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Math Help - is this logarithm undefined?

  1. #1
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    is this logarithm undefined?

    is this logarithm undefined where

    log 8 -4 + log 8 -2 = 1

    i know that you cannot take the log of a negiative but if you combine the logarithms to
    log 8 (-4x-2) = 1
    log 8 8 = 1
    1=1
    therefore it satisifies.

    is that an acceptable answer or is it incorrect?
    if so why is it not allowed?
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by way123 View Post
    is this logarithm undefined where

    log 8 -4 + log 8 -2 = 1

    i know that you cannot take the log of a negiative but if you combine the logarithms to
    log 8 (-4x-2) = 1
    log 8 8 = 1
    1=1
    therefore it satisifies.

    is that an acceptable answer or is it incorrect?
    if so why is it not allowed?

    log is defined for negative numbers but ln is not defined for negative numbers if the base is positive it is not defined for negative but for negatives it is ok like this

    log_{-2} -8 = 3 since (-2)^3 = -8

    the question is wrong since the base is positive

    log_{8} -4 = x \Rightarrow 8^x = -4 but 8^x is always positive for all real numbers

    so the question is wrong
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  3. #3
    Super Member Bacterius's Avatar
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    Logarithmic functions are defined for positive numbers in the standard way and for negative numbers with their analytic continuation (involves complex numbers). So for the real numbers, logarithmic functions are only defined for positive numbers.

    But I honestly don't understand your question. Could you express it in a "how can I solve this function for x " way ?
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  4. #4
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    sorry that wasnt the original question i just put what i was not sure about. ill write the question and just look at your answer.

    where i ran in to trouble was in in the check.

    solve and check

    log 8 (2-x) + log 8 (4-x) = 1
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  5. #5
    Super Member Bacterius's Avatar
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    Quote Originally Posted by way123 View Post
    sorry that wasnt the original question i just put what i was not sure about. ill write the question and just look at your answer.

    where i ran in to trouble was in in the check.

    solve and check

    log 8 (2-x) + log 8 (4-x) = 1
    You can use the property that \log_b{(xy)} = \log_b{(x)} + \log_b{(y)}, then it is pretty straightforward

    And to check, just plug in your result for x and see if the equation holds.

    Note : \log_8{(a)} = \frac{\log{(a)}}{\log{(8)}}

    Where \log is the standard logarithm key on your calculator
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  6. #6
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    no i know that.
    i did that and got x ^2 -6x
    therefore x is 0 and 6
    and we have to check with left side, right side charts
    so when i did the check for six i got

    L.S. |R.S.
    =log 8 -4 + log 8 -2 |=1

    and then what do i do :S
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  7. #7
    Super Member Bacterius's Avatar
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    To solve

    \log_8{(2-x)} + \log_8{(4-x)} = 1

    \log_8{\left [ (2-x)(4-x) \right ]} = 1

    8^{\log_8{\left [ (2-x)(4-x) \right ]}} = 8^1

    (2-x)(4-x)= 8

    (2-x)(4-x) - 8= 0

    8 - 2x - 4x + x^2 - 8= 0

    x^2 - 6x= 0

    x(x - 6)= 0

    x = 0 \ \ \textrm{and} \ \ x = 6

    To check

    \log_8{(2-6)} + \log_8{(4-6)} = 1

    \log_8{(-4)} + \log_8{(-2)} = 1

    \log_8{\left [ (-4)(-2) \right ]} = 1

    \log_8{(8)} = 1

    1 = 1

    Successful check !

    \log_8{(2)} + \log_8{(4)} = 1

    \log_8{\left [ (2)(4) \right ]} = 1

    \log_8{(8)} = 1

    1 = 1

    Successful check !
    ______________________________________

    Does it make sense now ? I do not know what you mean by left side/right side charts though. I guess the goal is to show that both sides are equal ...
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  8. #8
    Super Member Bacterius's Avatar
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    Oh I think I know why you had a problem, you couldn't evaluate the logarithm of a negative number. Well, here you can combine both logs to obtain a positive number. If no combination is possible, then the answer is undefined and you have to conclude that the solution you found to the quadratic equation is outside the domain of the original function (involving logs).
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  9. #9
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    Quote Originally Posted by Bacterius View Post
    To solve

    \log_8{(2-x)} + \log_8{(4-x)} = 1

    \log_8{\left [ (2-x)(4-x) \right ]} = 1

    8^{\log_8{\left [ (2-x)(4-x) \right ]}} = 8^1

    (2-x)(4-x)= 8

    (2-x)(4-x) - 8= 0

    8 - 2x - 4x + x^2 - 8= 0

    x^2 - 6x= 0

    x(x - 6)= 0

    x = 0 \ \ \textrm{and} \ \ x = 6

    To check

    \log_8{(2-6)} + \log_8{(4-6)} = 1

    \log_8{(-4)} + \log_8{(-2)} = 1

    \log_8{\left [ (-4)(-2) \right ]} = 1

    \log_8{(8)} = 1

    1 = 1

    Successful check !
    No, not a successful check! The original problem was \log_8{(2-x)} + \log_8{(4-x)} = 1, not \log_8((2-x)(4-x))
    x= 0 satisfies the second of those but NOT the first equation. log_8(2- 6)= log_8(-4) and log_8(4- 6)= log_8(-2) do not exist so x= 6 does NOT satisfy the given equation. The fact that x= 6 satisfies log_8((4-x)(2-x)) is not relevant.


    \log_8{(2)} + \log_8{(4)} = 1

    \log_8{\left [ (2)(4) \right ]} = 1

    \log_8{(8)} = 1

    1 = 1

    Successful check !
    ______________________________________

    Does it make sense now ? I do not know what you mean by left side/right side charts though. I guess the goal is to show that both sides are equal ...
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  10. #10
    Super Member Bacterius's Avatar
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    Oh I didn't know that, thanks HallsOfIvy
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