# Is solution unique for $\left(1+\frac{1}{x}\right)^x=c$ ($c$ is a positive number) when x can be negative?

A youtube video suggests an equation $$\left(1+\frac{1}{x}\right)^{x+1}=\left(1+\frac{1}{7}\right)^7$$ and solves it in the following way: $$1+\frac{1}{x} = \frac{x+1}{x} = \left(\frac{x}{x+1}\right)^{-1}=\left(1-\frac{1}{x+1}\right)^{-1}$$ Therefore: $$\left(1+\frac{1}{x}\right)^{x+1} = \left(1-\frac{1}{x+1}\right)^{-(x+1)}\$$ Assuming $$t=-(x+1)$$ gives: $$\left(1+\frac{1}{t}\right)^{t}=\left(1+\frac{1}{x}\right)^{x+1}=\left(1+\frac{1}{7}\right)^7$$ Hence $$t=7$$ and $$x=-8$$.

However can we really assume that $$t=7$$, given that $$t$$ may be either positive or negative?

Even if we take for granted that function $$f(t)=\left(1+\frac{1}{t}\right)^t$$ is monotonically increasing (which isn't really obvious), it isn't continuous being undefined for $$t\in[-1,0]$$ (where $$1+\frac1t$$ is undefined or non-positive), therefore I don't see a reason to claim that the solution is unique.

• you can easily prove that for $t<-1$, $f(t) > e$ and so this only leaves the case $t$ is positive, where $f(t)$ increases and tends to $e.$ Oct 5, 2022 at 23:25

Take $$\lim_{x\to -\infty}(1+\frac{1}{x})^x$$ and make substitution $$t=\frac{1}{x}$$. So $$\lim_{x\to -\infty}\left(1+\frac{1}{x}\right)^x = \lim_{t\to -0}{(1+t)^\frac{1}{t}}= e^{\lim_{t\to -0}\frac{\ln(1+t)}{t}}$$ Then by l'Hôpital's rule: $$\lim_{t \to -0}\frac{\ln(1+t)}{t}=\lim_{t\to-0}\frac{1}{t+1}=1$$ Therefore $$\lim_{x\to -\infty}\left(1+\frac{1}{x}\right)^x = e$$. Given that function $$\left(1+\frac{1}{x}\right)^x$$ is increasing and continuous for $$x \in (-\infty, -1)$$ , it follows that $$\left(1+\frac{1}{x}\right)^x>e$$ for any $$x \in (-\infty, -1)$$.
After comments and your own answer, just for uour curiosity, I give below the only explicit solution for the equation $$\left(1+\frac{1}{x}\right)^x=c$$ Let $$x=\frac 1t$$ to make $$(1+t)^{\frac{1}{t}}=c\quad \implies \quad t =-\frac{W\left(-\frac{\log (c)}{c}\right)}{\log (c)}-1\quad \implies \quad x=$$ where $$W(.)$$ is Lambert function.
In the real domain, the argument must be $$\geq -\frac 1e$$, which, for this case, implies $$c\geq e$$.