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  1. #1
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    Logs

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

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

    a) $\displaystyle log_3{270}$

    $\displaystyle = log_3{2^{7}} + log_3{5^{3}} + 17$


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

    $\displaystyle = log_3{64} - log_3{125}$

    $\displaystyle = log_3{2^{6}} - log_3{5^{3}}$

    $\displaystyle = 6log_3{2} - 3log_3{5}$
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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 $\displaystyle log_{3}2$ and $\displaystyle log_{3}5$

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

    a) $\displaystyle log_3{270}$

    $\displaystyle = log_3{2^{7}} + log_3{5^{3}} + 17$


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

    $\displaystyle = log_3{64} - log_3{125}$

    $\displaystyle = log_3{2^{6}} - log_3{5^{3}}$

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

    $\displaystyle log_3{2^{7}} + log_3{5^{3}} + 17$

    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 $\displaystyle log_{3}2$ and $\displaystyle log_{3}5$

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

    a) $\displaystyle log_3{270}$

    $\displaystyle = log_3{2^{7}} + log_3{5^{3}} + 17$


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

    $\displaystyle = log_3{64} - log_3{125}$

    $\displaystyle = log_3{2^{6}} - log_3{5^{3}}$

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

    see,

    A) $\displaystyle log_3 270$

    $\displaystyle = log_3(27 \times 2 \times 5)$

    $\displaystyle = log_3 27 + log_3 2 + log_3 5$

    $\displaystyle = log_3 3^3 + log_3 2 + log_3 5$

    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) $\displaystyle log_3 270$

    $\displaystyle = log_3(27 \times 2 \times 5)$

    $\displaystyle = log_3 27 + log_3 2 + log_3 5$

    $\displaystyle = log_3 3^3 + log_3 2 + log_3 5$

    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,

    $\displaystyle
    270 = 27 \times 10 = 3^3 \times 2 \times 5
    $

    I took 27 because it will be $\displaystyle 3^3$, 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,

    $\displaystyle
    270 = 27 \times 10 = 3^3 \times 2 \times 5
    $

    I took 27 because it will be $\displaystyle 3^3$, 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:
    $\displaystyle log_3{100}$ ... the answer is $\displaystyle 2 log_3{2} + 2 log_3{5}$ ... 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:
    $\displaystyle log_3{100}$ ... the answer is $\displaystyle 2 log_3{2} + 2 log_3{5}$ ... 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... $\displaystyle log_3{100}$, why is it $\displaystyle 5^2$ and $\displaystyle 2^2$ 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... $\displaystyle log_3{100}$, why is it $\displaystyle 5^2$ and $\displaystyle 2^2$ which equals to only 29 and not 100?
    you don't have to add,

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

    $\displaystyle 5^2\times 2^2 = 100$

    do you remember log laws:

    $\displaystyle \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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