When two values, $a$ and $b$, have the same value for their sum, product, and $a^b$...

Question

For what values of $a$ and $b$ do the following three expressions assume the same value? $$a+b,\qquad ab, \qquad a^b$$ and what is that value?

Clearly $\quad (a,b)=(2,2)\quad$ with a corresponding value of $4\quad$is one such answer.

But I stumbled upon another (yet more tricky) possible solution: $$(a,b)\approx (−3.1410415255\ldots, 0.758514858196\ldots)$$ that also, if you approximate as below, at least seemingly enables the three above expressions to take on the same value:

$\quad\left(\approx −2.3825266673\ldots\right)$.

I'm not sure what to make of this, but I am sure that negative odd rationals raised to an odd power, or an odd rational power, produces a negative number.

Also, somewhat curiously, though the first two are symmetric, the third isn't, yet our values for $a,b$ have $\quad a^b=\pm b^a,\quad$ so that we could say $$a+b=ab=a^b=\pm b^a$$ For these specific choices of $a$ and $b$.

Any amplifying information about anything regarding this, (other than $(-a)^b$ being undefined), would be much appreciated.

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UPDATE:

The overriding conviction here is that $a^b$, as described above, is undefined. Yet this value, whatever it's nature, can be approximated by odd rationals. Indeed we can all see that $$\boxed{(-\alpha)^\beta<0 \quad \text{if } \quad \alpha, \beta \quad \text{are ratios of odd natural numbers}}$$

We don't know what these numbers are, so why not approximate as below? $$(−3.1410415255\ldots)^{0.758514858196\ldots} \approx \left(-\frac{71}{23}-\frac{1}{19}-\frac{1}{691}\right)^{\left(\frac{17}{23}+\frac{1}{127}+\frac{1}{131}+\frac{1}{283}+\frac{1}{2953}\right)} $$
One can approximate any such number this way to any degree using a suitable arrangement of odd numbers of prime reciprocals.

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The origin of the question is just my own curiosity/interest...

Answer 1

The equation has $3$ forms, if you do not choose a shape that satisfies all forms, then you would get limited solutions because of symmetry... I disagree to @YvesDaoust's solution $$p = a+b = a^b = ab $$ Our trick is to take the equation in two's $ab = a+b$ , $ab = a^b$ , $a+b = a^b$ and solve to see what satisfies each of them then recombine to get a general solution

The symmetric part of the equation is $ab= a+b$ while $a^b$ is non-symmetric, this means we have a fixed choose here, now say that $x \in \{a,b\}$ $$x^2-(a+b)x+ab= 0$$ $$x^2-px+p = 0$$ $$ x = \frac{ p \pm \sqrt{ p^2-4p}}{2}$$ $x$ here is any one of $(a,b)$, so after we find the value of $p$ we can use the quadratic to get both $a$ and $b$ $$a^b = p$$ $$(\frac{ p-\sqrt{p^2-4p}}{2})^{(\frac{ p+\sqrt{p^2-4p}}{2})} = p$$ This is the equation that satisfies all three, but since we are not specific to which of the $x$ has $a$ or $b$, then we interchange to account for all solutions

$$(\frac{ p+\sqrt{p^2-4p}}{2})^{(\frac{ p-\sqrt{p^2-4p}}{2})} = p$$

Let's make it finer, by taking $\log()$

$$(\frac{ p-\sqrt{p^2-4p}}{2})\cdot \log(\frac{ p+\sqrt{p^2-4p}}{2}) = \log(p)$$ You can solve this with newton method to get value of $p$, just becareful of $\log()$

I plotted the function on Wolfram, then I saw some extra solutions, I set to imaginary plot and I saw some complex solutions, you can get the complex solutions also by substituting $p = \alpha + \omega i$, then separating the equation for real and imaginary parts

I saw that another negative $p$ and positive $p$ which lies about $≈6.177$, but because of limited resources I can't put the value here... Remember that if we get $p$, we can extract $a,b$ by solving $x^2-px+p$

$$(\frac{ p-\sqrt{p^2-4p}}{2})\cdot \log(\frac{ p+\sqrt{p^2-4p}}{2}) = \log(p)$$

$$(\frac{ p+\sqrt{p^2-4p}}{2})\cdot \log(\frac{ p-\sqrt{p^2-4p}}{2}) = \log(p)$$

Go ahead and plot the two functions

Answer 2

The power $a^b$, where $b$ can be any real number is only defined for $a>0$. You may also define it for $a=0$, if you prefer, but in this case the value at $b=0$ cannot be defined by continuity on both variables. You might choose to define $0^0=0$, but it would be a stretch.

Anyway, let's assume $a>0$. Then you have $$ b=\frac{a}{a-1},\qquad b-1=\frac{a-a+1}{a-1}=\frac{1}{a-1} $$ (clearly, $a\ne1$). From $ab=a^b$ we obtain $b=a^{b-1}$, so $$ \frac{a}{a-1}=a^{1/(a-1)} $$ Setting $t=1/a$ we obtain $$ \frac{1}{1-t}=\frac{1}{t^{t/(1-t)}} $$ and, in particular, $0<t<1$. Thus we can rewrite the equation as $$ (1-t)^{1-t}=t^t $$ or, as well $$ (1-t)\log(1-t)=t\log t $$ (natural logarithm). Consider the function $$ f(t)=t\log t-(1-t)\log(1-t) $$ Then \begin{align} f'(t)&=\log t+1+\log(1-t)+1 \\[6px] f''(t)&=\frac{1}{t}-\frac{1}{1-t}=\frac{1-2t}{t(1-t)} \end{align} Thus $f'$ has a maximum at $1/2$ and $$ f'(1/2)=2-2\log2>0 $$ Since $\lim_{t\to0}f(t)=\lim_{t\to1}f(t)=-\infty$, we conclude $f'$ vanishes on two points.

Noting that $f(1/4)=\log(16/27)<0$ and using the symmetry around $t=1/2$, we conclude that $f$ has a negative minimum in the interval $(0,1/2)$ and a positive maximum in the interval $(1/2,1)$. Finally, we can observe that $\lim_{t\to0}f(t)=\lim_{t\to1}f(t)=0$, so the final conclusion is that $f$ only vanishes at $1/2$.

So the unique solution is $a=b=2$.

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You might want to “extend” this to negative values of $a$, but the only ways I see is to either declare that you want to find $a<0$ such that $a+b$, $ab$ and $-|a|^b$ have the same value. Your idea of approximating with unit fractions can only work if you already know the value you want to approach. Anyway, for the changed problem, we can see that $b>0$ so the problem transforms into finding $b,c>0$ such that $b<c$ and $b-c$, $bc$ and $c^b$ have the same value.

Now you see that $b-c=bc$ implies $b=c/(1-c)$, so $c<1$. Then $$ b-1=\frac{2c-1}{c} $$ and you can proceed similarly as before.