how do you do 5a i)??? (1 Viewer)

[jamesfirst]

I got -Ln (5/3)


.... I just absolute valued it because I didn't know what to do... Can I still get 1 mark ?

 

[D94]

I got -Ln (5/3)


.... I just absolute valued it because I didn't know what to do... Can I still get 1 mark ?

No, you can't just absolute value it; it actually works out right. Depends on the working, but you should get 1 out of 2 for it.

 

[barbernator]

you should have got -ln(3/5), which changes to ln((3/5)^-1) which = ln(5/3)

 

[damo_g]

Do you mean 5b. i)?

k = -ln(3/5)
= -(ln(3) - ln(5))
= ln(5) - ln(3)
= ln(5/3)

 

[jamesfirst]

Do you mean 5b. i)?

k = -ln(3/5)
= -(ln(3) - ln(5))
= ln(5) - ln(3)
= ln(5/3)

wtf....


WTFFFFFFFFFFFFFF IT WAS 3/5?!?!?!?!?!?! FUCK!!!!!!!!!!!!!!!!!!!!!!!!!!!!

 

[Nympha]

wtf....


WTFFFFFFFFFFFFFF IT WAS 3/5?!?!?!?!?!?! FUCK!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Well, yeah, they told you to prove that k = ln5/3.
Note that, as others have said above, -ln3/5=ln5/3.

 

[jamesfirst]

I know that...


It's because I started to panic after I got stuck on parametrics...