# Max number of solutions to $\exp\bigg(\frac{\rho}{\log(x)}\bigg)=x$ for complex, quaternion and octonion numbers?

If $$\rho$$ is a positive real number and $$x$$ is a real variable, I have verified that the following equation should have exactly two solutions, related to each other by being reciprocals of each other.

$$\exp\bigg(\frac{\rho}{\log(x)}\bigg)=x$$

Questions:

What if $$\rho$$ is a complex number and $$x$$ is a complex variable? What is the maximum number of solutions that can occur for a given $$\rho$$? Do the solutions have any relation to each other as they do for a real variable?

What if $$\rho$$ is a quaternion number and $$x$$ is a quaternion variable? What is the maximum number of solutions that can occur for a given $$\rho$$? Do the solutions have any relation to each other as they do for a real variable?

What if $$\rho$$ is an octonion number and $$x$$ is an octonion variable? What is the maximum number of solutions that can occur for a given $$\rho$$? Do the solutions have any relation to each other as they do for a real variable?

• changed it michael – Ultradark Oct 7 at 20:45
• The logarithm function is a multiple-valued function when it takes complex arguments, so you'll have to be careful and define what you mean by $\log x$ in this case. I'm not even sure that well-defined exponential and logarithm functions exist when you go to the quaternions and octonions. – Michael Seifert Oct 7 at 20:56
• @MichaelSeifert One option is to define log and exp via power series when the series converge. You don't need associativity for this, just power associativity, so it is okay for octonions. In any case, you're completely right that the first thing one needs is to specify precise definitions for exp and log. – Kimball Oct 8 at 8:00

Because all the parameters involves are real, going to quaternions or octonions won't introduce any more solutions than the complex solutions with their imaginary parts rotated. (Since every unit imaginary number is a square root of $$-1$$, algebraically they are indistinguishable, and they will behave the same in any equation with real parameters.) That is, given any complex solution $$a+bi$$ and unit imaginary octonion $$u$$, we can say that $$a+bu$$ is also a solution, and conversely.
Let's introduce $$y=\log x$$ and instead solve $$\exp(\rho/y)=\exp(y)$$; this avoids the issue of defining $$\log$$.
Then by Euler's formula, the real parts of $$\rho/y$$ and $$y$$ are equal, and their imaginary parts are parallel. Given their imaginary parts are parallel, they must differ by a unit imaginary number times $$2\pi$$. Without loss of generality we might as well use $$i$$, so solve $$\rho/y+2\pi i n=y$$ to get $$y=n\pi i\pm\sqrt{\rho+n\pi^2}$$ by the quadratic formula. Note $$n=0$$ case corresponds to real solutions, which gives opposite $$y$$ values and hence reciprocal values for $$x=\exp(y)$$. How you deal with the $$\log$$ issues after that is up to you.