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Math Help - Logarithm problem

  1. #1
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    Logarithm problem

    Hi everyone,

    Need a lot of help on this:

    Find all real numbers x, if any, that satisfy the eq:

    (log x^2)^2 = log x^4

    NOTE: each log has base 10
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by sarahh View Post
    Hi everyone,

    Need a lot of help on this:

    Find all real numbers x, if any, that satisfy the eq:

    (log x^2)^2 = log x^4

    NOTE: each log has base 10
    I'm assuming the LHS is [log(x^2)]^2 and the RHS is log(x^4)?

    Then
    [2 log(x)]^2 = 4 log(x)

    Set y = log(x)

    [2y]^2 = 4y

    4y^2 = 4y

    y^2 = y

    y^2 - y = 0

    y(y - 1) = 0

    So y = 0 or y = 1.

    Thus log(x) = 0 or log(x) = 1
    log(x) = 0
    10^[log(x)] = 10^0
    x = 1

    and
    log(x) = 1
    10^[log(x)] = 10^1
    x = 10.

    Thus x = 1 or x = 10.

    -Dan
    Last edited by topsquark; October 5th 2006 at 02:49 PM.
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  3. #3
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by topsquark View Post
    log(x) = 1
    10^[log(x)] = 10^1
    x = 10.

    -Dan
    What is the rule here...
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Quick View Post
    What is the rule here...
    The function
    y = a^x
    has an inverse defined as y = log(a) x. In English this reads the "log to the base a of x." (The "a" is subscripted.)

    So we know that
    a^[log(a) x] = x
    and
    [log(a) a^x] = x

    Typically (though not universally, there's a thread on the forum somewhere on this) "log to the base 10" is usually shortened to the abreviation "log" (with no base mentioned) and "ln" is log(e) where e = 2.817...

    -Dan

    For general consumption, I fixed an error in my earlier post. (That's what I thought Quick was going to have posted about. )
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  5. #5
    dan
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    good explanation topsquark. that even sence to me!!!


    dan
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  6. #6
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    Quote Originally Posted by topsquark View Post
    The function
    y = a^x
    a>0
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    a>0
    But of course!

    -Dan
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  8. #8
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    Quote Originally Posted by ThePerfectHacker View Post
    a>0
    and a ≠ 1
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by earboth View Post
    and a ≠ 1
    (laughs) You know I never thought about that one...but you're right.

    -Dan
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  10. #10
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    Quote Originally Posted by earboth View Post
    and a ≠ 1
    You can have exponents for 1 just it will be trivial.

    You cannot have logarithms for 1, so that might be a reason.
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