74th Romanian Mathematical Olympiad

We notice that the equation has solutions $x_1=3$ and $x_2=27$. We show that the equation does not have any other solutions.

Continuation A. Using the properties of logarithms, the equation becomes $15\cdot 3^{{\log }_5(x-2)}=10+5^{{\log }_{3}x}$, or $15(x-2{)}^{{\log }_{5}3}=10+x^{{\log }_{3}5}$, with $x>2$. Since ${\log }_{3}5>1$ and $0<{\log }_{5}3<1$, the function $2<x\mapsto 15(x-2{)}^{{\log }_{5}3}$ is strictly concave, and the function $0<x\mapsto 10+x^{{\log }_{3}5}$ is strictly convex. Hence, the function $f:(2,\infty )\to \mathbb{R}$, $f(x)=15(x-2{)}^{{\log }_{5}3}-10-x^{{\log }_{3}5}$ is strictly concave, and the equation has at most two solutions.

Continuation B. From the existence condition of logarithms, we have $x>2$. We observe that the function $f:(2,\infty )\to (0,\infty )$, $f(x)=3^{{\log }_5(5x-10)}$ has inverse which is $f^{-1}(x)=\frac{5^{{\log }_{3}x}+10}{5}$, and the equation from the statement becomes $f(x)=f^{-1}(x)$. Since $f$ is strictly increasing, the equation from the statement is equivalent to $f(x)=x$, meaning $3^{{\log }_5(5x-10)}=x$, or ${\log }_5(5x-10)={\log }_{3}x$. If ${\log }_5(5x-10)={\log }_{3}x=t$, then $x=3^t=\frac{1}{5}(5^t+10)$, which is equivalent to solving the equation $5\cdot 3^t=5^t+10$, or $5={\left(\frac{5}{3}\right)}^t+10{\left(\frac{1}{3}\right)}^t$, with $t>{\log }_{3}2$. Since the function $g:({\log }_{3}2,\infty )\to \mathbb{R}$, $g(t)={\left(\frac{5}{3}\right)}^t+10{\left(\frac{1}{3}\right)}^t$ is strictly convex, being the sum of strictly convex functions, the equation $g(t)=5$ admits at most two solutions.