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Math Help - Elementary theory of logarithms

  1. #1
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    Elementary theory of logarithms

    Hi,
    I am wondering if anyone can help me with this exercise:
    Given that log2=0.3
    show that log5=0.7
    I tried the following approach:
    0.7=0.3+0.3+\frac{1}{3}\times0.3<br />
\\ =\log2+\log2+\log2^\frac{1}{3}
    =\log(4\times\sqrt[3]{2})
    \approx\log5

    Thanks for help

    Gibo
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  2. #2
    Junior Member SuperCalculus's Avatar
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    Quote Originally Posted by Gibo View Post
    Hi,
    I am wondering if anyone can help me with this exercise:
    Given that log2=0.3
    show that log5=0.7
    I tried the following approach:
    0.7=0.3+0.3+\frac{1}{3}\times0.3<br />
\\ =\log2+\log2+\log2^\frac{1}{3}
    =\log(4\times\sqrt[3]{2})
    \approx\log5

    Thanks for help

    Gibo
    That mostly seems fine, but that last step seems underdeveloped. What I would do is:
    0.7 = 0.3 + 0.3 + (\frac{1}{3}\*0.3) = Log2 + Log 2 + \frac{1}{3}\ Log2 = 2Log2 + \frac{1}{3}\ Log2 = Log2^2 + Log2^\frac{1}{3}\  <br />
0.7= Log(2^2*2^\frac{1}{3}\ ) = Log2^\frac{7}{3} \approx\log 5

    Dunno, just seems a little more "fleshed-out" if you will, to me.
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  3. #3
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    Hello, Gibo!

    \text{Given that }\log2\,=\,0.3,\,\text{ show that: }\log5\,=\,0.7

    Assuming that the logs are base-10 . . .

    \log(5) \;=\;\log\left(\dfrac{10}{2}\right) \;=\;\log(10) - \log(2) \;=\;1 - 0.3 \;=\;0.7
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  4. #4
    Junior Member SuperCalculus's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Gibo!


    Assuming that the logs are base-10 . . .

    \log(5) \;=\;\log\left(\dfrac{10}{2}\right) \;=\;\log(10) - \log(2) \;=\;1 - 0.3 \;=\;0.7
    ...-facepalm- How did I not think of that?!
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  5. #5
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    Thanks for your help.
    Regards,
    Gibo
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  6. #6
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    Thanks for yor help.
    Regards,
    Gibo
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