Question

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

Answer 1

log(162/343) + log(7/9)^2 - log(1/7)
=log[(162×7)/343] + log(49/81)
=log(162/49) +log(49/81)
=log[(162×49)/(49×81)]
=log2
proved

Answer 2

Answer:

Hence proved that $log(\frac{162}{343})+2log(\frac{7}{9})-log(\frac{1}{7})=log2$

Step-by-step explanation:

We have to prove

$log(\frac{162}{343})+2log(\frac{7}{9})-log(\frac{1}{7})=log2$

Left hand side,

$log(\frac{162}{343})+2log(\frac{7}{9})-log(\frac{1}{7})$

$log(\frac{162}{343})+log(\frac{7}{9})^2-log(\frac{1}{7})$      $[\because loga^b=bloga]$

$log(\frac{162}{343})+log(\frac{49}{81})-log(\frac{1}{7})$

$log(\frac{162}{343}\times \frac{49}{81})-log(\frac{1}{7})$      $[\because loga+logb=log(ab)]$

$log(\frac{2}{7})-log(\frac{1}{7})$

$log(\frac{\frac{2}{7}}{\frac{1}{7}})$      $[\because loga-logb=log(\frac{a}{b})]$

$log(\frac{2}{7}\times \frac{7}{1})$

$log(2)$

Hence proved that left hand side is equal to the right hand side.