Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

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

Basic/elementary level method/logic

Update:

Is it true that there is some pattern as doubling a same number increases a digit after 3/4 times?

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  • Anonymous
    9 years ago
    Favorite Answer

    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..

  • 9 years ago

    " 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.

  • 9 years ago

    y=2^(30)

    logy=30log2

    =30(.3010)

    =9.3

    10digits

  • 9 years ago

    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

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