Math question ?

Why are 0 and e within the domain of ln(1-lnx)?

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  • 3 weeks ago

    Hello,

    ► They aren't.

    The domain of the logarithmic function is (0; +∞).

    So ln(0) DOES NOT EXIST.

    ► In your function

       𝑓(𝑥) = ln[ 1 − ln(𝑥) ]

    If 𝑥=0, the enclosed logarithmic function will have a ln(0) value.

    If 𝑥=𝑒, 1−ln(𝑥) will become 0, leading to the external logarithm to have a ln(0) value.

    So both values ARE NOT in the domain of 𝑓.    ◄◄◄ANSWER

    ► So if you are interested in the domain of:

       𝑓(𝑥) = ln[ 1 − ln(𝑥) ]

    The domain of the logarithmic function is (0; +∞). So whatever is enclosed in the external brackets must be positive:

       1 − ln(𝑥) > 0

       1 > ln(𝑥)

       ln(𝑒) > ln(𝑥)

       𝑒 > 𝑥

    And obviously, whatever is enclosed in the internal bracket must be positive:

       𝑥 > 0

    So you end up with the mandatory condition for the function to be defined:

       0 < 𝑥 < 𝑒

    And the domain of 𝑓 is:

       𝐷𝑓 = (0; 𝑒)      ◄◄◄ANSWER

    Regards,

    Dragon.Jade :-)

  • 3 weeks ago

    They aren't.  The domain is:

    { x | 0 < x < e }

    EDIT to talk to comments.

    You can't get the log of zero.

    The problem:

    ln[1 - ln(x)]

    When x = 0, then you can't get ln(0) so x can't be zero.

    When x = e, the ln(e) resolves to 1.  The outside log simplifies to ln(0), which again we can't do.

    As a result, 0 and e cannot be in the input set, but any values in between can.

    You can't get the log of a negative number, so x < 0 values are out and x > e values are out since the inside ln(x) will resolve to a value larger than 1 resulting in a negative number for the outside log, which again you can't do.

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