how would this work out

Solve: ln(e) (-.3675)

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- Jan 4th 2009, 05:02 PMlucifer_xSeries and sequence question
how would this work out

Solve: ln(e) (-.3675) - Jan 4th 2009, 05:05 PMJester
- Jan 4th 2009, 05:08 PMlucifer_x
- Jan 4th 2009, 05:11 PMChris L T521
- Jan 4th 2009, 05:15 PMlucifer_x
is the ln(e)

cuz the e is suppose to be the under root of ln - Jan 4th 2009, 05:57 PMmr fantastic
- Jan 4th 2009, 06:10 PMlucifer_x
ok look

LOG(e) (-.3675)

the e is like the lower part of log - Jan 4th 2009, 06:16 PMmr fantastic
If you mean $\displaystyle log_e (-0.3675)$ then:

1. There is no real value.

2. Talk about "under the root" is confusing and misleading. The correct term is "base", as in "The base of the log is e".

3. What is meant to be assumed by the multiplication symbol here:

- Jan 4th 2009, 06:18 PMlucifer_x
yes i know i misread my homework and put ln instead of the log, that was my bad but everyone makes mistakes.

but yes that was the right question

http://www.mathhelpforum.com/math-he...491bcba7-1.gif - Jan 4th 2009, 06:31 PMmr fantastic
I'm not talking about ln versus log or whatever. I'm talking about your incorrect use of the multiplication symbol and your incorrect use of the expression "under the root".

And as I have already said, the expression http://www.mathhelpforum.com/math-he...491bcba7-1.gif has no real value.

Are you attempting to find the non-real value? If so, then you should have said so in the first place. Otherwise, there's nothing more to be said. - Jan 4th 2009, 07:17 PMlucifer_x
i really dont know what my teacher wants all he put was

solve http://www.mathhelpforum.com/math-he...491bcba7-1.gif

and thats it, i will probably put no real values