log-x-y-y-x-

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

\log_{(\frac{x}{y})} (\frac{y}{x}) = ?

Answered by LMTV last updated on 05/Apr/16

{(\frac{x}{y})}^{?} = \frac{y}{x}

? = - 1

Commented by FilupSmith last updated on 05/Apr/16

S = \log_{\frac{x}{y}} (\frac{y}{x})

S = \frac{\ln (y) - \ln (x)}{\ln (x) - \ln (y)}

S \ln (\frac{x}{y}) = \ln (\frac{y}{x})

∴ {(\frac{x}{y})}^{S} = \frac{y}{x}

if {(\frac{x}{y})}^{S} = - 1

∴ \frac{y}{x} = - 1

y = - x

∴ {(\frac{x}{- x})}^{S} = - 1

{(- 1)}^{S} = - 1

∴ S = 1

∴ \log_{\frac{x}{y}} (\frac{y}{x}) = 1

i f x > 0, y > 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 .

Commented by FilupSmith last updated on 06/Apr/16

\log_{\frac{x}{y}} (\frac{y}{x}) = \frac{\ln (\frac{y}{x})}{\ln (\frac{x}{y})} = \frac{\ln y - \ln x}{\ln x - \ln y} ? ? ?

Commented by FilupSmith last updated on 06/Apr/16

W here is my mistake ?

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

S o r r y, I m i s u n d e r s t o o d .

Answered by FilupSmith last updated on 05/Apr/16

= \frac{\ln y - \ln x}{\ln x - \ln y}

Commented by FilupSmith last updated on 07/Apr/16

you are correct!!

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

\frac{\ln y - \ln x}{\ln x - \ln y}

{C o u l d n}^{'} t w e p r o c e e d f u r t h e r

a s f o l l o w s

= \frac{- (\ln x - \ln y)}{\ln x - \ln y} = - 1

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

\log_{(\frac{x}{y})} (\frac{y}{x}) = ?

\log_{(\frac{x}{y})} {(\frac{x}{y})}^{- 1} = (- 1) \log_{(\frac{x}{y})} (\frac{x}{y}) = (- 1) (1) = - 1

[∵ \log_{a} a = 1]