How do you solve $13 sinh x - 7 cosh x+1=0$ ?

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Answer 1 · 2016-06-24T04:17:06

x = 0.5108, nearly.

Using $sinh x =(e^x-e^(-x))/2 and cosh x = (e^x+e^(-x))/2$ ,

the given equation reduces to

$3 e^(2x)+e^x-10=0$ . Solving this quadratic in $e^x$ ,

$e^x=-2 and 5/3$ ..

As #e^x>0, e^x = 5/3..

Inversely, $x = ln (5/3)=0.51082563$ , nearly.

How do you solve 13 sinh x - 7 cosh x+1=013 sinh x - 7 cosh x+1=0?