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Math Help - Difficult simplification for logarithm

  1. #1
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    Difficult simplification for logarithm

    How do you simplify log(2square root of 10)-1/3 log0.8-log10/3
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  2. #2
    MHF Contributor
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    Quote Originally Posted by kingkaisai2
    How do you simplify log(2square root of 10)-1/3 log0.8-log10/3
    Log[2sqrt(10)] -(1/3)Log(0.8) -Log(10/3) -------(i)

    Is that what you mean?
    If yes, do you want to combine the 3 logs into one log only?
    If yes again, then,
    = Log[2sqrt(10)] -Log[cubrt(0.8)] - Log(10/3)
    = Log{[2sqrt(10)] / [cubrt(0.8)] / (10/3)} ----------(ii)

    Umm, it's getting too complicated. Let us simplify the 3 logs separately, for simplicity.
    I assume, by "simplify", you mean no decimal points.

    Log[2sqrt(10)]
    = Log[2*(10)^(1/2)]
    = Log(2) +Log(10^1/2)
    = Log(2) +(1/2)Log(10)
    = Log(2) +(1/2)(1)
    = Log(2) +(1/2) ------**

    (1/3)Log(0.8)
    = (1/3)Log[8/10]
    = (1/3)[Log(8) -Log(10)]
    = (1/3)Log(8) -(1/3)Log(10)
    = (1/3)Log(2^3) -(1/3)(1)
    = Log[(2^3)^(1/3)] -(1/3)
    = Log[2] -(1/3) ----------**

    Log(10/3)
    = Log(10) -Log(3)
    = 1 -Log(3) ----------------**

    Therefore, (i) becomes
    = [Log(2) +(1/2)] -[Log(2) -(1/3)] -[1 -Log(3)]
    = 1/2 +1/3 -1 +Log(3)
    = (3 +2 -6)/6 +Log(3)
    = -1/6 +Log(3)
    = Log(3) -(1/6) ----------------------------answer.

    Check,
    Using a calculator,

    Log[2sqrt(10)] -(1/3)Log(0.8) -Log(10/3)
    = 0.80103 -(-0.03230) -0.52288
    = 0.31045

    Log(3) -1/6
    = 0.47712 -0.16667
    = 0.31045

    They are the same, so, OK.
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  3. #3
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    Hello, kingkaisai2!

    I assume I'm reading it correctly . . .

    Simplify: \log(2\sqrt{10}) - \frac{1}{3}\log(0.8) -\log\left(\frac{10}{3}\right)
    We have: \log(2\sqrt{10}) - \log(0.8^{\frac{1}{3}}) - [\log(10) - \log(3)]

    Then: \log(2\sqrt{10}) - \log(\sqrt[3]{0.8}) - \log(10) + \log(3)

    And: \log\left[\frac{2\sqrt{10}\cdot3}{\sqrt[3]{0.8}\cdot10}\right] \;= \;\log\left[\frac{3}{\sqrt{10}\cdot\sqrt[3]{0.8}}\right]

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    If you really want to impress/surprise/terrify your teacher, keep going.

    The denominator is: \sqrt{10}\cdot\sqrt[3]{8\cdot10^{-1}} \;= \;10^{\frac{1}{2}}\cdot\sqrt[3]{8}\cdot10^{-\frac{1}{3}}

    . . . = \;2\left(10^{\frac{1}{2}}\cdot10^{-\frac{1}{3}}\right) \;= \;2\cdot10^{\frac{1}{6}}

    The fraction becomes: \frac{3}{2\cdot10^{\frac{1}{6}}}

    Rationalize: \frac{3}{2\cdot10^{\frac{1}{6}}} \cdot \frac{10^{\frac{5}{6}}}{10^{\frac{5}{6}}} \;=\;\frac{3\cdot10^{5/6}}{20}
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