The Student Room Group

Logarithms expontential growth decay help

Someone pls help me with this question I can’t figure it out because we haven’t being given either k or a
B7E2EE57-BB20-4DA1-B009-FBE971D8FCC4.jpg.jpeg
Original post by Halpme9
Someone pls help me with this question I can’t figure it out because we haven’t being given either k or a


You're told the value of the car when it's new, i.e. when t=0.

What does k0k^0 equal?
Reply 2
Original post by ghostwalker
You're told the value of the car when it's new, i.e. when t=0.

What does k0k^0 equal?


ah ok its 1 so 12499 = A1 meaning A = 12499 can i apply this to my equation 7000 = AK^36 meaning 7000= 12499K^36
Original post by Halpme9
ah ok its 1 so 12499 = A1 meaning A = 12499 can i apply this to my equation 7000 = AK^36 meaning 7000= 12499K^36


Yes.
Reply 4
Original post by ghostwalker
Yes.

ok thanks after going through with (ii) i didnt get the right answer given what have i done wrong?
7000= 12499k^36
log7000 = log12499k^36
log7000 = 36log12499k
log7000 / 36log12499 = k
Original post by Halpme9
ok thanks after going through with (ii) i didnt get the right answer given what have i done wrong?
7000= 12499k^36
log7000 = log12499k^36
log7000 = 36log12499k
log7000 / 36log12499 = k


Forgot your rules of logs

logAkt=logA+tlogk\log Ak^t= \log A +t\log k

Corrected.
(edited 6 years ago)
Reply 6
Original post by ghostwalker
Forgot your rules of logs

logAkt=logA+klogt\log Ak^t= \log A +k\log t


ok i did actually try this but i got lost when i my answer ended up being
k = log 7000 - log 12499 / 36log
i dont know how to divide by 36log?
Original post by Halpme9
ok i did actually try this but i got lost when i my answer ended up being
k = log 7000 - log 12499 / 36log
i dont know how to divide by 36log?


Corrected my previous post.

So,

7000=12499×k367000=12499 \times k^{36}

Rearrange and taking logs:

log7000log12499=36logk\log 7000 - \log 12499 = 36 \log k

And logk=log7000log1249936\log k = \dfrac{\log 7000-\log 12499}{36}

Work it out and then take the antilog.

Quick Reply

Latest