What is the function $y = (1+1/x)^x$ solved for $x$?

Question

I came across this function in algebra ($e$ being its limit as $x$ goes to infinity) while studying compounded interest. Since this function is a little modified from the real interest formula $y=(1+1/n)^{nt}$ where $n$ is # of times compounded per year and $t$ is time. (I just replaced $n$ with $x$ and assumed $t$ is $1$ to get the modified function). And I wondered: is it possible to solve this function for $x$? (or $n$) Is it possible? If it is, how can it be done? If it isn't, why? I tried turning it into a logarithm which quickly became a dead end because it was $\log_{1+1/x}(y) = x$, for obvious reasons. How does one isolate $x$ (if one can)? Essentially I want to know if this function has an inverse. Thank you!

Answer 1

One may solve for $x$ in terms of the Lambert W function as follows:

Rearrange the equation a bit:

$$y=\left(1+\frac1x\right)^x\\y^{1/x}=1+\frac1x\\1=\left(1+\frac1x\right)y^{-1/x}\\\frac1y=\left(1+\frac1x\right)y^{-(1+1/x)}\\\frac1y=\left(1+\frac1x\right)e^{-\ln(y)(1+1/x)}\\-\frac{\ln(y)}y=-\ln(y)\left(1+\frac1x\right)e^{-\ln(y)(1+1/x)}\\W_k\left(-\frac{\ln(y)}y\right)=-\ln(y)\left(1+\frac1x\right)\\-\frac{W_k(-\ln(y)/y)}{\ln(y)}=1+\frac1x$$

$$x=-\left(1+\frac{W_k(-\ln(y)/y)}{\ln(y)}\right)^{-1}$$

where $e=\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n,\ln=\log_e,$ and $W_k$ is the $k$th branch of the Lambert W function.

Note that since $-\frac{\ln(y)}y=\ln(1/y)e^{\ln(1/y)}$, there is a choice of branch for each $y$ that would give a simpler solution, though it turns out this solution is extraneous.

There are other solutions here as well, but for actual calculation purposes, it is probably better to tackle the problem numerically:

For quick and lazy work, we can see that

$$x=\log_{1+1/x}(y),\quad x=\left(y^{1/x}-1\right)^{-1}$$

with the initial approximation of $x\approx\frac{2y-e}{2e-2y}$ to get an iterative approximation. Iterate the first equation for $y<e$ and the second for $y>e$. For example, with $y=1.5$, we use the first one, with the initial approximation of

$x\approx\frac{3-e}{2e-3}\approx0.11562$

and by substituting into the first equation,

$x\approx\log_{1+1/0.11562}(1.5)\approx0.17887$

and again,

$x\approx\log_{1+1/0.17887}(1.5)\approx0.21503$

etc.

a few calculator clicks later...

which gives the approximate solution: $x\approx0.253860169$ for $y=1.5$.

(Note that these can be improved, but require more background to understand)

Answer 2

I wondered: is it possible to solve this function for $x$ ? (or $n$) Is it possible? If it is, how can it be done? If it isn't, why?

So we have the equality $$y=\left(1+\frac 1x\right)^x,$$ and you would like to know whether it is possible to express $x$ explicitly in terms of $y,$ or when $x=n$ (by which I assume you mean $x$ is real in general whereas $n$ is a positive integer).

In both cases it is not -- changing $x$ to $n$ doesn't help significantly either, because (if you apply the binomial expansion formula with $x=n$), the last term $1/n^n$ gives an indication why it might be impossible.

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I'm curious, though. Why do you need this explicit rearrangement?