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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} \normalsize \text{Find the number of digits in }\ 60^{12}\hspace{.33em}\\~\\ \end{align}}\)

  2. mathstudent55
    • one year ago
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    |dw:1438031157075:dw|

  3. mathmath333
    • one year ago
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    but calclulting whole stuff is cumbersome

  4. ganeshie8
    • one year ago
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    dumb q, but are you allowed to use calculator for taking logarithm

  5. mathmath333
    • one year ago
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    no calculo tor

  6. ganeshie8
    • one year ago
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    i mean my question of asking whether you're allowed to use a calculator is dumb :)

  7. mathmath333
    • one year ago
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    i mean no cheating stuff allowed

  8. ganeshie8
    • one year ago
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    number of digits equals below expression, i don't see an easy way to simplify it though\[\large \lfloor \log_{10} 60^{12}\rfloor+1\]

  9. mathmath333
    • one year ago
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    where did u get this formula

  10. mathmath333
    • one year ago
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    what is the name of the formula

  11. ganeshie8
    • one year ago
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    Its not a formula as such, just work few examples till you get convinced, keeping in mind the definition of logarithm \[\large 10^x = N ~~\iff~~ x=\log_{10} N\]

  12. mathmath333
    • one year ago
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    how did u add +1

  13. ganeshie8
    • one year ago
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    how many digits are there in \(\large 10^1\) ?

  14. mathmath333
    • one year ago
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    \(\large \lfloor \log_{10} 60^{12}\rfloor\color{red}{+1}\)

  15. mathmath333
    • one year ago
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    2

  16. ganeshie8
    • one year ago
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    how many digits are there in \(\large 10^2\) ?

  17. mathmath333
    • one year ago
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    3

  18. mathmath333
    • one year ago
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    i see

  19. ganeshie8
    • one year ago
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    how many digits are there in \(\large 10^{300}\) ?

  20. mathmath333
    • one year ago
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    301

  21. ganeshie8
    • one year ago
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    how did you get 301 ?

  22. mathmath333
    • one year ago
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    add +1

  23. ganeshie8
    • one year ago
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    You got 301 by adding 1 to the logarithm, 300.

  24. ganeshie8
    • one year ago
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    It helps to keep in mind the actual definition of logarithm : logarithm = exponent

  25. ganeshie8
    • one year ago
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    \[\large 10^{\color{red}{300}} \] \(\color{red}{300}\) is called the logarithm of \( 10^{\color{red}{300}}\) relative to base \(10\)

  26. ganeshie8
    • one year ago
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    as you can see, the number of digits in the decimal expansion of \( 10^{\color{red}{300}}\) equals \[\large \lfloor \log_{10} 10^{\color{red}{300}}\rfloor + 1\]

  27. mathmath333
    • one year ago
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    that was easy

  28. mathmath333
    • one year ago
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    \(log _{10} 10\)

  29. ganeshie8
    • one year ago
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    \(\large 10 = 10^\color{red}{1}\) so the logarithm of \(10\) is \(\color{red}{1}\) (relative to base 10)

  30. mathmath333
    • one year ago
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    is it important to include floor function

  31. ganeshie8
    • one year ago
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    how many digits are there in \(\large 3^4\) ?

  32. mathmath333
    • one year ago
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    2

  33. ganeshie8
    • one year ago
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    how did u get 2 ?

  34. ganeshie8
    • one year ago
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    http://www.wolframalpha.com/input/?i=floor%28log%2810%2C3%5E4%29%29%2B1

  35. mathmath333
    • one year ago
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    81=3^4

  36. ganeshie8
    • one year ago
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    try using the formula

  37. ganeshie8
    • one year ago
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    \[\large \lfloor \log_{10} 3^{4}\rfloor+1\]

  38. ganeshie8
    • one year ago
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    u will see why floor function is necessary

  39. mathmath333
    • one year ago
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    ok

  40. mathmath333
    • one year ago
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    it seems i need to remember log values to use the formula

  41. ganeshie8
    • one year ago
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    not worth it, there must be a simpler way to solve it

  42. ganeshie8
    • one year ago
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    Remember, if you're struggling too hard, then it means only one thing : your method is wrong/not clever

  43. mathmath333
    • one year ago
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    ok thnx

  44. ganeshie8
    • one year ago
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    is this really supposed to be solved w/o calculator ?

  45. mathmath333
    • one year ago
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    yes lol

  46. ganeshie8
    • one year ago
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    i don't really see how one computes it without knowing the value of \(\log_{10}6\), or without working the actual number hmm

  47. mathmath333
    • one year ago
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    log6=log2+log3

  48. mathmath333
    • one year ago
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    i just need to rote log 2 and log 3

  49. ganeshie8
    • one year ago
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    i got better things to memorize than these log values haha

  50. mathmath333
    • one year ago
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    its matter of pattern , not rote learning

  51. mathmath333
    • one year ago
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    log 2=0.30 , log3=0.47 easy

  52. ganeshie8
    • one year ago
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    Okay, then lets pretend that we have access to log2 and log3 values :)

  53. ganeshie8
    • one year ago
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    log6 = log2 + log3 = 0.30 + 0.47 = 0.77 ?

  54. mathmath333
    • one year ago
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    yep correct

  55. ganeshie8
    • one year ago
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    good, plug them in the formula and see..

  56. mathmath333
    • one year ago
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    http://www.wolframalpha.com/input/?i=log_%7B10%7D6%3D

  57. ganeshie8
    • one year ago
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    \[\large{\begin{align} \lfloor \log_{10} 60^{12}\rfloor+1 &= \lfloor 12*\log_{10} 60\rfloor+1\\~\\ &= \lfloor 12*(\log_{10}10 +\log_{10} 6)\rfloor+1\\~\\ &\approx \lfloor 12*(1 +0.77)\rfloor+1\\~\\ &=21+1 \end{align}}\]

  58. mathmath333
    • one year ago
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    http://openstudy.com/users/mathmath333#/updates/55b3954fe4b071e6530e16e9 one more method for finding log

  59. ganeshie8
    • one year ago
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    Ahh you wanto use taylor series is it \[\ln(1+x)=-\sum_{n=1}^\infty\frac{(-x)^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\]

  60. mathmath333
    • one year ago
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    i just will remember values to save time

  61. ganeshie8
    • one year ago
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    do let me know if/when you find a better method

  62. mathmath333
    • one year ago
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    yep .

  63. phi
    • one year ago
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    you can make some reasonable approximations to do this problem \[ 60^{12}= 2^{12} \cdot 3^{12} \cdot 10^{12} \] 2^10 is 1024 or close to 10^3 3^12 is 81^3 and 81 is 2^3 * 10 thus \( 3^{12} \approx \left(2^3 \cdot 10^1\right)^3 = 2^9 \cdot 10^3\) putting this mess together \[ 2^{12} \cdot 3^{12} \cdot 10^{12} \\ 10^3 \cdot 2^2 \cdot 2^9 \cdot 10^3 \cdot 10^{12} \\ 2 \cdot 10^3 \cdot 10^{18} \\ 2 \cdot 10^{21} \] which means 22 digits.

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