# Solve $3^x + 28^x=8^x+27^x$

The equation $3^x + 28^x=8^x+27^x$ has only the solutions $x=2$ and $x=0$? If yes, how to prove that these are the only ones?

• Note that $27^x = (3^3)^{x}$, and that $28^x = 4^x \cdot 7^x$. We also have $8^x = 2^x \cdot 4^x$. – Newb Nov 26 '13 at 14:34
• Yes, this is clear! That's nothing new :D – user85046 Nov 26 '13 at 14:40
• What is the range of $x$ ? $x$ is real or integer ? – Ewan Delanoy Nov 26 '13 at 14:43
• It is relatively easy to show that $x=0, 2$ are the only integer solutions. On the other hand, to show that $x=0, 2$ are the only real solutions might require a bit of calculus. – Ivan Loh Nov 26 '13 at 14:51
• x is a real number. – user85046 Nov 26 '13 at 14:53

## 4 Answers

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:

1. $f(x)$ is $k$ times differentiable
2. For $i=0, 1, \ldots, k-1$, we have $f^{(i)}(0) \leq 0$ and $f^{(i)}(x)>0$ for all sufficiently large $x$.
3. $f^{(k)}(x)$ 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.

• +1 It's a good generalization. After the OP clarified the domain of $x$ to be real numbers, I got to thinking about this from a different point of view. – hardmath Nov 26 '13 at 20:08
• Couldn't your argument for showing that $f$ has at most one positive root be put in a separate lemma ? What you show is that, if $f^{(d)}>0$ everywhere and $f^{(j)}(0) \leq 0$ for every $j\leq d$, then $f$ has at most one positive root. – Ewan Delanoy Nov 26 '13 at 20:16
• @EwanDelanoy Good point. I also used the fact that $f^{(i)}(x)>0$ for sufficiently large $x$. – Ivan Loh Nov 26 '13 at 20:56
• Again, nice work! I always give you the 10th upvote. :P – Ahaan S. Rungta Nov 30 '13 at 22:05

Here's a partial proof. Let

$$f(x)=28^x-27^x-8^x+3^x$$

so that

$$f'(x)=\ln(28) 28^x-\ln(27)27^x-\ln(8)8^x+\ln(3)3^x$$

For $x\gt2$ we have

$$f'(x)\gt \ln(28)28^2-\ln(27)27^2-\ln(8)8^2\approx 76.7\gt0$$

so $f(x)$ is strictly increasing for $x\gt2$, and hence, since $f(2)=0$, has no further zeroes for $x\gt2$.

Since for all sufficiently large $x$ the expression $28^x + 3^x - 8^x - 27^x$ is positive and increasing, there cannot be roots $x$ beyond that point.

The question is how to pin down where the expression becomes positive and increasing.

Now $28^x$ grows fastest, by a factor of $28$ when $x$ is incremented by 1. This suggests a strategy of dividing through by $28^x$ to get a simpler expression to estimate.

Define $f(x) = 1 + \left(\frac{3}{28}\right)^x - \left(\frac{8}{28}\right)^x - \left(\frac{27}{28}\right)^x$. As noted in the Question, $f(2) = 0$. By standard calculus techniques:

$$f'(x) = -\left( \log \frac{28}{3} \right)\left(\frac{3}{28}\right)^x + \left( \log \frac{7}{2} \right)\left(\frac{2}{7}\right)^x + \left( \log \frac{28}{27} \right)\left(\frac{27}{28}\right)^x$$

If $x \ge 2$, we can prove the derivative is positive:

$$f'(x) = \left(\frac{2}{7}\right)^x \left[ \left(\log \frac{7}{2}\right) - \left(\log \frac{28}{3}\right)\left(\frac{3}{8}\right)^x \right] + \left(\log \frac{28}{27}\right)\left(\frac{27}{28}\right)^x$$

$$\ge \left(\frac{2}{7}\right)^x \left[ \left(\log \frac{7}{2}\right) - \left(\log \frac{28}{3}\right)\left(\frac{3}{8}\right)^2 \right] + \left(\log \frac{28}{27}\right)\left(\frac{27}{28}\right)^x$$

and by direct computation the quantity in square brackets is positive, approx. $0.938664$. Thus $f(x) \gt 0$ for all $x \gt 2$. After checking $f(1)$ we know the only nonnegative integer solutions are the two identified in the Question, $x=0,2$.

As clarified, however, we need to determine all real solutions $x$. We can do this almost by inspection, calling upon an extension of Descartes Rule of Signs due to Laguerre(1883). Setting $z=3^x \gt 0$ we can rewrite the problem as finding positive real roots of the function:

$$g(z) = z^{\log_3 28} - z^3 - z^{\log_3 8} + z$$

The number of sign changes is two, and by the version Thm. 3.1 in the PDF linked above, there are at most two positive real roots. Since we know $z=1,9$ are roots, these are the only positive real roots (from which it follows $x=0,2$ are the only real roots of the original problem).

It should be evident from this brief discussion that a much broader application of the principle can be made, e.g. like the generalization @IvanLoh has proved.

• Interesting; I had not seen that particular generalisation of Descartes Rule of Signs before. – Ivan Loh Nov 27 '13 at 15:16

You don't have to "solve" this equation. Instead of it you have to implement two steps. 1) Guesse the roots of this equation (success, you've already did this !=) 2) Proove that they are unique.

For the latter consider two intervals: $x>2$ and $x<-2$. Why these intrvals? Because, in the interval $[-2;2]$ there are roots. Then, for each x in the first two intervals proove that functions: $f(x)=3^x+28^x$ and $g(x)=8^x+27^x$ have tha same sign. After you have done this, you have prooved that there