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Math Help - Logs

  1. #1
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    Logs

    Use logarithm rules to rewrite each expression in terms of log_{3}2 and log_{3}5

    Here's what I did... but I think I didn't do them right... please help?

    a) log_3{270}

     = log_3{2^{7}} + log_3{5^{3}} + 17


    b) log_3{\frac{64}{125}}

    = log_3{64} - log_3{125}

    = log_3{2^{6}} - log_3{5^{3}}

    = 6log_3{2} - 3log_3{5}
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  2. #2
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    Quote Originally Posted by Macleef View Post
    Use logarithm rules to rewrite each expression in terms of log_{3}2 and log_{3}5

    Here's what I did... but I think I didn't do them right... please help?

    a) log_3{270}

     = log_3{2^{7}} + log_3{5^{3}} + 17


    b) log_3{\frac{64}{125}}

    = log_3{64} - log_3{125}

    = log_3{2^{6}} - log_3{5^{3}}

    = 6log_3{2} - 3log_3{5}
    How did you get

      log_3{2^{7}} + log_3{5^{3}} + 17

    in the first one?

    The second one is right.
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  3. #3
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    Quote Originally Posted by Macleef View Post
    Use logarithm rules to rewrite each expression in terms of log_{3}2 and log_{3}5

    Here's what I did... but I think I didn't do them right... please help?

    a) log_3{270}

     = log_3{2^{7}} + log_3{5^{3}} + 17


    b) log_3{\frac{64}{125}}

    = log_3{64} - log_3{125}

    = log_3{2^{6}} - log_3{5^{3}}

    = 6log_3{2} - 3log_3{5}
    Your A) part is wrong, but B) part is right.

    see,

    A) log_3 270

    = log_3(27 \times 2 \times 5)

    = log_3 27 + log_3 2 + log_3 5

    = log_3 3^3 + log_3 2 + log_3 5

    can you finish it now??
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  4. #4
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    Quote Originally Posted by Shyam View Post
    Your A) part is wrong, but B) part is right.

    see,

    A) log_3 270

    = log_3(27 \times 2 \times 5)

    = log_3 27 + log_3 2 + log_3 5

    = log_3 3^3 + log_3 2 + log_3 5

    can you finish it now??
    yes... I see it now. How do you know the right numbers that go to 270? Is it just trial and error or is there a method to figure it out?
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  5. #5
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    Quote Originally Posted by Macleef View Post
    yes... I see it now. How do you know the right numbers that go to 270? Is it just trial and error or is there a method to figure it out?
    you see,

     <br />
270 = 27 \times 10 = 3^3 \times 2 \times 5<br />

    I took 27 because it will be 3^3, because base of log is 3.

    Did you get it now??
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  6. #6
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    Quote Originally Posted by Shyam View Post
    you see,

     <br />
270 = 27 \times 10 = 3^3 \times 2 \times 5<br />

    I took 27 because it will be 3^3, because base of log is 3.

    Did you get it now??
    Yes I do when someone does it for me... I can't solve it myself...

    for example:
    log_3{100} ... the answer is 2 log_3{2} + 2 log_3{5} ... and I don't know how to get that on my own??
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  7. #7
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    Quote Originally Posted by Macleef View Post
    Yes I do when someone does it for me... I can't solve it myself...

    for example:
    log_3{100} ... the answer is 2 log_3{2} + 2 log_3{5} ... and I don't know how to get that on my own??
    You need more practice of these types of questions. If you have to make in terms of log 2 and log 5, then you have to think in terms of factors of 2 and 5, and their powers;

    100 = factors of 2 and 5
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  8. #8
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    ok for that question... log_3{100}, why is it 5^2 and 2^2 which equals to only 29 and not 100?
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  9. #9
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    Quote Originally Posted by Macleef View Post
    ok for that question... log_3{100}, why is it 5^2 and 2^2 which equals to only 29 and not 100?
    you don't have to add,

    you have to multiply.(only then logs get added)

    5^2\times 2^2 = 100

    do you remember log laws:

    \log (a\times b) = \log a + \log b

    can you finish that now??
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  10. #10
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    oh ok I get it now! I just realized what I've been doing wrong... I've been adding instead of multiplying...
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