Question

ln|-2| is a # or DNE?
integral from -inf to 0 of 1/(3x-2) dx

Answer 1

well, the absolute value bars makes everything a positive value, except 0

Answer 2

consider the derivaitve of ln(3x-2)

Answer 3

3(1/3x-2)

Answer 4

yep, and we are missing a 3 so lets divide it off

1/3 ln(3x-2) is our basic function that was derived

Answer 5

yes i got that far, 1/3 times the limit as t approaches negative inf of the integral of t to 0 of ln|3x-2|

Answer 6

ln(x) doesnt converge ... its monotonic increasing

Answer 7

but when i evaluate it. [ ln|3(0)-2) - ln|3t-2| ]

Answer 8

i would get [ln|-2| - 0 ]

Answer 9

yeah, and ln(inf) = inf

Answer 10

ln|-inf| being zero correct?

Answer 11

no, ln(1) = 0

Answer 12

i thought as ln approches negative inf it approches 0 from the right

Answer 13

or rather ln(x) . as x approches neg inf ln approches 0

Answer 14

ln|-inf| = ln|inf| does not approach zero



the derivative tells us that much ... its always positive, and never zero

Answer 15

It kinda sounds like you are getting ln's inverse confused with ln

Answer 16

ohhh yes you are right

Answer 17

on the graph x can never be zero thats why i kept thinking it DNE

Answer 18

notice that we are working with ln|x| which is symmetric about the y axis

Answer 19

@freckles you mean e^x?

Answer 20

thank you guys though

Answer 21

http://www.wolframalpha.com/input/?i=y+%3D+ln%7Cx%7C

Answer 22

so for the record abs(-inf) => +inf

Answer 23

|a| = |-a| so the limits as a to inf is equal to the limits as a to -inf .... which is just inf

Answer 24

its bad form to say inf is a number .... but yeah, you can mentally address it like that

Answer 25

ok then, so my final answer would be -inf since i am -ln|3t-2| t approaching -inf

Answer 26

1/3 (ln(2) - |inf|) looks that way to me or DNE depending on how your material defines it

Answer 27

http://www.wolframalpha.com/input/?i=1%2F3%28ln%7C2%7C-ln%7Cinf%7C%29

Answer 28

thank you

Answer 29

youre welcome :) good luck