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Math Help - log function....

  1. October 2nd 2009, 03:52 PM #1
    frozenflames
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    log function....



    how would i get x? the previous method does not work.....
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  2. October 2nd 2009, 04:00 PM #2
    pickslides
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    You already have x!
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  3. October 2nd 2009, 04:02 PM #3
    frozenflames
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    Quote Originally Posted by pickslides View Post
    You already have x!
    well i meant in its simplest form...
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  4. October 2nd 2009, 04:07 PM #4
    pickslides
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    It doesn't get much more simple

    x = 2^{\log_23+\log_28}

    x = 2^{\log_23+\log_22^3}

    x = 2^{\log_23+3\log_22}

    x = 2^{\log_23+3\times 1}

    x = 2^{\log_23+3}

    x = 2^3\times 2^{\log_23}

    x = 8\times 2^{\log_23}
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  5. October 2nd 2009, 04:11 PM #5
    frozenflames
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    i'm supposed to be getting a whole number i think
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  6. October 2nd 2009, 04:13 PM #6
    Matt Westwood
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    Well it's nearly there. What's 2^{\log_2 x} for any x?
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  7. October 2nd 2009, 04:16 PM #7
    frozenflames
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    Quote Originally Posted by Matt Westwood View Post
    Well it's nearly there. What's 2^{\log_2 x} for any x?
    honestly i'm not too sure, my guess would be 1?
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  8. October 2nd 2009, 04:17 PM #8
    Soroban
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    Hello, frozenflames!

    Simplify: . x \:=\:2^{\log_23 + \log_28}

    We have: . x \;=\;2^{\log_23+\log_28} \;=\; 2^{\log_2(3\cdot8)} \;=\;2^{\log_2(24)} \;=\;24

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  9. October 2nd 2009, 04:22 PM #9
    Matt Westwood
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    Quote Originally Posted by frozenflames View Post
    honestly i'm not too sure, my guess would be 1?
    This is the most fundamental truth about exponentials and logarithms that you really need to understand: one is the inverse of the other. That is, logarithms "undo" exponents, and exponents "undo" logarithms.

    a = \log_b c \iff c = b^a for any a, and any b, c positive. That's how logarithms are defined.

    So 2^{\log_2 x} = \log_2 (2^x) = x.
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