It's easy to guess that there are solutions such that $a, b, c, d$ are in the form of $n^{x}$, where $n$ is a rather small number. After a few attempts, we can see that we obtain simple equations when $n=2$ or $n=3$ : for $n=2$, the equation becomes in the form of $2^{t}+2^{t}+2^{t+1}=2^{t+2}$ for some non-negative integer $t$; for $n=3$, the equation becomes in the form of $3^{t}+3^{t}+3^{t}=3^{t+1}$ for some non-negative integer $t$. In the first case, we hope that $t$ is a multiple of two of $3,4,5$, that $t+1$ is a multiple of the last one, and that $t+2$ is a multiple of 11 . Therefore, $t \equiv 15,20,24(\bmod 60)$ and $t \equiv 9(\bmod 11)$. It's easy to check that the only solution that satisfies the given inequality is the solution with $t=20$, and $(a, b, c, d)=(128,32,16,4)$. In the case where $n=3$, we must have that $t$ is a multiple of 60 , which obviously doesn't satisfy the inequality restriction. Remark: By programming, we find that the only two solutions are $(a, b, c, d)=(128,32,16,4)$ and $(a, b, c, d)=(160,16,8,4)$, with the the former being the intended solution.