How would I find the value of $a$ from the below equation:
$$a\cosh(\frac a{50})-a=20$$
I have tried doing it by turning the $\cosh$ into its respective exponential form but I ended up in a dead end in the rearranging. Is there an alternative, simpler way?
On
The numerical solution of this transcendental equation is $A = 45.3732176802461...$.
I obtained it via the fixed point method after having transformed the initial equation
$$a\cosh(\frac a{50})-a=20$$
(using formula $\cosh(2a)=2 \sinh^2(a)+1$) into :
$$a(1+2 \sinh^2(\frac a{100}))-a=20 \ \ \iff \ \ a \ \sinh^2(\frac a{100})=10\tag{1}$$
Function $f$ defined by $f(a)=a \sinh^2(\frac a{100})$ is increasing because its derivative $f'(a)=\sinh^2(\frac a{100})+\tfrac{a}{100} \sinh \tfrac{2a}{100}$ is positive ; as f(a) tends to $+\infty$ when $a \to \infty$, $f$ takes value $10$ for a unique value $A$.
Let us now transform (1) into $a=g(a)$ where $g(a):=100 \ a \ \text{asinh}\left(\sqrt{\tfrac{10}{a}}\right)$
Fixed point recurrence $a_{n+1}=g(a_n)$ with (for example) $a_0=20$ gives value $A$.
As said in comments, there is no way to get an analytical solution to find $a$ such that $$a\cosh(\frac a{50})-a=20$$ and some numerical method would be required (Newton method will be the simplest).
Consider that you look for the zero of function $$f(a)=a\cosh(\frac a{50})-a-20$$Plotting it reveals that the solution is between $40$ and $50$; so, let us start using $a_0=50$ which will ensure that the solution will be reached without any overshoot (by Darboux theorem).
Below are given the successive iterates $$\left( \begin{array}{cc} n & a_n \\ 0 & 50.000000000000000000 \\ 1 & 45.836520166672586528 \\ 2 & 45.378402530655470850 \\ 3 & 45.373218337983547755 \\ 4 & 45.373217680246110845 \\ 5 & 45.373217680246100259 \end{array} \right)$$