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Math Help - other log problem

  1. #1
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    other log problem

    ok this my other log problem i having, i make attempt here
    express as a single log and evaluate

    2 log5 50-log520-2log510+log5100

    first i make this =log 5 50x20x100/100
    =log5 1000
    =log 5 10^3
    =10

    i not sure if this is correct ???

    Thanks you for any helping !!

    other problem i had was log 5 40 to calculate i get 5.3219 do i use calculator correctly ??
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  2. #2
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    2 log5 50-log520-2log510 +log5100

    Let me write "log to the base 5 of 50" as Log(5)[50], so,

    2Log(5)[50] -Log(5)[20] -2Log(5)[10] +Log(5)[100]

    = Log(5)[50^2] -Log(5)[20] -Log(5)[10^2] +Log(5)[100]
    = Log(5)[2500] -Log(5)[20] -Log(5)[100] +Log(5)[100]
    = Log(5)[2500] -Log(5)[20]
    = Log(5)[2500/20]
    = Log(5)[125] -------------------the single log
    = Log(5)[5*5*5]
    = Log(5)[5^3]
    = 3Log(5)[5]
    = 3*1
    = 3 ------------------answer.

    -----------------------------------------
    I forgot the Log(5)[40].

    My calculator cannot give directly antilogs that are not based on 10 or on "e", so I need to convert first the Log(5)[40] into based on 10 or based on "e".

    Conversion:
    Log(b)[a] = Log(c)[a] / Log(c)[b]

    In base 10,
    Log(5)[40]
    = Log(10)[40] / Log(10)[5]
    = 1.60206 / 0.69897
    = 2.292

    Or, in base "e" or natural log,
    Log(5)[40]
    = Ln[40] / Ln[5]
    = 3.688879 / 1.609438
    = 2.292 --------------same as in base 10.

    Therefore, Log(5)[40] = 2.292 ------------answer.
    Last edited by ticbol; August 26th 2007 at 04:25 AM.
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  3. #3
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    confused

    I am not sure which is right on the explanations could you please let me know
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  4. #4
    MHF Contributor red_dog's Avatar
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    Use the properties of logarithms:
    \log_ax+\log_ay=\log_a(xy)
    \displaystyle\log_ax-\log_ay=\log_a\frac{x}{y}
    \log_ax^n=n\log_ax
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  5. #5
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    Hello, gregorio!

    I will assume you know the basic properties of logs . . .


    Express as a single log and evaluate:

    . . 2\!\cdot\!\log_5(50) - \log_5(20) - 2\!\cdot\!\log_5(10) + \log_5(100)

    We have: . \log_5\!\left(50^2\right) - \log_5(20) - \log_5\!\left(10^2\right) + \log_5(100)

    . . = \;\log_5(2500) - \log_5(20)\, \underbrace{- \log_5(100) + \log_5(100)}_{\text{equals }0}

    . . = \;\log_5\!\left(\frac{2500}{20}\right) \;=\;\log_5(125) \;=\;\log_5\!\left(5^3\right)\;=\;3\!\cdot\!\under  brace{\log_5(5)}_{\text{equals }1} \;=\;3\cdot1 \;=\;\boxed{3}

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  6. #6
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    yes now i seeing this

    i not see before the 2 same that cancel out !!! thank you, i finding sometimes i making sillly mistakes, more concentrated needed!!!
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