Question

Xlogy-logz.ylogz-logx.zlogx-logy=1
how to prove this??

Answer 1

it was supposed to x to the power logy-logz y to the power logz-logx z to the power logx-logy

Answer 2

\(\huge x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}\)
like this ?

Answer 3

ah thank god..yes like that

Answer 4

consider
\(\huge \log[x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}]\)
can u simplify that ?

Answer 5

use, \(\log AB=\log A+\log B \\ \log A^{B}=B\log A\)

Answer 6

you should get that =0 = ln 1

Answer 7

i dont understand y u multiplied the whole by log????

Answer 8

to be able to simplify

Answer 9

else, nothing can't be really done to that expression

Answer 10

i have done that problem in this way
\[x ^{\log y/z}y ^{\log z/x}z ^{\log x/y}\]

Answer 11

and i cudnt figure out the next step..... :(

Answer 12

i know, i tried that also....since i could not proceed, i thought of taking log....

Answer 13

if u dont mind,can u show me the simplification process...??? :)

Answer 14

\(\huge \log[x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}] \\\large =\log[x^{\log y-\log z}]+\log[y^{\log z-\log x}]+\log[z^{\log x-\log y}]\)
got this ?

Answer 15

now use
\( \\ \log A^{B}=B\log A\)

Answer 16

@suss

Answer 17

oh sorry man i jus lost connection
leeme check it out

Answer 18

y did u keep the + sign ???

Answer 19

@hartnn

Answer 20

because
\(\log AB=\log A+\log B \\ \)

Answer 21

ah finally got it .....thx a ton man

Answer 22

sure ?
u getting that = 0 ?
then write 0=log 1

Answer 23

yes 100%..i hav to use formula of log ab bak ther ......i really got it man

Answer 24

good :)
welcome ^_^