Solving $(1 + ax)^{1/a} = (1 + bx)^{1/b}$ Here's the problem:
For positive odd integers $a,b$ with $a > b$, find the solution $x \leq -2/b$ to the equation
$$(1 + ax)^{1/a} = (1 + bx)^{1/b}.$$
I was able to prove that there is only one $x$ with $x < -2/b$ (see below) that satisfies the problem.
By using the substitution $u = 1+bx$, I obtained
$$\left(1 + \frac ab (u-1) \right)^{1/a} = u^{1/b},$$
which led me to the nicer looking equation
$$u^{a/b} - \frac abu + \frac ab -1 = 0.$$
I also noticed that using the substitution $v = u^{1/b}$ led me to
$$v^a - \frac ab v^b + \frac ab -1 = 0,$$
which also yielded
$$\sum_{r=0}^{a-1} v^r - \frac ab\sum_{r=0}^{b-1} v^r = 0,$$
which is an $a-1$ degree polynomial in $v$. Interestingly, the first $b-1$ terms of this polynomial have the same coefficient equal to $1 - a/b$ and the rest have a coefficient equal to $1$.
But, through either of these approaches, I still get stuck. Any help would be much appreciated!

Sorry for the late response! There were some mistakes on my part, and I hope to correct them here in this edit.
Proof of unique solution $x < -2/b$.
Begin by raising $a$ on both sides to obtain
$$1 + ax = (1 + bx)^{a/b} \implies 0 = (1 + bx)^{a/b} - ax - 1.$$
Define $f(x) = (1 + bx)^{a/b} - ax - 1$. Its derivative is given by
$$f'(x) = a(1 + bx)^{\frac ab - 1} - a = a((1+bx)^{\frac {a-b}b} - 1).$$
Now, since $a,b$ are both positive odd integers with $a  >b$, we have that $a-b$ is a positive even integer, and thus,
$$(1 + bx)^{\frac{a-b}b} \geq 0$$
holds over all real $x$.
The quantity $(1 + bx)^{\frac{a-b}b} \leq 1$ only when $|1 + bx| \leq 1$, or when $-2/b \leq x \leq 0$, and similarly $(1 + bx)^{\frac{a-b}b} \geq 1$ only when $|1 + bx| \geq 1$, or when either $x \geq 0$ or $x \leq -2/b$.
Finally, note that $(1 + bx)^{\frac{a-b}b} = 1$ when $|1 + bx| = 1$, or when $x = 0$ or $x = -2/b$.
From here, since $a > 0$, we have that
$$f'(x) > 0 \quad \text{when } x \in (-\infty, -2/b) \cup (0, \infty),$$
and
$$f'(x) < 0 \quad \text{when } x \in (-2/b, 0),$$
and
$$f'(x) = 0 \quad \text{when } x = 0, -2/b.$$
Here I've used interval notation to denote the possible values of $x$.
Notice $f(0) = 0$. Since $f$ strictly increases for $x > 0$, we have that $f(x) > f(0) = 0$ for all $x > 0$.
Now, $f$ strictly decreases for $-2/b < x < 0$, so $f(x) > f(0) = 0$, for $-2/b < x < 0$, which implies that there are no roots in the interval $(-2/b, 0)$.
Finally, we note the value $f(-2/b) = 2a/b - 2$, for which, since $a/b > 1$, we have $2a/b - 2 > 0$. The function $f$ strictly increases for $x < -2/b$, so $f(x) < f(-2/b)$, and in particular, $f$ is injective on this interval. That is, on this interval,
$$f(s) = f(t) \implies s = t, \quad \text{for } s,t < -2/b.$$
Since $f(-2/b) > 0$, there is exactly one $x$ with $x < -2/b$ such that $f(x) = 0$, signaling the end of the proof.
For a more visual representation of this proof, see this Desmos graph.

I should also mention some background on how I stumbled across this problem. I was investigating a generalization of Bernoulli's inequality, for which the classic version states that
$$(1 + x)^n \geq 1 + nx$$
for $n = 1,2,\dots$ and $x \geq -1$.
In particular, I was investigating the inequality
$$\left(1 + x\right)^{\frac ab} \geq 1 + \frac ab x$$
over real $x$ and positive integers $a,b$ and determining when the above holds. It is known that the above holds for $x \geq -1$, which can be shown with calculus over real $a/b$. (Here's the Wikipedia page if you're interested.)
However, I was more interested with the case in which $x < -1$. There must be additional restrictions, however, in particular $b$ must be odd, or $(1 + x)^{a/b}$ will not be a real number.
I was able to show that the inequality held for all real $x$ when $a$ was even, however, for odd $a$, I had determined that there was a negative nonzero solution to the equation
$$(1 + x)^{a/b} = 1 + \frac ab x.$$
At this solution, say $x_0$, I had that the inequality did not hold for $x < x_0$.
To obtain the original inequality I posted here, you raise both sides by $1/a$ and take the transformation $x \mapsto bx$, which gave the much more aesthetically pleasing equation
$$(1 + ax)^{1/a} = (1 + bx)^{1/b}.$$
From here, I think it would be appropriate to ask a few other questions other than "what's the solution?". In particular, for positive integers $n,m$, if $R(n,m)$ is the solution to the equation,
$$(1 + (2n-1)x)^{2n-1} = (1 + (2m-1)x)^{2m-1},$$
then...

*

*Can we approximate $R(n,m)$ to a precise degree? (This one is mostly solved by Newton's method which was mentioned by someone who for some reason I can't find now.)


*Can we create any identities involving $R(a,b)$?
I am particularly interested with question 2. If I find anything, I'll be sure to add it in another edit to this post. Thanks guys!
 A: I could only solve it with assumptions, like $a$ is an odd number multiply of $b$ which is let to be $\alpha$, and that when arriving at a point when the two sides of the equation are both a product of two terms from which one is $\alpha - 1$ which must be even, then I assumed $\alpha - 1$ is 2, so $\alpha$ is assumed to be 3, and that gave me $b=1$, $a=3$, $x=-3$, and that is a solution. This doesn't prove though that this is the only solution and I guess is almost nothing to do with (pre)calculus. But if needed I can share the steps.

Update
The statement in the question "I was able to prove that there is only one $x$" misses $x\in?$. Because if you are looking for integers only, it might be true, but for reals you can always write up a polynomial for fixed $(a,b)$ as sirous did. My approach is that as $x\le-2$ and $b\ge1$, therefore $bx\le-2$. Now using your definition of $u$, $u\le-1$. Let $-u=t$, $t\ge1$. Using $t$ instead of $u$ we have
$(-t)^{a/b} + \frac abt + \frac ab - 1 = 0$
and as $a$ is odd, $(-t)^{a/b} = -t^{a/b} $
using this and doing rearrangements we get
$\frac abt + \frac ab= t^{a/b} + 1$
$\frac ab = \frac {t^{a/b} + 1}{t + 1}$
as $a\gt b$, the LHS $\gt1$, and as $t\gt1$, $t^{a/b}\gt t \Longrightarrow t^{a/b} + 1 \gt t + 1$, so the RHS $\gt1$ also $\forall t$, except for $t=1$, but for $t\gt 1$ the numerator grows more than the denominator, so though I am unable to prove, but common sense tells me $\exists t\in\mathbb{R}$ $\forall \frac ab$, so there are infinitely many solutions if $x\in\mathbb{R}$. Perhaps one needs to show that the RHS is strictly monotonically increasing, continuous, its minimum value is $1$, does not have a global maximum, and therefore can produce any ratio $\gt1$.
A: Comment:
We may write:
$(1+ax)^b=(1+bx)^a$
$x^b(\frac{1}{x}+a)^b=x^a(\frac{1}{x}+b)^a$
$x^b=(\frac{1}{x}+b)^a$
$x^a=(\frac{1}{x}+a)^b$
This is the only possible case.
Now if $x<0$ then $(\frac{1}{x}+b)^a>0$, because $\frac{1}{x}$ is a negative fraction and $b$ is a positive integer. which results in $x>0$ which is a contradiction.Similar result comes out from other equation. I wonder how this equation can have negative solutions. Could you show that?
Now let ignore these manupolations and check this with known values for a and b.
Let $a=3$ and $b=1$ we have:
$(1+3x)^1=(1+x)^3$
$x^3+3x^2=0$
Which gives $x=-3$
For $a=5$ and $b=1$ we get $x=-2.85$.
Which shows there can be solutions $x<-2$. But I do not think we can find a parametric solution for $x$ for $a$ and $b$.
A: I strongly doubt that there is a general solution except in special cases, as
$$(1+ax)^b=(1+bx)^a$$
results in a polynomial of "high" degree.  A special case is $a=3b$; it's easily shown the negative solution is $x=-3/b$.
Furthermore it's quite obvious the negative solution is the zero of a polynomial of degree $a/\operatorname{gcd}(a,b)-2$.
