Here's the problem:
For positive odd integers $a,b$ with $a > b$, find the solution $x \leq -2$ to the equation
$$(1 + ax)^{1/a} = (1 + bx)^{1/b}.$$
I was able to prove that there is only one $x$ with $x \leq -2$ 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!
from Hot Weekly Questions - Mathematics Stack Exchange
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