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Math Help - Dazed and Confused

  1. #1
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    Dazed and Confused

    If log b 3=0.5
    log b 5= 0.7

    evaluate log b √3
    √6


    Log b √3=logb 3^1/2

    =1/2 logb 3

    1/2(0.5)
    =.25

    Am I on the right track? And if I am, the √6 is somewhat throwing me off. Can I get help with that please? Ty
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  2. #2
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    Re: Dazed and Confused

    Quote Originally Posted by Past45 View Post
    If log b 3=0.5
    log b 5= 0.7

    evaluate log b √3
    √6


    Log b √3=logb 3^1/2

    =1/2 logb 3

    1/2(0.5)
    =.25

    Am I on the right track? And if I am, the √6 is somewhat throwing me off. Can I get help with that please? Ty
    It is almost impossible to read what you posted.
    Learn to use basic LaTeX.
    [TEX]\log_b\sqrt{3}=\frac{\log_b(3)}{2}[/TEX] gives \log_b\sqrt{3}=\frac{\log_b(3)}{2}.
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  3. #3
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    Re: Dazed and Confused

    Now, I am confused in what you wrote, lol.

    So logb √3 equals to logb(3)/2?


    Let me try to re-write this problem. No clue what this LaTeX thing is, so please forgive me.

    If log b 3=0.5
    logb 5= 0.7


    Now I have to evaluate log √3 and log √6. My steps for √3 is already posted above this for that problem.
    Last edited by Past45; April 20th 2012 at 06:44 PM.
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  4. #4
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    Re: Dazed and Confused

    If, by "log b" you mean "log base b", log_b or log_b then

    Use the laws of logarithms: log_b(\sqrt{3})= log_b(3^{1/2})= (1/2)log_b(3) exactly as you say.

    log_b(\sqrt{6})= log_b(6^{1/2})= (1/2)log_b(6)= (1/2)log_b(2(3)) = (1/2)(log_b(2)+ log_b(3)).
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