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Math Help - Log Question

  1. #1
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    Log Question

    To nine decimal places, \log_{10}{2} = 0.301029996 and \log_{10}{3} = 0.477121255

    i) Calculate \log_{10}{5} and \log_{10}{6} to three decimal places. By taking logs, or otherwise, show that

    5 \times 10^{47} < 3^{100} < 6 \times 10^{47}
    Hence write down the first digit of 3^{100}

    ii) Find the first digit of each of the following numbers: 2^{1000} , 2^{10 \ 000} and 2^{100 \ 000}


    [OCR STEP I 2000, Question 1]
    I could do with a tiny bit of guidance on this. thanks in advance for any help

    - Bobak
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  2. #2
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    Hello, bobak!

    Here's the first one . . .


    Given: . \begin{array}{ccc}\log_{10}2 &=&0.30103 \\ \log_{10}3 &=&0.47712 \end{array}

    calculate (a)\;\log_{10}5\text{ and }(b)\;\log_{10}6 to three decimal places.

    (a)\;\;\log_{10}(5) \;=\;\log_{10}\left(\frac{10}{2}\right) \;=\;\log_{10}(10) \;- \;\log_{10}(2) \;=\;1 \;- \;0.30103 \;\approx\;0.699

    (b)\;\;\log_{10}(6) \;=\;\log_{10}(2\cdot3) \;=\;\log_{10}(2)\;+ \;\log_{10}(3) \;=\;0.30103 \;+\;0.47712 \;\approx \;0.778

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  3. #3
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    Quote Originally Posted by bobak View Post
    I could do with a tiny bit of guidance on this. thanks in advance for any help

    - Bobak
    And here's the next:

    It boils down to showing 47 + \log_{10} 5 < 100 \log_{10} 3 < 47 + \log_{10} 6.

    And you know the values of all the logs from the earlier part ......

    So clearly 3^{100} = 5.abcd..... \times 10^{47} - its first digit is now obvious.
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  4. #4
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    Quote Originally Posted by bobak View Post
    I could do with a tiny bit of guidance on this. thanks in advance for any help

    - Bobak
    For the final bit, try to apply the reasoning from earlier:

    Can it be shown that a \times 10^x < 2^{1000} < (a + 1) \times 10^x, where obviously a and x aren't known yet!

    Note that \log_{10} 2^{1000} = 1000 \log_{10} 2 = 301.029996. So clearly x = 301.

    So can you find an a such that a \times 10^{301} < 2^{1000} < (a + 1) \times 10^{301}. In other words, can you find an a such that

    <br />
301 + \log_{10} a < 301.029996 < 301 + \log_{10} (a + 1).

    Again, you'd use the known values of log from earlier. Warning: spoiler below.


























    Answer: a = 1 so the first digit is 1.
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  5. #5
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    well you cant use log values form earlier.

    but a is clearly 1

    thanks Mr F

    edit: you added the spoiler while i typed this
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  6. #6
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    On the topic of spoilers the forum really should have hide/show tags.
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