Question

log (p^2/qr) + log (q^2/pr) + log(r^2/pq) = 0​

Answer 1

It has been proved that

$\bf \: log \left( \frac{ {p}^{2} }{qr} \right) + log \left( \frac{ {q}^{2} }{pr} \right ) + log \left( \frac{ {r}^{2} }{pq} \right) = 0$

Given:

• $log \left( \frac{ {p}^{2} }{qr} \right) + log \left( \frac{ {q}^{2} }{pr} \right ) + log \left( \frac{ {r}^{2} }{pq} \right) = 0$

To find:

• Prove the equation.

Solution:

Formula to be used:

1. $\bf log \left( \frac{ m }{n} \right) = log(m) - log(n)$
2. $\bf log \left( mn \right) = log(m) + log(n)$
3. $\bf log( {n}^{m} ) = m log(n)$

Step 1:

Simplify the LHS by applying rule 1.

$log ({p}^{2}) - log(qr) + log( {q}^{2}) - log(pr) + log( {r}^{2} ) - log(pq) = 0$

Step 2:

Apply rule 1 and 3.

$2log ({p}) - ( log(q) + log(r) )+ 2log ({q}) - ( log(p) + log(q) ) + 2log ({r}) - ( log(p) + log(q) )= 0$

Open the brackets and simplify.

$2log ({p}) - log(q) - log(r) + 2log ({q}) - log(p) - log(q) + 2log ({r}) - log(p) - log(q) = 0$

Add the similar terms.

$2log ({p}) -2 log(p) + 2log ({q}) - 2log(q) + 2log ({r}) - 2 log(r) = 0$

Cancel the similar terms having opposite signs.

$\cancel{2log ({p})} -\cancel{2 log(p) } +\cancel{2log ({q}) } - \cancel{ 2log(q)} +\cancel{2log ({r}) } - \cancel{2 log(r) } = 0$

$0 = 0$

LHS= RHS

Thus,

It has been proved that $\bf log \left( \frac{ {p}^{2} }{qr} \right) + log \left( \frac{ {q}^{2} }{pr} \right ) + log \left( \frac{ {r}^{2} }{pq} \right) = 0$

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Answer 2

$\displaystyle \sf log\bigg( \frac{ {p}^{2} }{qr} \bigg) + log\bigg( \frac{ {q}^{2} }{pr} \bigg) + log\bigg( \frac{ {r}^{2} }{pq} \bigg) = 0 \: \: is \: proved$

Given :

$\displaystyle \sf log\bigg( \frac{ {p}^{2} }{qr} \bigg) + log\bigg( \frac{ {q}^{2} }{pr} \bigg) + log\bigg( \frac{ {r}^{2} }{pq} \bigg) = 0$

To find :

The expression

Formula Used :

We are aware of the formula on logarithm that

$\sf{\: log(ab) = log(a) + log(b) }$

Solution :

Step 1 of 2 :

Write down the given expression

Here the given expression is

$\displaystyle \sf log\bigg( \frac{ {p}^{2} }{qr} \bigg) + log\bigg( \frac{ {q}^{2} }{pr} \bigg) + log\bigg( \frac{ {r}^{2} }{pq} \bigg) = 0$

Step 2 of 2 :

Prove the expression

$\displaystyle \sf log\bigg( \frac{ {p}^{2} }{qr} \bigg) + log\bigg( \frac{ {q}^{2} }{pr} \bigg) + log\bigg( \frac{ {r}^{2} }{pq} \bigg)$

$\displaystyle \sf = log\bigg( \frac{ {p}^{2} }{qr} \times \frac{ {q}^{2} }{pr} \times \frac{ {r}^{2} }{pq} \bigg) \: \: \: \bigg[ \: \because \: log(ab) = log(a) + log(b)\bigg]$

$\displaystyle \sf = log\bigg( \frac{ {p}^{2} \times {q}^{2} \times {r}^{2} }{qr \times pr \times pq} \bigg)$

$\displaystyle \sf = log\bigg( \frac{ {p}^{2} \times {q}^{2} \times {r}^{2} }{{p}^{2} \times {q}^{2} \times {r}^{2} } \bigg)$

$\displaystyle \sf = log \: 1$

$= 0$

Hence the proof follows

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