Question

If $e^{x} = y + \sqrt{1+y^{2} }$ , then y is equal to?

Answer 1

${e}^{x} = y + \sqrt{1 + {y}^{2} } \\ = > {e}^{x} - y = \sqrt{1 + {y}^{2} } \\ = > {( {e}^{x} - y )}^{2} = 1 + {y}^{2} \\ = > { ({e}^{x} )}^{2} - 2( {e}^{x} )(y) + {y}^{2} = 1 + {y}^{2} \\ = > {e}^{2x} - 2y {e}^{x} = 1 \\ = > {e}^{2x} - 1 = 2y {e}^{x } \\ = > \frac{{e}^{2x} - 1}{2 {e}^{x} } = y \\ = > y = \frac{{e}^{2x} - 1}{2 {e}^{x} }$

This is the answer.

Hey; Sorry This might not really be the best time to get this answer...

BTW this is my way of saying thanks to you for Answering my question.

(the comments are banned so(-_-) i found no other way.)

i really got some good punchlines.

Anyways if you could:? could you add some current affairs and stats in that answer. I could really use them and that answer is a bit short...(^_^) no offence though...

Thanks!

Regards;

Leukonov/Olegion.

Answer 2

Answer:

I HOPE IT'S HELP YOU

SORRY IF IN MY ANSWER ANY MISTAKE