Question

calc problem

Answer 1

to begin with, I have a=e^2 & c=e^5 ? not sure if that is correct

Answer 2

thats correct keep going

Answer 3

For a, I have ln |e^7| =7 ?

Answer 4

your close but you need to evaluate it at 1 also

\[\int\limits_{1}^{e^7} \frac{ 1 }{ t }dt = \ln(e^7) - ?\]

Answer 5

well actaully yea nvm it goes to zero so your right its 7 haha

Answer 6

hehe, oh cool! for some reason i was stumped but now i see what i was doing wrong :)

Answer 7

haha np the other parts are pretty much the same process

Answer 8

yes now i understand! Thank you for responding anyway :)

Answer 9

depending on your definition, sometimes log is defined as
\[\ln(x)=\int_1^x\frac{dt}{t}\]

Answer 10

oh yes @satellite73 I believe thats the same formula i have here in my notes! I didn't see the "ln" in (ln a)=2 so i thought a was just 2!