Question

log√q(p^2).log√r(q^2).log√p(r^2)=64​

Answer 1

Answer:

log√q(p^2).log√r(q^2).log√p(r^2)=64

Answer 2

Answer:

sorry for answering so late but I hope it helps

Step-by-step explanation:

$lhs = log_{ \sqrt{q} }( {p }^{2} ) + log_{ \sqrt{r} }( {q }^{2} ) + log_{ \sqrt{p} }( {r }^{2} )$

$= \: \frac{ log \: {p}^{2} \: }{ log( \sqrt{q} ) } + \frac{ log \: {q}^{2} \: }{ log( \sqrt{r} ) } + \frac{ log \: {r}^{2} \: }{ log( \sqrt{p} ) }$

$= \frac{ log( {p}^{2} ) }{ log( \ \: {q}^{ \frac{1}{2} } ) } + \frac{ log( {q}^{2} ) }{ log( {r}^{ \frac{1}{2} } ) } + \frac{ log( {r}^{2} ) }{ log( {p}^{ \frac{1}{2} } ) }....... (because \: \sqrt{x } = {x}^{ \frac{1}{2} } )$

$= \: \frac{2 \: log \: p }{ \frac{1}{2} \times log(q) } \times \frac{2 log(q) }{ \frac{1}{2} \times log(r)} \times \frac{2 log(r) }{ \frac{1}{2} \times log(p)}$

$cutting \: log \: p \: from \: log \: p \: and \: same \: to \: others$

$= \: \frac{2}{ \frac{1}{2} } \times \frac{2}{ \frac{1}{2} } \times \frac{2}{ \frac{1}{2} }$

$= \: 4 \times 4 \times 4 = 64$

HENCE PROVED.......

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