How to solve for $V$, given the equation $943460=V^2\ln\left[\cosh\left(\frac{3735.2}{V}\right)\right]$

Question

I was working on a homework problem and I simplified the result down to:

$$943460=V^2\ln\left[\cosh\left(\frac{3735.2}{V}\right)\right]$$

According to Wolfram Alpha, this equality gives $V=266$, which is the correct answer from the back of the textbook. But I don't know how to go about numerically showing this to be true.

Any help with simplifying functions ln() and cosh() is much appreciated.

Answer 1

You may rearrange the equation by writing $\ln()$ in its exponential form and expressing $\cosh x = (e^x+e^{-x})/2$,

$$e^{943460/V^2}=\cosh\left(\frac{3735.2}{V}\right)=\frac 12 \left( e^{3735.2/V} + e^{-3735.2/V } \right)$$

Observe that the last term would be quite small and could be dropped. Then, the equation simplifies to

$$e^{943460/V^2}=\frac 12 e^{3735.2/V} = e^{3735.2/V-\ln(2)} $$

or,

$$943460/V^2 - 3735.2/V +\ln(2)=0$$

which is just a quadratic equation in $1/V$, whose acceptable solution is

$$\frac{1}{V} = 0.003764$$

or,

$$V=265.7$$

Answer 2

If you graphed the function $$f(V)=V^2\,\log⁡\left(\cosh\left(\frac{3735.2}{V}\right)\right)-943460$$ you probably noticed that it looks like a straight line. This is a very good information and then we can start Newton method even with a very small value $V_0$.

The sucessive iterates would then be $$\left( \begin{array}{cc} n & V_n \\ 0 & 0.1 \\ 1 & 252.596 \\ 2 & 265.650 \\ 3 & 265.685 \end{array} \right)$$