How do I go about solving this equation? $3^x + 10^x = 4^x + 9^x$ 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...
 A: 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!
A: 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 $a_2>a_1$!
A: 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 $0\lt a\lt b\lt1$.
