I'm completely rusty on this
How would be the way of determing the value of x in something like this
$\ 100 = \frac{50}{(1 + x)^a} + \frac{50}{(1 + x)^b} + \frac{50}{(1 + x)^c}$
a, b, c are known but are fractions themselves so I put just these letters for the sake of simiplicity.
I'm trying to do something like $\ 100 = 50 (-a log(1+ x) $ and so on but to be fair I'm absolutely lost
Thanks
Let us start with the function $$f(x)=\frac{1}{(1 + x)^a} + \frac{1}{(1 + x)^b} + \frac{1}{(1 + x)^c}-2=0$$ As said before, a first change of variable $y=\frac{1}{1+x}$ allows to rewrite it as $$f(y)=y^a+y^b+y^c-2=0$$ which could already be handled easily; but we can make it nicer looking at the exponents $a=\frac{90}{365}$,$b=\frac{60}{365}$,$c=\frac{45}{365}$ and notice that these exponents are in ratio $6:4:3$; so, for conveniency and to avoid fractional powers, let us define $$z=y^{\frac{15}{365}}$$ so the equation write now $$f(z)=z^6+z^4+z^3-2=0$$ A good candidate for solving this polynomial is Newton method which, starting from a reasonable guess $z_0$ will update it according to $$z{n+1}=z_n-\frac{f(z_n)}{f'(z_n)}$$ For your case, this gives the iterative scheme $$z_{n+1}=\frac{5 z_n^6+3 z_n^4+2 z_n^3+2}{z_n^2 \left(6 z_n^3+4 z_n+3\right)}$$ We shall admit that we look for a positive solution of $x$, then of $z$; since the sum of three positive terms has to be equal to $2$, we can easily bracket the solution using $3z^6<2$ and $3z^2>2$ which means that the solution we look for is such that $$\sqrt{\frac{2}{3}}=0.816497<z<\sqrt[6]{\frac{2}{3}}=0.934655$$ which is a quite narrow range. Since these numbers are rather close to $1$, let us be lazy and start iterating at $z_0=1$; the following iterates are so obtained : $0.9230769231$, $0.9095860491$, $0.9092211895$, $0.9092209302$ which is the solution for ten significant figures.
Now, back to $x$; the effective change of variable we made is $$z=\frac{1}{(1 + x)^{\frac{15}{365}}}=\frac{1}{(1 + x)^{\frac{3}{73}}}$$ So, the simplest way is going to logarithms; from the value of $z$, we have the value of $\log(1+x)$ and then by exponentiation the value of $x$. For his case, we get $$x=9.13236132490548$$
One last point to mention : equation $f(z)=0$ has two real roots while equation $f(x)=0$ has only one root. This is relevant from the change of variables and the negative root of $f(z)=0$ must be discarded.
If there have been no such relations between coefficients $a,b,c$, the simplest would have been to solve $$f(y)=y^a+y^b+y^c-2=0$$ using Newton method starting from an appropriate guess deduced from the inequalities used here.