Results 1 to 6 of 6

Thread: Log Question

  1. #1
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591

    Log Question

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

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

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

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


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

    - Bobak
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, bobak!

    Here's the first one . . .


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

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

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

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

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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 $\displaystyle 47 + \log_{10} 5 < 100 \log_{10} 3 < 47 + \log_{10} 6$.

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

    So clearly $\displaystyle 3^{100} = 5.abcd..... \times 10^{47}$ - its first digit is now obvious.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    9
    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 $\displaystyle a \times 10^x < 2^{1000} < (a + 1) \times 10^x$, where obviously a and x aren't known yet!

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

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

    $\displaystyle
    301 + \log_{10} a < 301.029996 < 301 + \log_{10} (a + 1)$.

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


























    Answer: a = 1 so the first digit is 1.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    On the topic of spoilers the forum really should have hide/show tags.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum