It's easy to see that $v_2(a)=\dfrac{a}{b}$ and $v_3(a)=\dfrac{3a}{b}$ and $\dfrac{a}{b}$ is integer So $a$ is power of 54

$54^a=a^b \Rightarrow \frac{a}{b}=\log_{54}{a}$ Intuitively both sides must be positive integers, so the result follows.

S.Das93 wrote: Intuitively As an FYI, saying "intuitively" in a math proof is a death trap -- that's often where the mistakes lie. The equation does tell you that $\log_{54}a$ is rational, but how do you pass that on to saying $\log_{54}a$ is an integer?

is this right? clearly, a is divisible by 54. let a=54k plugging back we get k is divisible by 54 and then again we let k=54s then s is divisible by 54 etc. so, a is power of 54

a_507_bc wrote: Let $a, b$ be positive integers such that $54^a=a^b$. Prove that $a$ is a power of $54$. Suppose that it's not true: $\Rightarrow a=\frac{54^c}{p}$, $p,c$ are integer $\Rightarrow 54^{bc-a}=p^b$ $\Rightarrow p=54^d$ $\Rightarrow a=54^{c-d} _\blacksquare$

It is easy to see that $a$ and $54$ have the same prime factors, so $a = 2^{\alpha}\times 3^{\beta}$ $$2^{3^{\alpha} . 2^{\beta}}\times 3^{3^{\alpha +1} . 2^{ \beta}}=3^{\alpha . b} \times 2^{\beta . b}$$ $$\alpha b=3^{\alpha +1}\cdot 2^{\beta}$$$$\beta b = 3^{\alpha} \cdot 2^{\beta}$$$\Rightarrow \frac{\alpha}{\beta}=3$ $\Rightarrow a= 3^{3k}\cdot 2^{k}=54^{k}\blacksquare$ QED

There isn't information about a (that's prime or not). So, a must include all of the divisors of 54. And that means, $a=54^k$ Also from the first equality that 's given us $54^a=(54^k)^b$. And $a=k*b$. $k*b=54^k$ . And that provs.