Find $n$ satisfying the equation $[\log_21]+[\log_22]+[\log_23]+\dots[\log_2n]=1538 $

For all $j$ such that $2^k\leq j<2^{k+1}$ for some integer $k$, we have $[\log j]=k$. Further note that there are exactly $2^k$ $j$s that $[\log[j]=k$ holds. Hence from $$0+1×2+2×2^2+3×2^3+4×2^4+5×2^5+6×2^6+7×2^7=1538$$, we can easily obtain $$\left([\log_21]\right)+\left([\log_22]+[\log_23]\right)+\cdots+\left([\log_2(2^7)+\cdots+\log_2(2^8-1)]\right)=1538$$

I'm supposeing that the square brackets are the INT function.

The second equation supposes that all values are the same size. The correct form of it would be along the lines of the following. For eg 153, the first adds 7 to the total, while the second equation adds just 1. The equation as written below is a restatement of the equation above.

$$n=\sum_{i=0}^72^ii=1538$$