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    • Thread Starter
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    hi, i'm really stuck



    where do i go from here?

    The answer is


    thanks.
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    If you leave it in base 2, stuff cancels out
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    ^what he said leave g(x) as it is
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    i used a calculator to give me





    i can see the pattern emerging, but can anyone explain why this is? is it a rule or is there a way to come to this conclusion without a calculator?
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    (Original post by buj)
    i can see the pattern emerging, but can anyone explain why this is? is it a rule or is there a way to come to this conclusion without a calculator?
    logn(x) and nx are inverse functions of each other, so they cancel.
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    You need to use the rule b\log a = \log a^b to get rid of the minus sign. Then use the fact that k^{\log_k x} = x.
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    First of all deal with the negative power. x^-a = 1 / x^a

    So 2^(-log2(x)) = 1 / 2^(log2(x))

    Taking logs is the opposite of raising something to a power so they sort of cancel out. If you think of base 10:

    log10=1
    so 10^(log10) = 10^1=10

    log 100 = 2
    so 10^(log100) = 10^2 = 100

    Generally,
    10^(log x) = x

    This works for any base so 2^(log2(x)) = x

    So 2^(-log2(x)) = 1/x

    Putting this into f (x) should give the required result.
 
 
 
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Updated: March 18, 2011
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