Determine functions f(x)=3^x+7^x-2*5^x, g(x)=3^x+9^x-2*6^x And use Power-Mean Inequality, we have f>0, g>0 for x<0, f=g=0 for x=0, f<0, g<0 for 0<x<1, f=g=0 for x=1, f>0, g>0 for x>1

Let $f(a) = a^x$. Noting that $(9, 7, 3, 3) \succ (6, 6, 5, 5)$, Karamata says that \[2f(5) + 2f(6) \le 2f(3) + f(7) + f(9)\]if $x \ge 1$ or $x \le 0$, and the reverse if $0 < x < 1$. Moreover, this is sharp only when $x = 0$ or $x = 1$, since I can otherwise smooth $(9, 7, 3, 3)$ to $(6, 6, 5, 5)$, which tightens the inequality, so the solution set is only $\boxed{x = 0, 1}$. Alternatively, you can consider $9^x + 3^x \ge 2\cdot 6^x$ and $7^x + 3^x\ge2\cdot 5^x$ via Jensen's inequality.

Define $f(a) = a^x$ for some fixed $x$ and $a>2$. Notice that \[f''(a) = x(x-1)a^{x-2}\]Thus $f(a)$ is convex for $x \ge 1$. For $x \ge 1$ notice that Karamata's gives \[f(9)+f(7)+f(3)+f(3) \ge f(6)+f(6)+f(5)+f(5)\]For equality to hold, we need $f''(a) = 0$, so $x = 1$. On the other hand, if $0 < x < 1$, then $f(a)$ is concave, so Karamata's holds in the opposite direction. For equality, we still want $f''(a) = 0$, but there are no $x$ that make it $0$ since $0<x<1$. Lastly, if $x \le 0$, the same analysis gives $x = 0$ as a solution. Thus the anwser is \[\boxed{x = 0 \text{ or } 1}\]

Let $f(y)=y^x.$ When $x<0$ or $x>1$ we see $f$ is convex so by Karamata $7^x+9^x+2\cdot 3^x > 2(8^x+3^x)>2(5^x+6^x)$ and when $0<x<1$ we see $f$ is concave so by Karamata $7^x+9^x+2\cdot 3^x<2(8^x+3^x)<2(5^x+6^x).$ We may check that $0,1$ are solutions so they are the only solutions.

Define $f(a) = a^x$ for some fixed $x$. Notice that $(9,7,3,3) \succ (6,6,5,5)$. Also, $f(a)$ is strictly convex for $x > 1$ or $x<0$. Therefore, for $x\in(-\infty, 0)\cup(1,\infty)$ Karamata's inequality gives \[f(9)+f(7)+f(3)+f(3) > f(6)+f(6)+f(5)+f(5)\](the strict inequality holds, as $9\neq6, 7\neq6, 3\neq5$) On the other hand, if $x\in(0,1)$, then $f(a)$ is strictly concave, so the opposite inequality holds. So, equality cannot hold for $x\in\mathbb{R}-\{0,1\}$. For $x=0$ and $x=1$ we can check that equality holds. Hence the only solution to the equation are $\boxed{x=0 \text{ or } 1}$.

Romania NMO 2023 Let $f(a)=a^x$ for some $x$, then we have that $f$ is convex, hence by Karamata $7^x+9^x+2\cdot 3^x > 2(8^x+3^x)>2(5^x+6^x)$ Now if $x \in (0,1)$ implying that $f$ is concave, hence by Karamata again $7^x+9^x+2\cdot 3^x<2(8^x+3^x)<2(5^x+6^x).$ Now manually we can check that the answers are $1,0 $only.

Omg I have an unique solution for once.

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Let's define the function $f(a) = a^x$. When $x\ge1$ or $x\le0$, $f$ is convex, and when $0\le x\le 1$, $f$ becomes concave. Hence, by Jensen's Inequality, when $x \in (-\infty, 0] \cup [1, \infty)$, \[ 2\cdot f(5) \le f(3) + f(7) \qquad \text{and} \quad 2\cdot f(6) \le f(3) + f(9) \]and when $x \in [0, 1]$, the inequality signs are flipped around. The equation given in the problem implies the equality cases for both of the two inequalities since it states that $2\cdot f(5) + 2\cdot f(7) = 2\cdot f(3) + f(7) + f(9)$. This happens if and only if $\boxed{x=0}$ or $\boxed{x=1}$.