$\left(1 + \frac{1}{x}\right)^{x+1} = \left(\frac{8}{7}\right)^7$.

Find $x$.

“Let’s make the power 7!((Accidental factorial))((Accidental exponent))” But if $x=6$ then the bracket is $\frac{7}{6}$.

“Let’s make the bracket $\frac{8}{7}$, then!” But if $x = 7$, the power is 8.

It doesn’t look like it’s going to be a positive integer – so that leaves us with the negative integers and other reals.

What if $x$ is negative? Then we might be able to make the left hand side into $\left(\frac{7}{8}\right)^{-7}$ – and indeed, $x = -8$ does that nicely.

Done and dusted, right? Well, not quite.

### Is it the only one?

Whether this is the only solution is – possibly – a matter of interpretation and domain.

If we’re sticking to the reals, which is a fairly reasonable assumption, then yes, it’s the only one.

A sketch of a proof:

• The function $f(x) = \left(1 + \frac{1}{x}\right)^{x+1}$ is nicely defined ((we’ll come back to this)) for $x < -1$ or $x \ge 0$.
• On both of those domains, the gradient is strictly negative
• In the limit as $x \to \infty$ and as $x \to -\infty$, $f(x)$ approaches $e$.
• The negative branch is strictly decreasing from $e$ to 0, and only permits one solution.
• The positive brach is strictly decreasing from $+\infty$ to $e$ and permits no solution.

But wait.

I could quite happily evaluate $f\left(-\frac{1}{3}\right)$, for example – that gives $(-2)^{\frac{2}{3}}$, which one could reasonably interpret as $\sqrt{4}$. And in general, rational numbers with odd denominators certainly allow real evaluations. If the numberator is also odd, then the evaluation is positive, and the rational numbers of that form are dense.

They also get arbitarily large – if $x$ is a small negative number, we get a large negative number to a power close to 1, which (given that the numerator of the power is even) is a large positive number.

Which means, we can get arbitrarily close to another “solution” in the missing section of the domain.

If we solve $|1+\frac{1}{x}| = \left(\frac{8}{7}\right)^{\frac{7}{x+1}}$, we get a second solution near -0.22922379.

Arbitrarily close, yes, but this doesn’t turn out to be an actual solution – the number that comes out is (almost certainly) irrational and gives a complex answer. Is there a hint that there might be a complex value of $z$ that works?

Well, Wolfram|Alpha can’t find it (its “solutions” to this appear to be numerical errors rather than actual answers). My complex analysis game – and my complex numerical analysis game – isn’t strong enough to find it, either. How about yours?

If you find a complex solution – or a proof that no such thing exists – let me know!