Finding the smallest set of values that make an exponent of a known value.

Question

I need some help with figuring out an equation or process? Not sure what the correct term is?

Full disclosure: Unfortunately I’m not even remotely a mathematician so I’m grateful for your patience. If I misuse any vocabulary or am unclear please let me know and I’ll try tor clarify what I’m trying to convey.

The Question.

The question is simple. I want to figure out the exponent of a value. But there is caveat:

• Only the total is known.
• And out of all the possible exponents that could be used: I want to find the smallest possible base with the smallest possible power.

That is to say: If I was to take the digit of the base number $x$ and the digit of the exponent power $y$, that both these numbers would be as small as possible without making the other larger.

Essentially the resulting pair of numbers would essentially be the smallest difference of values and/ or the smallest sum total.

Moreover I need to “prove” that it is the smallest combination of numbers possible. Unfortunately it can’t be “close enough.” It needs to be certain that this pair of numbers are the smallest pair possible values that can form the exponent total.

Here is an example that I hope illustrates what I am trying to achieve:

(But remember, I’m not just trying to find the answer. Rather I want to know if there is a mathematical process to finding the smallest possible pair in the first place?)

Let’s say that the equation is $x^{y} = 64$.

The most simple exponent is $64^{1}$. But it is also the largest number I could use as a base and so would not be considered a “successful” answer, (even if the exponent is only $1$).

$2^{6}$ would be considered more “successful” exponent. As the numbers are $2$ and $6$. It doesn’t matter if the base is smaller than the exponent. The only criteria is that they are made of the smallest numbers as viewed as a decimal value.

$4^{3}$ I believe to be the most successful. The base may have significantly increased but we have reduced the exponent by half its value.

$8^{2}$ would be less of a successas it has still gained a higher number. Whilst it is the smallest exponent possible aside from $1$, the base has risen to $8$.

Now, the first question would be how do I find $4^{3}$ out of all the exponents I could possibly use? Then secondly how can I be certain that there is no smaller combination of numbers that could be used to create an exponent that results in $64$?.

Or put another way: Is there a method of mathematically finding the smallest pair of numbers without simply wading through every possible exponent that makes the original sum and comparing the total sum and/or difference of every pair?

Thank you in advance! I know I’m probably over complicating this but unfortunately I don’t know what I don’t know.

edit: Here is the link @SohamSaha post using positive integers in case anyone should need it: Minimising $x+y$ in $x^y=a$

Answer 1

You are asking to minimize $x + y$ given some fixed constant $a > 0$ such that $x^y = a$. Since we are optimizing in two variables, we solve this using Lagrange multipliers. We let $f(x, y) = x + y$ and $g(x, y) = x^y - a$. Then, we need to find a constant $\lambda$ such that $$\begin{align*}\frac{\partial f(x, y)}{\partial x} &= \lambda \frac{\partial g(x, y)}{\partial x} \\ \frac{\partial f(x, y)}{\partial y} &= \lambda \frac{\partial g(x, y)}{\partial y} \\ g(x,y) &= 0.\end{align*}$$

The operation $\dfrac{\partial f}{\partial x}$ means to take the derivative (or if you want, the rate of change) of $f$ with respect to $x$ while treating all other variables (in this case, $y$) as a constant.

The first two equations become $$\begin{align*}1 &= yx^{y - 1} \\ 1 &= x^y \ln x\end{align*}$$ which, by transitivity, is just $yx^{y - 1} = x^y \ln x$. Dividing both sides by $x^{y - 1}$, we get $y = x \ln x$. Note that we didn't even need to get $\lambda$ here.

Hence, we just need to solve $x^{x \ln x} = a$. Then, $x$ will be the base, and $x \ln x$ will be your exponent.

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I don't have any idea how to solve $x^{x\ln x} = a$ as I believe it will need to use some techniques involving the Lambert W function.

Answer 2

This is not an independent answer and continues the one given by @soupless

Proceeding with the answer given by @soupless, we can solve their equation like so:

$$ \begin{eqnarray} &&x^{x\ln x}=a \\ &\Rightarrow&\ln a=x(\ln x)^2=e^{\ln x}(\ln x)^2 \\ &\Rightarrow&\frac{\sqrt{\ln a}}{2}=\exp\left({\frac{\ln x}{2}}\right)\frac{\ln x}{2} \\ &\Rightarrow&\frac{\ln x}{2}=W\left(\frac{\sqrt{\ln a}}{2}\right) \\ &\Rightarrow&x=\exp\left({2W\left(\frac{\sqrt{\ln a}}{2}\right)}\right) \\ \end{eqnarray} $$

Also, solving for $y$: $$y=\frac{\ln(a)}{2W\left(\frac{\sqrt{\ln a}}{2}\right)}$$

So, $(x,y)=\left(\exp\left({2W\left(\frac{\sqrt{\ln a}}{2}\right)}\right),\frac{\ln(a)}{2W\left(\frac{\sqrt{\ln a}}{2}\right)}\right)$ is the final solution.

Answer 3

Your question is a bit unclear to me, but to me, the most reasonable interpretation is the following:

Given a real number $a>1$, find positive real numbers $x$ and $y$ such that $x^y=a$ and $x+y$ is minimal.

I'll explain the restrictions to positive $x$ and $y$, and $a>1$, after solving this question.

From $x^y=a$ it follows that $y=\frac{\ln a}{\ln x}$. Because $a>1$ and $y>0$, it follows that $\ln a>0$ and so $\ln x>0$, meaning that $x>1$. Now you want to minimize $$f(x)=x+\frac{\ln a}{\ln x}.$$ Because $f$ is continuous on $\Bbb{R}_{>1}$, the set of reals greater than $1$, we must have $f'(x)=0$ if $f(x)$ is minimal. Some elementary calculus shows that $$f'(x)=1-\frac{\ln a}{x(\ln x)^2},$$ and so if $f(x)$ is minimal then we must have $x(\ln x)^2=\ln a$. This implies $$y=\frac{\ln a}{\ln x}=\frac{x(\ln x)^2}{\ln x}=x\ln x.$$ It also follows that $\sqrt{\ln a}=\sqrt{x}\ln x$, and so for $w:=\sqrt{x}$ we have $$\sqrt{\ln a}=w\ln w^2=2w\ln w,$$ meaning that $w\ln w=\tfrac12\sqrt{\ln a}$. This equation, in general, does not have a solution in terms of elementary functions. But for every $a>1$ there exists a unique real number $w$ such that $w\ln w=\tfrac12\sqrt{\ln a}$, and this real number $w$ is denoted by $$w:=W(\tfrac12\sqrt{\ln a}),$$ where $W$ denotes the Lambert $W$ function. It follows that the unique minimal pair $(x,y)$ with $x^y=a$ is $$x=W(\tfrac12\sqrt{\ln a})^2,\qquad y=x\ln x=2W(\tfrac12\sqrt{\ln a})^2\ln W(\tfrac12\sqrt{\ln a}).$$

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If we do not require $x$ and $y$ to both be positive, then we still find from $x^y=a$ that $y=\frac{\ln a}{\ln x}$, but now we see that as $x$ approaches $1$ from below the value of $y$ goes off to negative infinity. That is to say $$\lim_{x\uparrow 1}y = \lim_{x\uparrow 1}\frac{\ln a}{\ln x}=-\infty.$$ This is also clearly seen in the graph of $y$, here for $a=64$, as plotted by WolframAlpha:

This means we can make $x+y$ arbitrarily small, all the way to negative infinity, while keeping $x^y=a$. So there is no minimum of $x+y$, and so it makes sense to require that $x$ and $y$ must be positive.

Similarly, if we allow $a\leq1$, the situation gets messy:

• For $a=1$ we may take $x=1$ and then we can again take $y$ arbitrarily small, all the way to negative infinity. So again there is no minimum.
• For $0<a<1$ we can have $x$ approach $1$ from above, and as before have $y$ be arbitrarily small, all the way to negative infinity. So again, there is no minimum.
• For $a\leq0$ we must certainly have $x\leq0$, but then it becomes unclear what $x^y$ should even mean if $y$ is not an integer.

This is why I chose to restrict the question to $a>1$ and $x,y>0$.