Question

if f(x) = log(a-x/a+x), then find the value of f(x) + f(-x)​

Answer 1

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

$\rm :\longmapsto\:f(x) = log\bigg(\dfrac{a - x}{a + x} \bigg)$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:f(x) + f( - x)$

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

Given that

$\rm :\longmapsto\:f(x) = log\bigg(\dfrac{a - x}{a + x} \bigg)$

So,

$\rm :\longmapsto\:f( - x) = log\bigg(\dfrac{a - ( - x)}{a + ( - x)} \bigg)$

$\rm :\longmapsto\:f( - x) = log\bigg(\dfrac{a + x}{a- x} \bigg)$

We know that

$\boxed{ \bf{ \: \frac{1}{x} = {x}^{ - 1} }}$

So, using this identity,

$\rm :\longmapsto\:f( - x) = {\bigg(\dfrac{a - x}{a + x} \bigg) }^{ - 1}$

We know that,

$\boxed{ \bf{ \: log {x}^{y} = y \: logx}}$

So, using this property, we have

$\rm :\longmapsto\:f( - x) = - \: log\bigg(\dfrac{a - x}{a + x}\bigg)$

$\rm :\longmapsto\:f( - x) = - \: f(x)$

$\red{\bigg \{ \because \: f(x) = log\bigg(\dfrac{a - x}{a + x}\bigg) \bigg \}}$

$\bf\implies \:f(x) + f( - x) = 0$

Additional Information :-

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

$\boxed{ \bf{ \: log(x \div y) = logx - logy}}$

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

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

$\boxed{ \bf{ \: log_{x}(y) = \frac{logx}{logy}}}$

$\boxed{ \bf{ \: {e}^{logx} = 1}}$

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

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