Question

Soln of ques 36 pls?​

Answer 1

Answer:-

option(c)

Explanation:-

Given:-

p² + q² = 12pq

To Find:-

value of log (p+2q)

Solution:-

${p}^{2}+ {q}^{2} = 12pq$

By adding 3q² and 4q on both sides

${p}^{2}+ {q}^{2} + 3{q}^{2} + 4q= 12pq + 4q + 3 {q}^{2}$

${p}^{2}+ 4{q}^{2}+ 4q = 12pq + 4q + {q}^{2}$

${(p+2q)}^{2} = 16pq + 3 {q}^{2}$

By taking LOG on both sides

$log {(p+2q)}^{2} = log [16pq + 3 {q}^{2}]$

$2 log(p+2q) = log [16pq(1 +\dfrac{ 3q}{16p})]$

$log(p+2q) = \dfrac{1}{2}[log\: 16pq+ log (1 +\dfrac{ 3q}{16p})]$

$log(p+2q) = \dfrac{1}{2} [4 log \: 2 + log\: p + log\: q+ log (1 +\dfrac{ 3q}{16p})]$

Note:-

Assume Value of

$\dfrac{3q}{16p} < < < 1$

Hence

$log (1 +\dfrac{ 3q}{16p}) = log 1$

Now, we get

$log(p+2q) = \dfrac{1}{2} [4 log \: 2 + log\: p + log\: q+ log 1 ]$

$log(p+2q) = \dfrac{1}{2} [4 log \: 2 + log\: p + log\: q+ 0]$

$log(p+2q) = \dfrac{1}{2} [4 log \: 2 + log\: p + log\: q]$