How to determine the value of a variable in a equation with powers I'm completely rusty on this How would be the way of determing the value of x in something like this 100 =(50)/((1 + x)^a) + (50)/((1 + x)^b) + (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

By setting $\frac{1}{1+x}=z$, you are simply looking for a root of $z^a+z^b+z^c=2$

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
$zn+1=z_n-<!--\; -\; -->\frac{f(z_n)}{f^{\prime }(z_n)}$
For your case, this gives the iterative scheme
$z_{n+1}=\frac{5z_n^6+3z_n^4+2z_n^3+2}{z_n^2(6z_n^3+4z_n+3)}$
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.