Question

10.25 = (1/B)*Ln( Cosh( 1.695*(9.8*B)^(1/2) ) )

Solve for B

Ln is the natural log and cosh is the hyperbolic function of cos.

I have tried working this problem multiple times and have not been able to get the right answer.

Answer 1

For this particular equation I'm going to use expansion series of cosh and exp

Let's start

Transfer B towards left hand side,

Take exponential of both sides, that will cancel out natural logarithm on right hand side

Now using expansion of exp(x) on LHS and expansion of Cosh(x) on RHS, note that x in both parantheses are different.

I'm just limiting myself upto second order.

by solving above expression one can get

I hope this will help you to get your result.