Find all real solutions of $6^x+1=8^x-27^{x-1}$

Question

Find all real solutions of $6^x+1=8^x-27^{x-1}$.

Things I tried: We want solutions of $$2^x3^x+1 = (2^x)^3-\frac{(3^x)^3}{27}.$$ Write $a=2^x$ and $b=3^x$. This gives $$ab+1 = a^3-\frac{b^3}{27}$$ or $$27ab+27=27a^3-b^3$$

How to continue?

I can't prove it yet but interestingly the only solutions are 1 and 2. — user88595, May 18 '14 at 13:53
You got to $3ac + 1 = a^3-c^3$. What happens if you try to put $a-c = p$ and $ac = q$? — Improve, May 18 '14 at 14:12
@Improve The problem with $3ac+1=a^3-c^3$ was that I assumed the solution to be an integer. — Jolien, May 18 '14 at 14:15
Ok, but for fun, just assume $b = 3c$ for some number $c$. (You can do that, especially when b is a real number) — Improve, May 18 '14 at 14:16
Hm it is awful: I get $a = \frac{p\pm\sqrt{p^2-4q}}{2}$ and I'm not sure which one to pick. — Jolien, May 18 '14 at 14:21
Well you are not supposed to try and solve it for a and c. The idea is to gain a simpler equation in terms of p and q. $3ac + 1 = a^3-c^3$ and remember $a^3-c^3 = (a-c)(a^2+ac+c^2)$ — Improve, May 18 '14 at 14:24

score 7 · Accepted Answer · edited Apr 13 '17 at 12:21

As you did, let $a=2^x, b=3^x$. We have $$ab+1=a^3-\frac{b^3}{27}$$

Note the standard identity $$u^3+v^3+w^3-3uvw=(u+v+w)\left(\frac{(u-v)^2+(v-w)^2+(w-u)^2}{2}\right)$$

Take $u=a, v=-\frac{b}{3}, w=-1$. Then $u^3+v^3+w^3-3uvw=a^3-\frac{b^3}{27}-1-ab=0$.

Thus $u+v+w=0$ or $(u-v)^2+(v-w)^2+(w-u)^2=0$. In the latter case, we get $u=v=w$ so $a=-\frac{b}{3}=-1$, so $a=-1, b=3$. But clearly $2^x=a \not =-1$, a contradiction.

Thus $u+v+w=0$.

Thus $a-\frac{b}{3}-1=0$, i.e. $2^x=3^{x-1}+1$.

Write $x=y+1$, and rewrite as $1^y+3^y=2^y+2^y$. Finally use the theorem I prove here. (Quoted below)

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 $ad-bc=0$

with $(a, b, c, d)=(1, 2, 2, 3)$, $ad-bc=-1<0$, and $y$ in place of $x$.

Thus there are exactly two solutions for $y$, given by $y=0$ and $y=t>0$, some $t$. By inspection, $y=1$ is a solution. Thus all real solutions for $y$ are $y=0, 1$.

Thus all real solutions for $x$ are $x=1, 2$.

Alternative: $ab+1=a^3-\frac{b^3}{27}$ You get by putting $b=3c$ that $3ac +1 = a^3-c^3$ Notice that $a$ and $c$ are positive, so $a>c$. Then put $k = a-c$, $p = ac$ and then $3ac +1 = a^3-c^3$ is reduced to $ 3p + 1 = k(k^2+3p)$ and further $3p(1-k) = (k-1)(k^2+k+1)$ Divide by $k-1$ and you get that $k^2+k+1 + 3p = 0$ which is impossible since $k$ and $p$ are positive. So the only solution is if $k-1=0$, or equivalently $a = c+1$, arriving at the same place. — Improve, May 18 '14 at 14:44

score 5 · Answer 2 · answered May 18 '14 at 14:36

If you make a plot of the function $f(x) = 27^{x-1} - 8^x + 6^x + 1$, you will notice it has two trivial roots at $x = 1$ and $2$. This sort of function is called Dirichlet polynomial. In 1883, Laguerre has proved a theorem:

Generalized Descartes' rule of signs

Given a Dirichlet polynomial $P(x) = \sum\limits_{j=1}^n a_j b_j^x$ where $a_j, b_j \in \mathbb{R}$ satisfies: $$ a_1, a_2, \ldots, a_n \ne 0,\quad\text{ and }\quad b_1 > b_2 > \cdots > b_n > 0$$ If $N$ is the number of sign changes in the coefficients $a_j$, then the number of real roots of $P(x)$ is bounded by $N$.

Since our Dirichlet polynomial $f(x)$ has two sign changes, the two roots $1$ and $2$ are all the real roots it has.

Notes

For more info, you can either look up the original paper (in French)

E Laguerre, Sur la théorie des équations numériques, J. Math. Pures et Appl. 9 (1883)

or a modern introduction of same subject

G.J.O Jameson, Counting zeros of generalized polynomials: Descartes' rule of signs and Laguerre's extensions, (Math. Gazette 90, no. 518 (2006), 223-234).
An online copy can be found here.

Ah yes I was introduced to this by hardmath in the same question I referred to, see here. (+1) — Ivan Loh, May 18 '14 at 14:46
@IvanLoh I'm sure I have seen similar question on M.SE more than a year ago. In fact, I got an english translation of Laguerre's paper from here. Unluckily, I can no longer locate where it is, the link I provided is the same as the one on the answer you referred to. — achille hui, May 18 '14 at 14:56

score 2 · Answer 3 · answered May 18 '14 at 14:40

The equation $27ab+27=27a^3-b^3$ you have found may be written as $$27a^3-27ab-27-b^3=0,$$ which may be viewed as a cubic in $b$. Its discriminant is $$\Delta_b(a)=-4\cdot(-1)\cdot(-27a)^3-27\cdot(-1)^2=-(a^3+1)^2,$$ which is negative for all $a$ except for $a=-1$, when it is zero. This does not correspond to a real value of $x$, hence for all possible values of $a$ there is a unique real $b$ satisfying the polynomial, which is $$b=3(a-1).$$ By definition we have $b=3^{\log_2a}$. So it remains to solve $3^{\log_2a}=3(a-1)$. By inspection we find the solutions $a=2$ and $a=4$ corresponding to $x=1$ and $x=2$. These are the only solutions, because such an exponential function intersects any line in at most two points.

Find all real solutions of $6^x+1=8^x-27^{x-1}$

3 Answers3

Linked