Question

please answer fast (class XI logarithm)

Answer 1

Hi ,

x = ( e^y - e^-y )/( e^y + e^-y )----( 1 )

i )1 + x = 1 + [ (e^y - e^-y)/(e^y+e^-y)]

= [(e^y +e^-y+e^y-e^-y )/(e^y+e^-y)]

= ( 2e^y )/[ e^y + e^-y ] ---( 2 )

Similarly ,

ii ) 1 - x = ( 2e^-y )/( e^y + e^-y ) ---( 3 )

Now ,

iii )( 1 + x )/( 1 - x ) = ( 2 )/ ( 3 )

= ( 2e^y )/ (2 e^-y )

= ( e^y)/ ( e^-y )

= e^2y ---( 4 )

Take log both sides

log [ ( 1+x)/(1 +x )] = log e^2y

= 2y log e

1/2 log [ (1 + x )/( 1 - x ) ] = y log e

Therefore ,

[ since log e ( base e ) = 1 ]

log √[( 1+x )/( 1 - x ) ] = y

Hence proved.

I hope this helps you.

: )