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Math Help - log function....

  1. #1
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    log function....



    how would i get x? the previous method does not work.....
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  2. #2
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    You already have x!
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  3. #3
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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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    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. #5
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    i'm supposed to be getting a whole number i think
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  6. #6
    Super Member Matt Westwood's Avatar
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    Well it's nearly there. What's 2^{\log_2 x} for any x?
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  7. #7
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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. #8
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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. #9
    Super Member Matt Westwood's Avatar
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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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