How do I go about solving this equation? $3^x + 10^x = 4^x + 9^x$

Question

How do I go about solving this equation? $$3^x + 10^x = 4^x + 9^x.$$

I noticed that $1$ and $0$ are solutions, so maybe a way to prove that they are the unique solutions. Taking the derivative does not seem to lead anywhere...

Answer 1

Hint. Use the Mean Value theorem with the function $f(a)=a^x$.

For a given $x$, we have that:

$4^x - 3^x=xa_1^{x-1}$ for some $a_1\in (3,4)$ and $10^x - 9^x=xa_2^{x-1}$ for some $a_2\in (9,10)$.

Therefore, if $3^x + 10^x = 4^x + 9^x$ then $$xa_1^{x-1}=4^x - 3^x=10^x - 9^x=xa_2^{x-1}.$$ What may we conclude? Note that $a_2>a_1$!

Answer 2

It's obvious that $0$ and $1$ are roots of the equation.

We'll prove that these unique roots.

Indeed, let $f(x)=x^a$, where $a>1$ or $a<0$.

Hence, $f$ is a convex function and since $3+10=4+9$, by Karamata we obtain: $$f(3)+f(10)>f(4)+f(9)$$ or $$3^a+10^a>4^a+9^a.$$ Let $0<a<1$.

Thus, $f$ is a concave function.

Hence, by Karamata again $$3^a+10^a<4^a+9^a$$ and we are done!

Answer 3

Let's show in general that if $m\gt0$ and $0\lt a\lt b\lt1$, then

$$(m-ma)^x+(m+ma)^x=(m-mb)^x+(m+mb)^x\iff x\in\{0,1\}$$

One direction is clear: If $x=0$ then both sides of the equation are $2$, while if $x=1$ then both sides are $2m$. So suppose $x\not\in\{0,1\}$. We can clearly pull out and cancel an $m^x$ from both sides, so we need only show that $(1-a)^x+(1+a)^x\not=(1-b)^x+(1+b)^x$ if $a\not=b$.

Consider the function $f(u)=(1+u)^x+(1-u)^x$. For any fixed $x$, we have

$$f'(u)=x((1+u)^{x-1}-(1-u)^{x-1})$$

If $x\not\in\{0,1\}$, then we cannot have $f'(u)=0$ for $0\lt u\lt1$. Thus, for any fixed $x\not\in\{0,1\}$, $f(u)$ is either strictly increasing or strictly decreasing for $0\lt u\lt1$. In either case $f(a)\not=f(b)$ if $0\lt a\lt b\lt1$.