Log and Floor

Consider the general case:

$\begin{aligned} x & = \sum_{k=1}^x \left \lfloor \log_n k \right \rfloor \\ & = \blue{\sum_{k=1}^{n-1} \left \lfloor \log_n k \right \rfloor} + \sum_{k=n}^x \left \lfloor \log_n k \right \rfloor & \small \blue{\text{Note that for }k < n, \log_n k < 1} \\ & = \blue 0 + \red{\sum_{k=n}^{n^2-1} \left \lfloor \log_n k \right \rfloor} + \sum_{k=n^2-1}^x \left \lfloor \log_n k \right \rfloor & \small \red{\text{and for }k < n^2, \log_n k < 2} \\ & = \red{n^2-1-(n-1)} + 2(x-(n^2-1)) & \small \blue{\text{There is always an }x, n^2-1 < x \le n^3,} \\ & = 2x - n^2 - n + 2 & \small \blue{\text{satisfying the equation.}} \\ \implies x & = n^2 + n - 2 & \small \blue{\text{Putting }n = 20} \\ & = \boxed{418} \end{aligned}$

The solution to this sum is pretty simple: $[log_{20}{1}] + [log_{20}{2}] + [log_{20}{3}] +..........[log_{20}{x}] =x$

Now to explain this solution bit easily, we can explain as follows:

1) G.I.F of Logarithm of 1,2,3.....is 0 till 19.

2) Thereafter G.I.F of logarithm of 20,21,22,23......399 is 1.

3) Then G.I.F of logarithm of 400 onwards till 8000 would be 2.

If we write $x=1+1+1+1+1+1+1+1+1+1+1+1....$ x times

By comparing, for the first 19 terms on the L.H.S, the value is 1 less than the first 19 terms on R.H.S. As a result, we are left with a deficit of 19 in the L.H.S.

Then, from the 20th to the 399th term on the L.H.S, the value is equal to their corresponding terms of R.H.S.

Then, From 400th term onwards the value of the terms on L.H.S are 1 greater than their corresponding terms on R.H.S

So we can meet our deficit of 19 by including 19 terms after the 399th term. Thus the value of x would be = 399 + 19 = 418.

Bonus: When the logarithm base is n , then the answer would be $n^{2}-1 + n-1 = n^{2} + n-2$ and the process of thinking is same as above.

Hope this helps. :)