Question

LOGARITHM SUM IMAGE ATTACHED PLEASE CHECK​

Answer 1

Step-by-step explanation:

We have,

$x = \frac{ {e}^{y} - {e}^{ - y} }{ {e}^{y} + {e}^{ - y} }$

$\implies \frac{1}{x} = \frac{ {e}^{y} + {e}^{ - y} }{ {e}^{y} - {e}^{ - y} }$

Using componendo and dividendo

$\implies \frac{1 + x}{1 - x} = \frac{ {e}^{y} + {e}^{ - y} + {e}^{y} - {e}^{ - y} }{ {e}^{y} + {e}^{ - y} - {e}^{y} + {e}^{ - y} }$

$\implies \frac{1 + x}{1 - x} = \frac{2 {e}^{y} }{2 {e}^{ - y} }$

$\implies \frac{1 + x}{1 - x} = \frac{ {e}^{y} }{ {e}^{ - y} }$

Taking log both sides,

$\implies log( \frac{1 + x}{1 - x} ) = log( \frac{ {e}^{y} }{ {e}^{ - y} } )$

$\implies log( \frac{1 + x}{1 - x} ) = log( \frac{ {e}^{y} }{ {e}^{ - y} } )$

$\implies log( \frac{1 + x}{1 - x} ) = log( {e}^{y} ) - log( {e}^{ - y} )$

$\implies log( \frac{1 + x}{1 - x} ) = y log(e) + y log(e)$

$\implies log( \frac{1 + x}{1 - x} ) = 2y$

$\implies y = \frac{1}{2} log( \frac{1 + x}{1 - x} )$