Question

Prove that 2log(15/18)-log (25/162)+log(4/9)=log2

Answer 1

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

Given Logarithmic expression is

$\rm :\longmapsto\:2log\bigg[\dfrac{15}{18} \bigg] - log\bigg[\dfrac{25}{162} \bigg] + log\bigg[\dfrac{4}{9} \bigg]$

We know,

$\boxed{ \tt{ \: log {x}^{y} = ylogx \: \: }}$

So, using this identity, we get

$\rm \: = \:log {\bigg[\dfrac{15}{18} \bigg]}^{2} - log\bigg[\dfrac{25}{162} \bigg] + log\bigg[\dfrac{4}{9} \bigg]$

can be rewritten as

$\rm \: = \:log {\bigg[\dfrac{15}{18} \bigg]}^{2} + log\bigg[\dfrac{4}{9} \bigg] - log\bigg[\dfrac{25}{162} \bigg]$

We know,

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

and

$\boxed{ \tt{ \: logx - logy = log \frac{x}{y} \: \: }}$

So, on using these results, we get

$\rm \: = \:log\bigg[\dfrac{ {15}^{2} }{ {18}^{2} } \times \bigg[\dfrac{4}{9} \bigg] \div \bigg[\dfrac{25}{162} \bigg]\bigg]$

$\rm \: = \:log\bigg[\dfrac{ {(3 \times 5)}^{2} }{ {( {3}^{2} \times 2)}^{2} } \times \bigg[\dfrac{ {2}^{2} }{ {3}^{2} } \bigg] \times \bigg[\dfrac{162}{25} \bigg]\bigg]$

$\rm \: = \:log\bigg[\dfrac{ \cancel {3}^{2} \times\cancel {5}^{2} }{\cancel {3}^{4} \times\cancel {2}^{2} } \times \bigg[\dfrac{\cancel {2}^{2} }{\cancel {3}^{2} } \bigg] \times \bigg[\dfrac{\cancel {3}^{4} \times 2 }{\cancel {5}^{2} } \bigg] \bigg]$

$\rm \: = \:log2$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: 2log\bigg[\dfrac{15}{18} \bigg] - log\bigg[\dfrac{25}{162} \bigg] + log\bigg[\dfrac{4}{9} \bigg] = log2 \: }}$

More to know :

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

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

$\boxed{ \tt{ \: log_{ {x}^{z} }( {x}^{y} ) = \frac{y}{z} }}$

$\boxed{ \tt{ \: {e}^{logx} = x}}$

$\boxed{ \tt{ \: {e}^{ylogx} = {x}^{y} }}$

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

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