Isolate Variable in Fraction I can approximate u with a calculator by guessing or using excel but I want to isolate it.
$100 = \dfrac{1 + \dfrac{1}{(1+u)^6}}{u}$
Can not seem to do it by hand myself. Is it possible using only simple algebra?
 A: The solutions that you are looking for are the roots of a degree 7. Polynomial. 
$$100u(1+u)^6-(1+u)^6-1 = 0$$
There is no shorthand formula for finding the roots of a high degree polynomial. Numerical approximation is probably the way to go, or use a graphing calculator and read the roots off.
A: This looks very much as a finance problem.
Let us rewrite it as
$$\frac{u}{1+\frac{1}{(1+u)^6}}=a$$
Develop the lhs as a Taylor series to get
$$\text{lhs}=\frac{1}{2}u+\frac{3 }{2}u^2-\frac{3 }{4}u^3-4 u^4+\frac{51 }{8}u^5+O\left(u^6\right)$$ and use series reversion to get
$$u=2 a-12 a^2+156 a^3-2392 a^4+40560 a^5+O\left(a^6\right)$$ Making $a=\frac 1 {100}$ would give
$$u=\frac{2367017}{125000000}=0.0189361$$ while the exact solution is $\approx 0.0189355$
Edit
We could make the problem more general considering
$$\frac{u}{1+\frac{1}{(1+u)^n}}=a$$
$$\text{lhs}=\frac{1}{2}u+\frac{n }{4}u^2-\frac{n }{8}u^3-\frac{n
   \left(n^2-4\right)}{48}  u^4+\frac{n \left(n^2-2\right)}{32}  u^5+O\left(u^6\right)$$ from which
$$u=2 a-2  na^2+2 n (2 n+1)a^3-\frac{2}{3} n (2 n+1) (7 n+4) a^4+$$ $$2  n (2 n+1)^2 (3   n+2)a^5+O\left(a^6\right)$$
