Question

solve this as fast as possible​

Answer 1

Question :- Prove that

$\rm \: log {\bigg[1 - \bigg \{1 - {(1 - {x}^{2}) }^{ - 1} \bigg\}^{ - 1} \bigg]}^{ - \dfrac{1}{2} } = logx$

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

Consider LHS

$\rm \: log {\bigg[1 - \bigg \{1 - {(1 - {x}^{2}) }^{ - 1} \bigg\}^{ - 1} \bigg]}^{ - \dfrac{1}{2} }$

$\rm \: = log {\bigg[1 - \bigg \{1 - \dfrac{1}{1 - {x}^{2} } \bigg\}^{ - 1} \bigg]}^{ - \dfrac{1}{2} }$

$\rm \: = log {\bigg[1 - \bigg \{ \dfrac{1 - {x}^{2} - 1}{1 - {x}^{2} } \bigg\}^{ - 1} \bigg]}^{ - \dfrac{1}{2} }$

$\rm \: = log {\bigg[1 - \bigg \{ \dfrac{ - {x}^{2}}{1 - {x}^{2} } \bigg\}^{ - 1} \bigg]}^{ - \dfrac{1}{2} }$

$\rm \: = log {\bigg[1 - \bigg \{ \dfrac{{x}^{2}}{{x}^{2} - 1} \bigg\}^{ - 1} \bigg]}^{ - \dfrac{1}{2} }$

$\rm \: = log {\bigg[1 -\dfrac{ {x}^{2} - 1}{ {x}^{2} } \bigg]}^{ - \dfrac{1}{2} }$

$\rm \: = log {\bigg[\dfrac{ {x}^{2} - ({x}^{2} - 1)}{ {x}^{2} } \bigg]}^{ - \dfrac{1}{2} }$

$\rm \: = log {\bigg[\dfrac{ {x}^{2} - {x}^{2} + 1}{ {x}^{2} } \bigg]}^{ - \dfrac{1}{2} }$

$\rm \: = log {\bigg[\dfrac{1}{ {x}^{2} } \bigg]}^{ - \dfrac{1}{2} }$

$\rm \: = log {\bigg[ {x}^{ - 2} \bigg]}^{ - \dfrac{1}{2} }$

$\rm \: = log {\bigg[ x\bigg]}^{( - 2) \times \dfrac{( - 1)}{2} }$

$\rm \: = \: logx$

Hence,

$\red{\rm\implies \: log {\bigg[1 - \bigg \{1 - {(1 - {x}^{2}) }^{ - 1} \bigg\}^{ - 1} \bigg]}^{ - \dfrac{1}{2} } = logx}$

$\rule{190pt}{2pt}$

Additional information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ log_{x}(x) = 1}\\ \\ \bigstar \: \bf{ log_{x}( {x}^{y} ) = y}\\ \\ \bigstar \: \bf{ log_{ {x}^{z} }( {x}^{w} ) = \dfrac{w}{z} }\\ \\ \bigstar \: \bf{ log_{a}(b) = \dfrac{logb}{loga} }\\ \\ \bigstar \: \bf{ {e}^{logx} = x}\\ \\ \bigstar \: \bf{ {e}^{ylogx} = {x}^{y}}\\ \\ \bigstar \: \bf{log1 = 0}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$