a-b-2p-b-c-2q-c-a-2r-prove-that-pqr-1-8-

Question Number 59727 by muneshkumar last updated on 14/May/19

a = b^{2 p} b = c^{2 q} c = a^{2 r} p r o v e t h a t p q r = \frac{1}{8}

Answered by MJS last updated on 13/May/19

\ln a = 2 p \ln b

\ln b = 2 q \ln c

\ln c = 2 r \ln a

\ln a = 8 p q r \ln a

8 p q r = 1

q = \frac{1}{8 p r}

\Rightarrow {it}^{'} s wrong

Commented by $@ty@m last updated on 14/May/19

q = \frac{1}{8 p r} \equiv \frac{1}{8} = p q r

{w h a t}^{'} s w r o n g ?

Commented by MJS last updated on 14/May/19

the question has been changed after I gave

my answer (look at my other comment)

Commented by $@ty@m last updated on 14/May/19

O h I s e e .

I a l r e a d y w e n t t h r o u g h y o u r o t h e r

c o m m e n t b u t (s i n c e t h e q n . h a s

b e e n c h a n g e d) I c o u l d f i n d i t s

r e l e v e n c y .

Answered by MJS last updated on 14/May/19

the given formulas are symmetric but

q = \frac{2 p r}{p + r} is not . (q = \frac{2 p r}{p + r} \Leftrightarrow p = \frac{q r}{2 r - q} \Leftrightarrow r = \frac{p q}{2 p - q})

\Rightarrow it cannot be universally true

Answered by tanmay last updated on 14/May/19

c = a^{2 r}

c = {(b^{2 p})}^{2 r}

c = b^{4 p r}

c = {(c^{2 q})}^{4 p r}

c = c^{8 p q r}

8 p q r = 1

p q r = \frac{1}{8}

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