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Math Help - Logarithm question

  1. #1
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    Logarithm question

    Simplfy a^logaX^3 and use it to find the maximum possible value of 8(27^log6X)+27(8^log6X)-X^3, given that x is a real number.
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  2. #2
    Senior Member pacman's Avatar
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    Simplfy a^logaX^3 and use it to find the maximum possible value of 8(27^log6X)+27(8^log6X)-X^3, given that x is a real number.

    a^logaX^3 = x^3,

    Let y = 8(27^log_6 X)+27(8^log_6 X)-X^3

    y = 8(27^log_6 X)+27(8^log_6 X) - 6^(log_6 x^3)

    = 216 when x = 6,

    THUS, global maximum is when x = 6.


    i plot it first and by observation, the logarithmic function gets a maximum @ about x = 6. So i guess, max when x = 6.

    ok
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  3. #3
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by yeoky View Post
    Simplfy a^logaX^3 and use it to find the maximum possible value of 8(27^log6X)+27(8^log6X)-X^3, given that x is a real number.
    Logarithms are popular this time of year ...

    I presume the equation you are trying to solve is:
    a^{\log_a (X^3)} .

    Remember how logarithms are defined. For any given base (in this case it's a), the "logarithm" function and the "exponentiation" functions are inverses of each other.

    So to "undo" a logarithm, you take the exponent. To "undo" an exponent, you take its logarithm.

    So a^{\log_a x} = \log_a (a^x).

    Let's do it in more detail. Suppose a = b^c. Then by definition of logarithm, \log_b a = c.

    Because c = \log_b a you can slot it straight into the equation a = b^c to get a = b^{\log_b a}.
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  4. #4
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    Quote Originally Posted by yeoky View Post
    Simplfy a^logaX^3 and use it to find the maximum possible value of 8(27^log6X)+27(8^log6X)-X^3, given that x is a real number.
    Thanks and appreciated and I fully understand the answer.

    By the way, is there way to prove that the max is 216 besides using the graph or guess and check method.
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  5. #5
    Senior Member pacman's Avatar
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    y = 8(27^log_6 X)+27(8^log_6 X) - 6^(log_6 x^3),

    y = 8(27^(log_6 x)) + 27(8^(log_6 x)) - 6^3(log_6 x)

    y = 8(27^(log_6 x)) + 27(8^(log_6 x)) - 216^(log_6 x)'

    y = ?
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  6. #6
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    Quote Originally Posted by pacman View Post
    y = 8(27^log_6 X)+27(8^log_6 X) - 6^(log_6 x^3),

    y = 8(27^(log_6 x)) + 27(8^(log_6 x)) - 6^3(log_6 x)

    y = 8(27^(log_6 x)) + 27(8^(log_6 x)) - 216^(log_6 x)'

    y = ?
    Thanks again. I understand this. But it is like gussing x=1, then y=216, but we will not know unless the graph is plotted or a few values of x are tested. Is there any way to prove or show?
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