Question

show that log 162/343 + 2 log 7/9 - log 1/7= log 2

​

Answer 1

TO PROVE :–

$\\ \bf \implies\log \left(\dfrac{162}{343} \right) + 2 \log \left(\dfrac{7}{9} \right) - \log\left(\dfrac{1}{7} \right) = \log(2)$

SOLUTION :–

• Let's take L.H.S –

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \right) + 2 \log \left(\dfrac{7}{9} \right) - \log\left(\dfrac{1}{7} \right)$

• Using identity –

$\\ \large\to \red{ \boxed{ \bf m\log \left(n\right) = \log( {n}^{m} )}}$

• So that –

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \right) +\log \left[\left(\dfrac{7}{9} \right)^{2} \right]- \log\left(\dfrac{1}{7} \right)$

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \right) +\log \left(\dfrac{49}{81} \right)- \log\left(\dfrac{1}{7} \right)$

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \right) +\log \left(\dfrac{49}{81} \right) + \log\left[\left(\dfrac{1}{7} \right)^{ - 1}\right]$

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \right) +\log \left(\dfrac{49}{81} \right) + \log(7)$

• Using identity –

$\\ \large\to \red{ \boxed{ \bf \log (p) + log(q) + log(r) = \log(pqr)}}$

• So that –

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \times \dfrac{49}{81} \times 7\right)$

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \times \dfrac{343}{81}\right)$

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{81}\right)$

$\\ \bf \: \: = \: \: \log(2)$

$\\ \bf \: \: = \: \:R.H.S.$

$\\ \bf \longrightarrow \: \:\red { \underbrace{Hence \: \: Proved}}$

Answer 2

TO PROVE :–

$\\ \bf \implies\log \left(\dfrac{162}{343} \right) + 2 \log \left(\dfrac{7}{9} \right) - \log\left(\dfrac{1}{7} \right) = \log(2)$

SOLUTION :–

• Let's take L.H.S –

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \right) + 2 \log \left(\dfrac{7}{9} \right) - \log\left(\dfrac{1}{7} \right)$

• Using identity –

$\\ \large\to \red{ \boxed{ \bf m\log \left(n\right) = \log( {n}^{m} )}}$

• So that –

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \right) +\log \left[\left(\dfrac{7}{9} \right)^{2} \right]- \log\left(\dfrac{1}{7} \right)$

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \right) +\log \left(\dfrac{49}{81} \right)- \log\left(\dfrac{1}{7} \right)$

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \right) +\log \left(\dfrac{49}{81} \right) + \log\left[\left(\dfrac{1}{7} \right)^{ - 1}\right]$

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \right) +\log \left(\dfrac{49}{81} \right) + \log(7)$

• Using identity –

$\\ \large\to \red{ \boxed{ \bf \log (p) + log(q) + log(r) = \log(pqr)}}$

• So that –

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \times \dfrac{49}{81} \times 7\right)$

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{343} \times \dfrac{343}{81}\right)$

$\\ \bf \: \: = \: \: \log \left(\dfrac{162}{81}\right)$

$\\ \bf \: \: = \: \: \log(2)$

$\\ \bf \: \: = \: \:R.H.S.$

$\\ \bf \longrightarrow \: \:\red { \underbrace{Hence \: \: Proved}}$