Question

Simplify: log3 to the base 4x log64 to the
base 243​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\bf :\longmapsto\: log_{4}(3) \times log_{243}(64)$

Consider,

$\rm :\longmapsto\: log_{4}(3)$

$\rm \: \: = \: \: log_{ {2}^{2} }(3)$

$\rm \: \: = \: \: \dfrac{1}{2} log_{2}(3)$

$\: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: log_{ {a}^{p} }( {b}^{q} ) = \dfrac{q}{p} log_{a}(b) \bigg \}}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{\rm :\longmapsto\: \bf \: log_{4}(3) = \dfrac{1}{2} log_{2}(3)}}$

Now,

Consider,

$\rm :\longmapsto\: log_{243}(64)$

$\rm \: \: = \: \: log_{ {3}^{5} }( {2}^{6} )$

$\rm \: \: = \: \: \dfrac{6}{5} log_{3}(2)$

$\: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: log_{ {a}^{p} }( {b}^{q} ) = \dfrac{q}{p} log_{a}(b) \bigg \}}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \red{ \boxed{\rm :\longmapsto\: \bf \: log_{243}(64) = \dfrac{6}{5} log_{3}(2)}}$

Now,

$\bf :\longmapsto\: log_{4}(3) \times log_{243}(64)$

$\rm \: \: = \: \: \dfrac{1}{2} log_{2}(3) \times \dfrac{6}{5} log_{3}(2)$

$\rm \: \: = \: \: \dfrac{3}{5}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: log_{x}(y) \times log_{y}(x) = 1\bigg \}}$

$\boxed{\bf :\implies\: log_{4}(3) \times log_{243}(64) = \dfrac{3}{5}}$

Additional Information :-

$\boxed{ \sf \: log_{x}(x) = 1 }$

$\boxed{ \sf \: log_{x}(y) = \dfrac{1}{ log_{y}(x) } }$

$\boxed{ \sf \: log(1) = 0 }$

$\boxed{ \sf \: {a}^{ log_{a}(x)} = x}$

$\boxed{ \sf \: {a}^{y log_{a}(x)} = {x}^{y} }$

$\boxed{ \sf \: {a}^{ log_{e}(b) } = {b}^{ log_{e}(a) } }$

$\boxed{ \sf \:logx + logy = log(xy) }$

$\boxed{ \sf \:logx - logy = log( \dfrac{x}{y} ) }$

$\boxed{ \sf \: log( {x}^{y} ) = ylogx }$