# Math question ?

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

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• Hello,

► They aren't.

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

So ln(0) DOES NOT EXIST.

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:

Regards,

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• They aren't.  The domain is:

{ x | 0 < x < e }

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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