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Math Help - Logarithms

  1. #1
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    Logarithms

    Find the value oflog 2^1/2+1(3-2^1/2)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Joyce
    Find the value of log 2^1/2+1(3-2^1/2)
    You need to use brackets to make this unambiguous.

    This could mean:

    <br />
\log(2^{1/2}+1)\times  (3-2^{1/2})<br />
,

    <br />
\log[(2^{1/2}+1)\times  (3-2^{1/2})]<br />

    or

    <br />
\log((2^1/2+1)(3-2^{1/2}))<br />

    or any number of other things

    RonL
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  3. #3
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    Hello,Joyce!

    I'd like to help, but I can't read the problem . . .


    Find the value oflog 2^1/2+1(3-2^1/2)

    Did you even look at what showed up on the screen?


    By the way, there should be no exponent in your log statement.

    You probably meant something like: \log_2\left(\frac{1}{2}\right)

    If you see \log_28, do not write it as: log 2^8, which means \log 2^8

    The "2" is a subscript, written 'below the line'.
    . . You can write: log_2(8) . . . okay?

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  4. #4
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    I think he meant
    <br />
\log_{\sqrt{2}+1}(3-\sqrt{2})<br />
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by shubhadeep
    I think he meant
    <br />
\log_{\sqrt{2}+1}(3-\sqrt{2})<br />
    Good idea, I would give you +rep for that but I can't give you more
    +rep today

    Then by the change of logarithm base rule, this can be written:

    <br />
\log_{\sqrt{2}+1}(3-\sqrt{2})=\frac{\log_e(3-\sqrt{2})}{\log_e(\sqrt{2}+1)} <br />
\approx 0.5231<br /> <br /> <br />

    RonL
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  6. #6
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    I too like Shubadeep's interpretation.

    But I hope that there are typos in the original problem.

    If it was: . \log_{1+\sqrt{2}}(3 + 2\sqrt{2}) . . . the answer is 2.


    Otherwise, I see no point to the problem
    . . other than using the Base-Change Formula
    . . which could have been practiced with \log_{1.586}(2.414)

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