Re-arranging cosh to make x the subject

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To rearrange the equation y = 38cosh(x/40) to solve for x, first isolate cosh by dividing both sides by 38, resulting in cosh(x/40) = y/38. Then, apply the inverse hyperbolic cosine function, arccosh, to both sides, giving x/40 = arccosh(y/38). Finally, multiply both sides by 40 to solve for x, yielding x = 40 * arccosh(y/38). For y = 50, substitute this value into the equation to calculate x. Understanding the use of arccosh is crucial for solving similar problems.
zfleas
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Hi everyone, my first post so forgive me if its rubbish!
I have been set a question in my HNC course and have hit a wall with where to proceed with it. I would just post the question but rather than that and you giving me the answer I would rather learn how to do it myself, so I will answer a similar one.

I need to calculate x in the following when y = 50
y=38cosh(x/40)

How would I go about making x the subject so I can calculate it? Or is it even more simple than that?
Any help is greatly appreciated
Cheers
:)
 
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There is a defined function for this, arccosh.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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