Question Number 5039 by Rasheed Soomro last updated on 05/Apr/16
$$\mathrm{log}_{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)} \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=? \\ $$
Answered by LMTV last updated on 05/Apr/16
$$\left(\frac{{x}}{{y}}\right)^{?} =\frac{{y}}{{x}} \\ $$$$?=−\mathrm{1} \\ $$
Commented by FilupSmith last updated on 05/Apr/16
$${S}=\mathrm{log}_{\frac{{x}}{{y}}} \left(\frac{{y}}{{x}}\right) \\ $$$${S}=\frac{\mathrm{ln}\left({y}\right)−\mathrm{ln}\left({x}\right)}{\mathrm{ln}\left({x}\right)−\mathrm{ln}\left({y}\right)} \\ $$$${S}\:\mathrm{ln}\left(\frac{{x}}{{y}}\right)=\mathrm{ln}\left(\frac{{y}}{{x}}\right) \\ $$$$\therefore\:\left(\frac{{x}}{{y}}\right)^{{S}} =\frac{{y}}{{x}} \\ $$$$ \\ $$$$\mathrm{if}\:\left(\frac{{x}}{{y}}\right)^{{S}} =−\mathrm{1} \\ $$$$\therefore\frac{{y}}{{x}}=−\mathrm{1} \\ $$$${y}=−{x} \\ $$$$ \\ $$$$\therefore\:\left(\frac{{x}}{−{x}}\right)^{{S}} =−\mathrm{1} \\ $$$$\left(−\mathrm{1}\right)^{{S}} =−\mathrm{1} \\ $$$$\therefore\:{S}=\mathrm{1} \\ $$$$ \\ $$$$\therefore\:\mathrm{log}_{\frac{{x}}{{y}}} \left(\frac{{y}}{{x}}\right)=\mathrm{1} \\ $$$$ \\ $$$${if}\:{x}>\mathrm{0},\:{y}>\mathrm{0} \\ $$
Commented by Rasheed Soomro last updated on 05/Apr/16
$$\mathrm{In}\:\mathrm{step}\:\mathrm{2}\:\mathrm{you}\:\mathrm{changed}\:\mathrm{the}\:\mathrm{base}\:\mathrm{of} \\ $$$$\mathrm{log}\:\mathrm{without}\:\mathrm{following}\:\mathrm{the}\:\mathrm{base}\:\mathrm{changing} \\ $$$$\mathrm{law}. \\ $$
Commented by FilupSmith last updated on 06/Apr/16
$$\mathrm{log}_{\frac{{x}}{{y}}} \left(\frac{{y}}{{x}}\right)=\frac{\mathrm{ln}\left(\frac{{y}}{{x}}\right)}{\mathrm{ln}\left(\frac{{x}}{{y}}\right)}=\frac{\mathrm{ln}\:{y}\:−\:\mathrm{ln}\:{x}}{\mathrm{ln}\:{x}\:−\:\mathrm{ln}\:{y}}??? \\ $$$$ \\ $$
Commented by FilupSmith last updated on 06/Apr/16
$${W}\mathrm{here}\:\mathrm{is}\:\mathrm{my}\:\mathrm{mistake}? \\ $$
Commented by Rasheed Soomro last updated on 06/Apr/16
$${Sorry},\:{I}\:{misunderstood}. \\ $$
Answered by FilupSmith last updated on 05/Apr/16
$$=\frac{\mathrm{ln}\:{y}\:−\:\mathrm{ln}\:{x}}{\mathrm{ln}\:{x}\:−\:\mathrm{ln}\:{y}} \\ $$
Commented by FilupSmith last updated on 07/Apr/16
$$\mathrm{you}\:\mathrm{are}\:\mathrm{correct}!! \\ $$
Commented by Rasheed Soomro last updated on 06/Apr/16
$$\frac{\mathrm{ln}\:{y}\:−\:\mathrm{ln}\:{x}}{\mathrm{ln}\:{x}\:−\:\mathrm{ln}\:{y}} \\ $$$${Couldn}'{t}\:{we}\:{proceed}\:{further} \\ $$$${as}\:{follows} \\ $$$$=\frac{−\left(\mathrm{ln}\:{x}\:−\:\mathrm{ln}\:{y}\right)}{\mathrm{ln}\:{x}\:−\:\mathrm{ln}\:{y}}=−\mathrm{1} \\ $$
Answered by Rasheed Soomro last updated on 05/Apr/16
$$\mathrm{log}_{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)} \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=? \\ $$$$\mathrm{log}_{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{−\mathrm{1}} =\left(−\mathrm{1}\right)\mathrm{log}_{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\left(−\mathrm{1}\right)\left(\mathrm{1}\right)=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because\mathrm{log}_{\mathrm{a}} \mathrm{a}=\mathrm{1}\right] \\ $$