Without use of log, can you prove that 2^30 has 10 digits

Basic/elementary level method/logic

I found that 2^n increases the number of digits like this:
1,2,3 (1); 1,2,3 (2); 1,2,3 (3); 1,2,3,4 (4); 1,2,3 (5); 1,2,3 (6); 1,2,3 (7); 1,2,3,4 (8); 1,2,3 (9)....
see:
2,4,8 (1); 16,32,64 (2); 128,256,512 (3); 1024,2048,4096,8192(4); 16384...
So you have to power 3 times, to get to the next digit. But after 3 digits, you have to power 4 times, to get to the next digit. After that, you get back to the old queue.
30 = 3+3+3+4+3+3+3+4+3+1
........|...|...|...|...|...|...|...|.… <-there are 10 digits

Thats what I found, knowing the power of two to the 13th degree.
It works for the power of two :)

Nevertheless, I would still use logarithms..

2,4,8,16,32,64,128,256,512,1024,
2048,4096,8192,16384,32768,65536,
131072,262144,524288,1048576,
2097152,4194304,8388608,16777216,
33554432,67108864,134217728,
268435456,536870912,
1073741824
and that is how you prove it

" Look ma, no logs!"...

Ok, I answered your question using logs earlier, here's my answer without logs:

2^10 = 1024, i.e just over a thousand. Hence 2^(30) = ( 2^(10) )^3 must be just over a billion, which means 10 digits.

y=2^(30)
logy=30log2
=30(.3010)
=9.3
10digits