1}−\frac{{A}}{\mathrm{2}}} \right)} \left({x}\right) \\ $$$$=−{A}\left(\frac{\mathrm{2}}{\mathrm{2}−{A}}\right)\mathrm{log}\:_{{x}} \left({x}\right) \\ $$$$=\frac{\mathrm{2}{A}}{{A}−\mathrm{2}}\:. \\ $$
Commented by mathdanisur last updated on 29/Jul/21
$${Thank}\:{you}\:{Ser} \\ $$