a=b^(2p ) b=c^(2q ) c=a^(2r) prove that pqr=(1/8)


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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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