An exponential equation over positive real numbers $4^x+14^x+3^x=11^x+10^x$

Question

Solve the following equation over the positive reals: $$4^x+14^x+3^x=11^x+10^x.$$

By inspecting the graph, the solutions must be $x \in \{1,2\}$

I tried using inequalities like $3^x+4^x<5^x$ for $x>2$ but I couldn’t work it forward. Thanks in advance!

Answer 1

We'll solve this equation for any real value of $x$.

Indeed, let $x>2$ and $x=1+t$.

Thus, $t>1$ and since $f(x)=x^t$ is a convex function, by Jensen we obtain: $$4^x+3^x+14^x=(3\cdot3^t+8\cdot14^t)+\left(4\cdot4^t+6\cdot14^t\right)\geq$$ $$\geq11\left(\frac{3\cdot3+8\cdot14}{11}\right)^t+10\left(\frac{4\cdot4+6\cdot14}{10}\right)^t=11^{t+1}+10^{t+1}=11^x+10^x.$$ The equality does not occur, which says that our equation has no roots for $x>2$.

Similarly, for $x<1$ let $x=1+t,$ where $t<0$ and since $g(x)=x^t$ is a convex function,

by the same Jensen we obtain that our equation has no roots for $x<1$.

Now, let $1\leq x\leq 2$ and $$h(x)=\left(\frac{3}{11}\right)^x+\left(\frac{4}{11}\right)^x+\left(\frac{14}{11}\right)^x-\left(\frac{10}{11}\right)^x-1$$ and we'll prove that $h$ is a convex function.

Indeed, $$h''(x)=\left(\frac{3}{11}\right)^x\ln^2\frac{11}{3}+\left(\frac{4}{11}\right)^x\ln^2\frac{11}{4}+\left(\frac{14}{11}\right)^x\ln^2\frac{14}{11}-\left(\frac{10}{11}\right)^x\ln^2\frac{11}{10}\geq$$ $$\geq\left(\tfrac{3}{11}\right)^2\ln^2\tfrac{11}{3}+\left(\tfrac{4}{11}\right)^2\ln^2\tfrac{11}{4}+\tfrac{14}{11}\ln^2\tfrac{14}{11}-\tfrac{10}{11}\ln^2\tfrac{11}{10}>\tfrac{10}{11}\ln^2\tfrac{14}{11}-\tfrac{10}{11}\ln^2\tfrac{11}{10}>0,$$ which says that our equation has maximum two roots on $[1,2].$

But $1$ and $2$ are roots and we are done!

Answer 2

HINT.-It is evident that $x=1$ is a solution and because of $14^3\gt 10^3+11^3$ the other only solution (intersection of two convex functions) can be computable by numerical methods. In this particular case the first integer $x=2$ to be tried is the other solution.

Answer 3

The equation at hand can be rewritten as $$14^x - 11^x - 10^x + 4^x + 3^x = 0$$ In general, given any non-zero real numbers $\alpha_1,\alpha_2,\ldots,\alpha_n$ and positive numbers $\beta_1 > \beta_2 > \cdots > \beta_n > 0$, functions of the form $$\alpha_1 \beta_1^x + \alpha_2 \beta_2^x + \cdots + \alpha_n \beta_n^x$$ is known as a Dirichlet polynomial (in $x$). In $1883$, Laguerre has proved a theorem:

Generalized Descartes' rule of signs

The number of real roots of Dirichlet polynomials is no more than number of sign changes in the finite sequence $(\alpha_1, \ldots, \alpha_n)$.

In addition, counting multiplicity, the difference between the number of real roots and number of sign changes is an even number.

For the Dirichlet polynomial at hand, the number of sign changes is $2$. This means counting multiplicity, it has either $2$ or $0$ roots. Since you already find two roots $1$ and $2$, these are all the roots it has.

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.