a-b-log-c-b-c-log-a-c-a-log-b-1-make-logc-c-the-subject-of-formula-

Question Number 23566 by NECx last updated on 01/Nov/17

{(\frac{a}{b})}^{\log c} . {(\frac{b}{c})}^{\log a} . {(\frac{c}{a})}^{\log b} = 1 .

m a k e \frac{l o g c}{c} t h e s u b j e c t o f f o r m u l a

Commented by NECx last updated on 01/Nov/17

p l e a s e h e l p w i t h w o r k i n g s .

Commented by mrW1 last updated on 02/Nov/17

for any a, b, c > 0 we have

\log a \times \log b - \log a \times \log c + \log b \times \log c - \log b \times \log a + \log c \times \log a - \log c \times \log b = 0

\log a \times (\log b - \log c) + \log b \times (\log c - \log a) + \log c \times (\log a - \log b) = 0

\log {(\frac{b}{c})}^{\log a} + \log {(\frac{c}{a})}^{\log b} + \log {(\frac{a}{b})}^{\log c} = 0

\log [{(\frac{b}{c})}^{\log a} \times {(\frac{c}{a})}^{\log b} \times {(\frac{a}{b})}^{\log c}] = 0

{(\frac{b}{c})}^{\log a} \times {(\frac{c}{a})}^{\log b} \times {(\frac{a}{b})}^{\log c} = 1

Answered by abwayh last updated on 02/Nov/17

\log [{(\frac{a}{b})}^{\log c} . {(\frac{b}{c})}^{\log a} . {(\frac{c}{a})}^{\log b}] = \log 1

\log {(\frac{a}{b})}^{\log c} + \log {(\frac{b}{c})}^{\log a} + \log {(\frac{c}{a})}^{\log b} = 0

\log c \log (\frac{a}{b}) + \log a \log (\frac{b}{c}) + \log b \log (\frac{c}{a}) = 0

\log c [\log (\frac{a}{b}) + \frac{\log a}{\log c} \log (\frac{b}{c}) + \frac{\log b}{\log c} \log (\frac{c}{a}) = 0

\log c [\log (\frac{a}{b}) + \log \underset{c}{} a \log (\frac{b}{c}) + \log \underset{c}{} b \log (\frac{c}{a}) = 0

\frac{\log c}{c} [\log (\frac{a}{b}) + \log_{c} a \log (\frac{b}{c}) + \log \underset{c}{} b \log (\frac{c}{a}) = \frac{0}{c}

∴ \frac{\log c}{c} = 0

Commented by mrW1 last updated on 02/Nov/17

\frac{\log c}{c} = 0 means \log c = 0 or c = 1, but

{(\frac{a}{b})}^{\log c} . {(\frac{b}{c})}^{\log a} . {(\frac{c}{a})}^{\log b} = 1 is valid

for any values of a, b, c > 0 .

Commented by NECx last updated on 04/Nov/17

t h a n k s b o s s f o r t h e a p r o a c h . I t

r e a l l y h e l p

Commented by NECx last updated on 04/Nov/17

i t h i n k t h e v a l u e s f o r a b c a r e

1000 100 a n d 10

Commented by mrW1 last updated on 04/Nov/17

no, a b c can be any positive values .

this is proved .

if you want an expression for \frac{\log c}{c},

then it is

\frac{\log c}{c} = r, 0 < r ⩽ \frac{1}{e \times \ln 10}

Commented by NECx last updated on 04/Nov/17

t h e n h o w p o s s i b l e i s i t t o g e t r ?

Commented by mrW1 last updated on 04/Nov/17

c is any value, that means r is any value

in the range (0, \frac{1}{e \times \ln 10}]

Commented by NECx last updated on 05/Nov/17

o k \dots t h a n k s

Facebook Tweet Pin