Question

Use int dt/t from a to ab = L(a) + L(b)

Answer 1

i think he means intigrate 1/t dt from a to b?

Answer 2

I am sorry. Use Dan int dt/t t= a to t= ab = log(a) + log(b)

Answer 3

\[ \int_{a}^{ab} \frac{1}{t} dt = \ln b\]

Answer 4

Scuse agin; int dt/t t= 1 to t = ab

Answer 5

\[ \int_{1}^{ab} \frac{1}{t} dt = \int_{1}^{a} \frac{1}{t} dt + \int_{a}^{ab} \frac{1}{t} dt = \ln a + \ln b \]

Answer 6

\[ \int_{1}^{ab} \frac{1}{t}dt = \ln(ab) - \ln 1 = \ln a + \ln b - 0\]

Answer 7

Get from dt/t from a to ab = log(b). Why does int dt/t from a to ab yield log(b). I know it's the answer; I don't know why. Substitution?

Answer 8

Dr. Jerison at MIT. Shows int dt/t from a to ab by using a change of variable step.

Answer 9

let y = at
dy = a dt
limit a=1, ab=b
\[ \int_{a}^{ab} \frac{1}{t}dt = \int_{1}^{b} \frac{1}{y}dy = \ln b \]

Answer 10

OK, thank you, but what motivates that substitution?

Answer 11

??

Answer 12

How did you know that a substitution was needed, and why y= at? I know it works. What inspires that substitution for this problem?

Answer 13

to set the lower limit to 1, and since we know ln1 = 0, makes things lot easier!!

Answer 14

OK. Thank you again. Tims

Answer 15

If you don't want to substitute
\[ \ln (ab) - \ln (a) = \ln(\frac{ab}{a}) = \ln b\]

Answer 16

What experience idles one need to understand Axiom of Choice?

Answer 17

?? with logs ??

Answer 18

No, the AC is supposed to do for the continuum what mathematical induction does for proving stuff in N.

Answer 19

.... i am not understanding the question

Answer 20

I'll write it down formally later and see if it makes sense to you. I may not be phrasing it correctly. I saw the theorem in a book, Naive Set Theory by Dr. Paul Halmos.

Answer 21

Sorry ... i'm not done algebra and analysis ... stuck on calc 3

Answer 22

Thank you any way. Tims

Answer 23

you are welcome

Answer 24

Is there a way to copy and paste from this site?