# Thread: exponential functions and depreciation

1. ## exponential functions and depreciation

A car depreciates at a rate of 30% per year. How much will a $20,000 car be worth after 5 years? 2. Originally Posted by math123456 A car depreciates at a rate of 30% per year. How much will a$20,000 car be worth after 5 years?
In the problem you are given the rate of decrease.

Since the decay equation takes on the form $\displaystyle P=P_0e^{-kt}$, we can take note that $\displaystyle k=.3$, $\displaystyle P_0=20000$ and $\displaystyle t=5$. Plugging this into the equation, we have $\displaystyle P=20000e^{-.3(5)}\approx 4462.60$

So after 5 years, the car will be worth about $4,462.60 Does this make sense? 3. I would use the model$\displaystyle V=A.P^t$where V = current value, A = initial value, P = percentage change and t = time. Therefore$\displaystyle V=20000\times (1-.3)^t\displaystyle V=20000\times .7^t$then make t = 5$\displaystyle V=20000\times .7^5= 3361.40$4. Originally Posted by Chris L T521 In the problem you are given the rate of decrease. Since the decay equation takes on the form$\displaystyle P=P_0e^{-kt}$, we can take note that$\displaystyle k=.3$,$\displaystyle P_0=20000$and$\displaystyle t=5$. Plugging this into the equation, we have$\displaystyle P=20000e^{-.3(5)}\approx 4462.60$So after 5 years, the car will be worth about$4,462.60

Does this make sense?
I don't want to pick at you but in my opinion the equation you used is only applicable for rates of decrease smaller than 10%.

With the given data you know that the car's value after 1 year is $14.000. If you now calculate the constant k you'll get:$\displaystyle k\approx -0.356675$Therefore your result will be too large. 5. Originally Posted by earboth I don't want to pick at you but in my opinion the equation you used is only applicable for rates of decrease smaller than 10%. With the given data you know that the car's value after 1 year is$14.000. If you now calculate the constant k you'll get: $\displaystyle k\approx -0.356675$

Therefore your result will be too large.
Thank you for pointing that out to me. I didn't notice that.

6. Originally Posted by earboth
I don't want to pick at you but in my opinion the equation you used is only applicable for rates of decrease smaller than 10%.
Earboth, what is the reason for this?

7. Originally Posted by pickslides
Earboth, what is the reason for this?
If you use Chris L T 521's formula:

$\displaystyle A = A_0 \cdot e^{k \cdot t}$

then $\displaystyle k = \ln(1-p)$ where p is the percentage of decrease.

Now compare:

decrease by 2% ====> ln(0.98) = -0.020202...
decrease by 5% ====> ln(0.95) = -0.051293...
decrease by 10% ====> ln(0.9) = -0.10536...
...
decrease by 20% ====> ln(0.8) = -0.22314...

You see that the approximation of k = -p is only sufficient for p < 10%.

8. Hello, math123456!

A car depreciates at a rate of 30% per year.
How much will a $20,000 car be worth after 5 years? Each year the car is worth only 70% of its value during the previous year. . . Its value after$\displaystyle n$years is: .$\displaystyle V \:=\:20,\!000(0.7)^n$In 5 years, its value is: .$\displaystyle V \;=\;20,\!000(0.7)^5 \;=\;\$3,\!361.40$

9. Seeing the way that Chris approached the problem,
I am obligated to ask: Is someone teaching that the $\displaystyle e$-formula
. . is the only way to handle depreciation?
If so, they are doing a great disservice.

While $\displaystyle P \:=\:P_oe^{-kt}$ is highly applicable to decay and population problems,
I feel that it requires more work when applied to depreciation problem,
. . especially when they give us the depreciation rate.

Consider the original problem . . .
. . Initial value: $20,000 . . Depreciation rate: 30% (annually) We start with: .$\displaystyle P \:=\:P_oe^{-kt}$When$\displaystyle t = 0,\:P = 20,\!000\!:\quad20,\!000 \:=\:P_oe^0 \quad\Rightarrow\quad P_o = 20,\!000$. . Hence, the function (so far) is: .$\displaystyle P \;=\;20,\!000\,e^{-kt}$When$\displaystyle t = 1,\;P = 14,\!000\!:\quad 14,\!000 \:=\:20,\!000e^{-k} \quad\Rightarrow\quad e^{-k} \:=\:0.7$. .$\displaystyle -k \:=\:\ln(0.7) \quad\Rightarrow\quad k \:=\:-\ln(0.7) \:=\:-0.356674944$Hence, the function is: .$\displaystyle \boxed{P \;=\;20,\!000\,e^{-0.356674944t}} $This equation is correct . . . I have no complaint. But simplify the equation: .$\displaystyle P \;=\;20,\!000\left(e^{-0.356674944}\right)^t $Since$\displaystyle e^{-0.356674944} \:=\:0.7$, we have: .$\displaystyle \boxed{P \;=\;20,\!000\,(0.7)^t} $We had this equation way back then, just "eyeballing" the problem. 10. Originally Posted by earboth If you use Chris L T 521's formula:$\displaystyle A = A_0 \cdot e^{k \cdot t}$then$\displaystyle k = \ln(1-p)$where p is the percentage of decrease. Now compare: decrease by 2% ====> ln(0.98) = -0.020202... decrease by 5% ====> ln(0.95) = -0.051293... decrease by 10% ====> ln(0.9) = -0.10536... ... decrease by 20% ====> ln(0.8) = -0.22314... You see that the approximation of k = -p is only sufficient for p < 10%. I see the error is becomes quite significant, so why isn’t my solution (when given the depreciation level) always considered the better?$\displaystyle V=20000\times (1-.3)^t\displaystyle V=20000\times .7^t$then make t = 5$\displaystyle V=20000\times .7^5= 3361.40$11. Originally Posted by pickslides I see the error is becomes quite significant, so why isn’t my solution (when given the depreciation level) always considered the better?$\displaystyle V=20000\times (1-.3)^t\displaystyle V=20000\times .7^t$then make t = 5$\displaystyle V=20000\times .7^5= 3361.40$The difficulty lies in the word "always" ... With the given problem the variable t is a discret number, that means actually you are dealing with a geometric sequence:$\displaystyle a_n=20000 \cdot \left(\dfrac7{10} \right)^n\$

If a value is increasing or decreasing continuously your method can't be applied, then you have to use the equation which was posted by ChrisLT521

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### exponential depreciation

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