Solve $3^x + 28^x=8^x+27^x$
By the end of this post, we shall prove a more general result:
Theorem: Let $a, b, c, d$ be real numbers such that $0<a<b \leq c<d$. Then the equation $$a^x+d^x=b^x+c^x$$ has
- Exactly two solutions, $x=0$ and $x=t>0$ for some $t$, if $ad-bc<0$
- Exactly two solutions, $x=0$ and $x=t<0$ for some $t$, if $ad-bc>0$
- Exactly one solution, $x=0$, if $ad-bc=0$
We first prove the following lemma:
Lemma $1$: Suppose that $f(x)$ is a function with the following properties:
- $f(x)$ is $k$ times differentiable
- For $i=0, 1, \ldots, k-1$, we have $f^{(i)}(0) \leq 0$ and $f^{(i)}(x)>0$ for all sufficiently large $x$.
- $f^{(k)}(x)$ has at most one positive real root.
Then $f(x)$ has at most one positive real root.
Proof: We prove by induction on $n$, $0 \leq n \leq k$, that $f^{(k-n)}(x)$ has at most one positive real root.
When $n=0$, this follows from the given conditions.
Suppose that the statement holds for $n=j \leq k-1$, i.e. $f^{(k-j)}(x)$ has at most one positive real root.
Assume on the contrary that $f^{(k-j-1)}(x)$ has at least two positive real roots.
Suppose that $u, v$ are positive real roots of $f^{(k-j-1)}(x)$ with $0<u<v$. By Rolle's theorem applied to $f^{(k-j-1)}(u)=f^{(k-j-1)}(v)=0$, $f^{(k-j)}(s)=0$ for some $s \in (u, v)$.
We now consider three cases:
Case $1$: $f^{(k-j-1)}(s)=0$
By Rolle's theorem applied to $f^{(k-j-1)}(u)=f^{(k-j-1)}(s)$, we have $f^{(k-j)}(t)=0$ for some $t \in (u, s)$. Clearly $t \not =s$. Therefore $f^{(k-j)}(x)$ has at least two positive real roots $s, t$, contradicting the induction hypothesis.
Case $2$: $f^{(k-j-1)}(s)>0$
Note that $f^{(k-j-1)}(x)$ is positive for sufficiently large $x$, so in particular, $f^{(k-j-1)}(w)>0$ for some $w>v$. Clearly there exists $\delta>0$ s.t. $$0<\delta<\min(f^{(k-j-1)}(s),f^{(k-j-1)}(w))$$ Now by Intermediate Value Theorem applied to $[s, v]$ and $[v, w]$, there exists $u_1, v_1$ with $u<s<u_1<v<v_1<w$ and $$f^{(k-j-1)}(u_1)=f^{(k-j-1)}(v_1)=\delta$$ By Rolle's theorem, we now get $f^{(k-j)}(t)=0$ for some $t \in (u_1, v_1)$. It is clear that $t \not =s$. Therefore $f^{(k-j)}(x)$ has at least two positive real roots $s, t$, contradicting the induction hypothesis.
Case $3$: $f^{(k-j-1)}(s)<0$
We have $f^{(k-j-1)}(0) \leq 0$.
Case 3a) $f^{(k-j-1)}(0)<0$
Clearly there exists $\delta<0$ s.t. $$\max(f^{(k-j-1)}(s),f^{(k-j-1)}(0))<\delta<0$$ Now by Intermediate Value Theorem applied to $[0, u]$ and $[u, s]$, there exists $u_1, v_1$ with $0<u_1<u<v_1<s<v$ and $$f^{(k-j-1)}(u_1)=f^{(k-j-1)}(v_1)=\delta$$ By Rolle's theorem, we now get $f^{(k-j)}(t)=0$ for some $t \in (u_1, v_1)$. It is clear that $t \not =s$. Therefore $f^{(k-j)}(x)$ has at least two positive real roots $s, t$, contradicting the induction hypothesis.
Case 3b) $f^{(k-j-1)}(0)=0$
By Rolle's theorem applied to $[0, u]$, we now get $f^{(k-j)}(t)=0$ for some $t \in (0, u)$. It is clear that $t \not =s$. Therefore $f^{(k-j)}(x)$ has at least two positive real roots $s, t$, contradicting the induction hypothesis.
Therefore we get a contradiction in all three cases, so $f^{(k-j-1)}(x)$ has at most one positive real root.
We are thus done by induction, so $f^{(k-n)}(x)=0$ has at most one positive real root for $n=0, 1, \ldots , k$. In particular, setting $n=k$ gives that $f(x)$ has at most one positive real root, as desired.
Lemma $2$: Let $p, q, r$ be real numbers such that $1<p \leq q<r$. Then the equation $$1+r^x=p^x+q^x$$ has
- No positive real solutions if $r \geq pq$
- Exactly one positive real solution if $r<pq$
Proof: If $r \geq pq$, then for $x>0$ we have $$1+r^x-p^x-q^x \geq 1+(pq)^x-p^x-q^x=(p^x-1)(q^x-1)>0$$ so indeed there are no positive real solutions.
If $r<pq$, let $$f(x)=1+r^x-p^x-q^x$$ Note that for $n \geq 1$ we have $$f^{(n)}(x)=(\log r)^nr^x-(\log p)^np^x-(\log q)^nq^x$$
We have $$\lim_{x \to 0+}{\frac{f(x)}{x}}=f'(0)=\log r-\log p-\log q<0$$
Thus $\exists u>0$ s.t. $f(u)<0$.
On the other hand, clearly $f(x)$ is positive for sufficiently large $x$, so $\exists v>u$ s.t. $f(v)>0$. Since $f(x)$ is continuous, we may conclude using the Intermediate Value Theorem that $f(x)$ has a positive real root.
We now show that $f(x)$ has at most one positive real root.
Consider $$g(y)=(\log r)^y-(\log p)^y-(\log q)^y$$
It is clear that $g(y)$ is non-negative for sufficiently large $y$, so there exists $k \in \mathbb{Z}^+$ s.t. $g(k) \geq 0$. Take the minimal such $k$, so $g(i)<0$ for $i=1, 2, \ldots ,k-1$.
We have $f^{(i)}(0)=g(i)<0$ for $i=1, 2, \ldots , k-1$, and $f(0)=0$. For $i=0, 1, \ldots k-1$, we clearly have $f^{(i)}(x)>0$ for sufficiently large $x$.
Now for $x>0$ we have
\begin{align} f^{(k)}(x)& =(\log r)^kr^x-(\log p)^kp^x-(\log q)^kq^x \\ & \geq \left((\log p)^k+(\log q)^k\right)r^x-(\log p)^kp^x-(\log q)^kq^x \\ &=(\log p)^k(r^x-p^x)+(\log q)^k(r^x-q^x) \\ &>0 \end{align}
Thus $f^{(k)}(x)$ has no positive real roots. By Lemma $1$, we conclude that $f(x)$ has at most one positive real root.
Therefore when $r<pq$, we have exactly one positive real solution. This concludes the proof of the lemma.
We now proceed to prove the main theorem:
Theorem: Let $a, b, c, d$ be real numbers such that $0<a<b \leq c<d$. Then the equation $$a^x+d^x=b^x+c^x$$ has
- Exactly two solutions, $x=0$ and $x=t>0$ for some $t$, if $ad-bc<0$
- Exactly two solutions, $x=0$ and $x=t<0$ for some $t$, if $ad-bc>0$
- Exactly one solution, $x=0$, if $ad-bc=0$
Proof: Clearly $x=0$ is always a solution.
For $x>0$, we may rewrite the equation as $$1+\left(\frac{d}{a}\right)^x=\left(\frac{b}{a}\right)^x+\left(\frac{c}{a}\right)^x$$ We may now take $p=\frac{b}{a}, q=\frac{c}{a}, r=\frac{d}{a}$ and use Lemma $2$ to get that there are
- No positive real solutions (for $x$) if $\frac{d}{a} \geq \frac{bc}{a^2}$, i.e. $ad \geq bc$
- Exactly one positive real solution (for $x$) if $\frac{d}{a}<\frac{bc}{a^2}$, i.e. $ad<bc$
For $x<0$, we may take $y=-x$ and rewrite the equation as $$\left(\frac{a}{d}\right)^x+1=\left(\frac{b}{d}\right)^x+\left(\frac{c}{d}\right)^x$$
$$1+\left(\frac{d}{a}\right)^y=\left(\frac{d}{b}\right)^y+\left(\frac{d}{c}\right)^y$$
We may now apply Lemma $2$ with $p=\frac{d}{c}, q=\frac{d}{b}, r=\frac{d}{a}$ and $y$ as the variable to get
- No positive real solutions (for $y$) if $\frac{d}{a} \geq \frac{d^2}{bc}$, i.e. $bc \geq ad$
- Exactly one positive real solution (for $y$) if $\frac{d}{a}<\frac{d^2}{bc}$, i.e. $bc<ad$
This translates to
- No negative real solutions (for $x$) if $bc \geq ad$
- Exactly one negative real solution (for $x$) if $bc<ad$
Combining, we get the statement of the theorem.