c4 last growth/decay question of the night!

Hi guys,

last question on growth/decay now!

2 equations for 2 different cars depreciating value:

1)$v=15000e^{-kt} k=0.155$

2)$w=18000e^{-0.175t}$

If both cars sold as new on the 1st Jan '05, I need to calculate the year they depreciated to the same value.

So, in the two equations above, t is the unknown to be determined.

$15000e^{-kt}=18000e^{-0.175t}$


$\frac {e^{-kt}}{e^{-0.175t}}=1.2$

I'm not sure if at this stage I'm going the right way about it....but this is what I did:

$e^{-kt} \times e^{(-0.175t)-1} = 1.2$

The line above makes me think I have done the wrong thing, because the -t in the second e cancels the first t from the first e.

Could anyone give me a few clues?
Thanks

Hi guys,

last question on growth/decay now!

2 equations for 2 different cars depreciating value:

1)$v=15000e^{-kt} k=0.155$

2)$w=18000e^{-0.175t}$

If both cars sold as new on the 1st Jan '05, I need to calculate the year they depreciated to the same value.

So, in the two equations above, t is the unknown to be determined.

$15000e^{-kt}=18000e^{-0.175t}$


$\frac {e^{-kt}}{e^{-0.175t}}=1.2$

I'm not sure if at this stage I'm going the right way about it....but this is what I did:

$e^{-kt} \times e^{(-0.175t)-1} = 1.2$

The line above makes me think I have done the wrong thing, because the -t in the second e cancels the first t from the first e.

Could anyone give me a few clues?
Thanks

Hi guys,

last question on growth/decay now!

2 equations for 2 different cars depreciating value:

1)$v=15000e^{-kt} k=0.155$

2)$w=18000e^{-0.175t}$

If both cars sold as new on the 1st Jan '05, I need to calculate the year they depreciated to the same value.

So, in the two equations above, t is the unknown to be determined.

$15000e^{-kt}=18000e^{-0.175t}$


$\frac {e^{-kt}}{e^{-0.175t}}=1.2$

I'm not sure if at this stage I'm going the right way about it....but this is what I did:

$e^{-kt} \times e^{(-0.175t)-1} = 1.2$

The line above makes me think I have done the wrong thing, because the -t in the second e cancels the first t from the first e.

Could anyone give me a few clues?
Thanks

I think you need to revise your laws of indices

$\dfrac{1}{e^{-u}} = e^u$

Time for a break?

write 15,000 and 18,000 as e^(ln15000) and e^(ln18000) use indice laws and compare both sides of the equation and rearrange to get t

What would I do with the e terms when I raise 15,000 & 18,000 to e?

a^ba^c=a^(b+c) so you just add the indices once you got everything to the power e on both sides and then you can simply remove the power e or ln both sides and then rearrange to get t.

I meant, if I'm raising the numbers to base e, then doesn't this affect the balance of the equation, so don't I also have to raise the e terms to base e too? Much like if I wanted to divide some terms by something, I'd have to divide all the terms by that something. Maybe I misunderstood your earlier post but I interpreted it as you saying I should raise some terms to base e, but not all of them.