Simplifying $\frac{\log(x)}{x}=y$. I am trying to find the value of $r$ where the Rule of 72 will accurately estimate an investment's doubling time. Put simply, the Rule of 72 requires that 72 be divided by the interest percentage per period to obtain the approximate number of periods required for doubling.
As a threshold matter, future value doubles present value where $(1+r)^t = 2$:
$FV = PV(1+r)^t$
To solve for $t$, I first took the logarithm of both sides:
$(1+r)^t = 2$
$\log((1+r)^t)=\log(2)$
$t = \frac{\log(2)}{\log(1+r)}$
Then, I determined that the Rule of 72 works perfectly where:
$\frac{72}{(r)(100)} = \frac{\log(2)}{\log(1+r)}$
$\frac{\log(1+r)}{r} = \frac{(100)(\log(2))}{72}$
And I simplify that into:
$$\frac{\log(1+r)}{r}=0.4181...$$
How do I solve for $r$ if I know $\frac{\log(1+r)}{r}=0.4181...$?
 A: It is useful to know that equation 
\begin{align}
\frac{\ln(1+x)}{x}=y
\end{align}
has a real solution for $y>0$ in terms of
Lambert W 
function:
\begin{align}
x&=
\begin{cases}
-1-\displaystyle\frac{\operatorname{W_{-1}}(-y\exp(-y))}{y},&\text{if $0<y<1$} \quad(1)\\
\quad 0,&\text{if $y=1$} \quad(2)\\
-1-\displaystyle\frac{\operatorname{W_{0}}(-y\exp(-y))}{y},&\text{if $y>1$} \quad(3)
\end{cases}
\end{align}
where $\operatorname{W_{0}}(t)$ 
and $\operatorname{W_{-1}}(t)$ are two real branches of 
Lambert W 
function for $-\exp(-1)\le t<0$.
Thus to find $r$ for the case
\begin{align}
\frac{\ln(1+r)}{r}&={100\over 72}\ln(2) 
\approx 0.96270
\end{align}
we need to use the case (1)
since the value of $y=\frac{25}{18}\ln(2)<1$:
\begin{align}
r&=-1-\displaystyle\frac{\operatorname{W_{-1}}(-y\exp(-y))}{y}.
\end{align}
Wolfram Alpha
gives the answer $r=0.07846871453015381610\dots$
A: Assume that $x,y$ are restricted to the real numbers; then
$${\log(1+x)\over x}=y\implies(x+1)^\frac 1x=e^y$$
further restricts $x\gt -1$, with a discontinuity at $x=0.$  $z=e^y$ is just a positive real number for all real $y$, so we only have to worry about $(x+1)^\frac 1x$ (which is plenty of course).  Note $\frac 1x-\frac 1{x+1}=\frac 1{x(x+1)}$, so we have
$$(x+1)^\frac 1x=(x+1)^\frac 1{x+1}(x+1)^\frac 1{x(x+1)}=\prod_{i=1}^\infty(x+1)^\frac 1{(x+1)^i}$$
We have one possible value $r$ where
$$\frac{\log(1+r)}{r}={100\over 72}\log 2\approx 0.4181...$$
because $\log(1+r)\over r$ is generally decreasing (except at the discontinuity).  This value is somewhere in the range $r\in [0.077,0.079]$ (determined by a graphing program) and is (more exactly) $r\approx 0.0784687145301538...$ (determined by Wolfram Alpha) or a bit less than $8$%.
