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Question Number 5039 by Rasheed Soomro last updated on 05/Apr/16

log_(((x/y))) ((y/x))=?

$$\mathrm{log}_{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)} \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=? \\ $$

Answered by LMTV last updated on 05/Apr/16

((x/y))^? =(y/x) ?=−1

$$\left(\frac{{x}}{{y}}\right)^{?} =\frac{{y}}{{x}} \\ $$$$?=−\mathrm{1} \\ $$

Commented by FilupSmith last updated on 05/Apr/16

S=log_(x/y) ((y/x)) S=((ln(y)−ln(x))/(ln(x)−ln(y))) S ln((x/y))=ln((y/x)) ∴ ((x/y))^S =(y/x) if ((x/y))^S =−1 ∴(y/x)=−1 y=−x ∴ ((x/(−x)))^S =−1 (−1)^S =−1 ∴ S=1 ∴ log_(x/y) ((y/x))=1 if x>0, y>0

$${S}=\mathrm{log}_{\frac{{x}}{{y}}} \left(\frac{{y}}{{x}}\right) \\ $$$${S}=\frac{\mathrm{ln}\left({y}\right)−\mathrm{ln}\left({x}\right)}{\mathrm{ln}\left({x}\right)−\mathrm{ln}\left({y}\right)} \\ $$$${S}\:\mathrm{ln}\left(\frac{{x}}{{y}}\right)=\mathrm{ln}\left(\frac{{y}}{{x}}\right) \\ $$$$\therefore\:\left(\frac{{x}}{{y}}\right)^{{S}} =\frac{{y}}{{x}} \\ $$$$ \\ $$$$\mathrm{if}\:\left(\frac{{x}}{{y}}\right)^{{S}} =−\mathrm{1} \\ $$$$\therefore\frac{{y}}{{x}}=−\mathrm{1} \\ $$$${y}=−{x} \\ $$$$ \\ $$$$\therefore\:\left(\frac{{x}}{−{x}}\right)^{{S}} =−\mathrm{1} \\ $$$$\left(−\mathrm{1}\right)^{{S}} =−\mathrm{1} \\ $$$$\therefore\:{S}=\mathrm{1} \\ $$$$ \\ $$$$\therefore\:\mathrm{log}_{\frac{{x}}{{y}}} \left(\frac{{y}}{{x}}\right)=\mathrm{1} \\ $$$$ \\ $$$${if}\:{x}>\mathrm{0},\:{y}>\mathrm{0} \\ $$

Commented by Rasheed Soomro last updated on 05/Apr/16

In step 2 you changed the base of log without following the base changing law.

$$\mathrm{In}\:\mathrm{step}\:\mathrm{2}\:\mathrm{you}\:\mathrm{changed}\:\mathrm{the}\:\mathrm{base}\:\mathrm{of} \\ $$$$\mathrm{log}\:\mathrm{without}\:\mathrm{following}\:\mathrm{the}\:\mathrm{base}\:\mathrm{changing} \\ $$$$\mathrm{law}. \\ $$

Commented by FilupSmith last updated on 06/Apr/16

log_(x/y) ((y/x))=((ln((y/x)))/(ln((x/y))))=((ln y − ln x)/(ln x − ln y))???

$$\mathrm{log}_{\frac{{x}}{{y}}} \left(\frac{{y}}{{x}}\right)=\frac{\mathrm{ln}\left(\frac{{y}}{{x}}\right)}{\mathrm{ln}\left(\frac{{x}}{{y}}\right)}=\frac{\mathrm{ln}\:{y}\:−\:\mathrm{ln}\:{x}}{\mathrm{ln}\:{x}\:−\:\mathrm{ln}\:{y}}??? \\ $$$$ \\ $$

Commented by FilupSmith last updated on 06/Apr/16

Where is my mistake?

$${W}\mathrm{here}\:\mathrm{is}\:\mathrm{my}\:\mathrm{mistake}? \\ $$

Commented by Rasheed Soomro last updated on 06/Apr/16

Sorry, I misunderstood.

$${Sorry},\:{I}\:{misunderstood}. \\ $$

Answered by FilupSmith last updated on 05/Apr/16

=((ln y − ln x)/(ln x − ln y))

$$=\frac{\mathrm{ln}\:{y}\:−\:\mathrm{ln}\:{x}}{\mathrm{ln}\:{x}\:−\:\mathrm{ln}\:{y}} \\ $$

Commented by FilupSmith last updated on 07/Apr/16

you are correct!!

$$\mathrm{you}\:\mathrm{are}\:\mathrm{correct}!! \\ $$

Commented by Rasheed Soomro last updated on 06/Apr/16

((ln y − ln x)/(ln x − ln y)) Couldn′t we proceed further as follows =((−(ln x − ln y))/(ln x − ln y))=−1

$$\frac{\mathrm{ln}\:{y}\:−\:\mathrm{ln}\:{x}}{\mathrm{ln}\:{x}\:−\:\mathrm{ln}\:{y}} \\ $$$${Couldn}'{t}\:{we}\:{proceed}\:{further} \\ $$$${as}\:{follows} \\ $$$$=\frac{−\left(\mathrm{ln}\:{x}\:−\:\mathrm{ln}\:{y}\right)}{\mathrm{ln}\:{x}\:−\:\mathrm{ln}\:{y}}=−\mathrm{1} \\ $$

Answered by Rasheed Soomro last updated on 05/Apr/16

log_(((x/y))) ((y/x))=? log_(((x/y))) ((x/y))^(−1) =(−1)log_(((x/y))) ((x/y))=(−1)(1)=−1 [∵log_a a=1]

$$\mathrm{log}_{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)} \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=? \\ $$$$\mathrm{log}_{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{−\mathrm{1}} =\left(−\mathrm{1}\right)\mathrm{log}_{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)=\left(−\mathrm{1}\right)\left(\mathrm{1}\right)=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\because\mathrm{log}_{\mathrm{a}} \mathrm{a}=\mathrm{1}\right] \\ $$

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