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    hey how do you find the value of:

    loga(1)???
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    (Original post by kickboxer 98)
    hey how do you find the value of:

    loga(1)???
    Use the definition of the logarithm, find x such that a^x = 1.
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    (Original post by EricPiphany)
    Use the definition of the logarithm, find x such that a^x = 1.
    i thought the answer was 1 but i dont know?
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    Anything to the power of 0 = 1
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    (Original post by kickboxer 98)
    i thought the answer was 1 but i dont know?
    The answer isn't 1. What do you know about the log function?
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    (Original post by kickboxer 98)
    hey how do you find the value of:

    loga(1)???
    a^x=1

    What is x?

    EDIT: They beat me to it
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    is the answer 0?
    anything to the power of 0 is 1?
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    (Original post by Bobjim12)
    Anything to the power of 0 = 1
    thats what i thought
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    (Original post by kickboxer 98)
    is the answer 0?
    anything to the power of 0 is 1?
    Yes, precisely! :-)
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    okay, now... if i had:

    2(loga 6) -1

    how would that expand to remove the brakets?
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    (Original post by kickboxer 98)
    okay, now... if i had:

    2(loga 6) -1

    how would that expand to remove the brakets?
    2\log_a 6 = \log_a 6^2
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    (Original post by Zacken)
    2\log_a 6 = \log_a 6^2
    what happened to the -1?

    question is:
    loga X = 2(loa 6)-1

    find X
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    (Original post by kickboxer 98)
    what happened to the -1?

    question is:
    loga X = 2(loa 6)-1

    find X
    I was only simplifying the log term for you.

    You want to get both sides in terms of \log_a so you can 'cancel' it out.

    \log_a x = \log_a 6^2 - \log_a a = \log_a \frac{6^2}{a}
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    I suggest you familiarise yourself with the log rules a bit before you attempt the questions you are currently working on
 
 
 
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