Logarithm Depreciation Word Problem

[Euler2718]

Homework Statement

A vehicle purchase for $32,000 depreciates at a rate of 75% every 6 years. Another vehicle purchased for $16,000 depreciates at a rate of 50% every 4 years. Create an exponential function for each situation, and use the functions to algebraically determine the amount of time it would take for the vehicles to be equal in value.2. Homework Equations

A = P(1 \pm i )^{n}

A is the future amount
P is the present/principal amount
i = interest rate per compounding period in decimal form
n = number of compounding periods

The Attempt at a Solution

Algebraically:

A_{1} = A_{2}

P_{1}(1-i_{1})^{n} = P_{2}(1-i_{2})^{n}

\log P_{1} + n\log (1-i_{1}) = \log P_{2} + n\log (1-i_{2})

n\log (1-i_{1}) - n\log (1-i_{2}) = \log P_{2} - \log P_{1}

n = \frac{ \log P_{2} - \log P_{1}}{\log (1-i_{1}) - \log (1-i_{2}) }

My issue is I don't know how to use the 6 years and the 4 years. I've done it with semi-annually, annually, quarterly, etc, but this one deals with years > 1 . For instance, if this was semi, it would be (75% / 2) /100 , to get i.

 

[SteamKing]

Homework Statement

A vehicle purchase for $32,000 depreciates at a rate of 75% every 6 years. Another vehicle purchased for $16,000 depreciates at a rate of 50% every 4 years. Create an exponential function for each situation, and use the functions to algebraically determine the amount of time it would take for the vehicles to be equal in value.

I think you have overlooked the key suggestion in the problem statement: Create an exponential function which describes the depreciation in value of each car.
This means that the value of each car is constantly declining, not every six months, not every month, not even every day, but all the time.

Using the standard formulas for compounding interest over discrete time periods does not meet the criterion of establishing an exponential function to describe this process.

You want a function which relates the final value of each car V(t) to its initial value V(0) over a certain time; in other words,

V(t) = V(0) * e^-rt

The rate r at which the value depreciates can be determined from the initial and the final values of each car and the time over which the depreciation occurs.

 

[Euler2718]

I think you have overlooked the key suggestion in the problem statement: Create an exponential function which describes the depreciation in value of each car.
This means that the value of each car is constantly declining, not every six months, not every month, not even every day, but all the time.

Using the standard formulas for compounding interest over discrete time periods does not meet the criterion of establishing an exponential function to describe this process.

You want a function which relates the final value of each car V(t) to its initial value V(0) over a certain time; in other words,

V(t) = V(0) * e^-rt

The rate r at which the value depreciates can be determined from the initial and the final values of each car and the time over which the depreciation occurs.

While I have no doubt that is correct, the given equation is the extent to this curriculum (it's a special word problem in the logarithm chapter). Is there anyway to do it using the standard formula?

 

[SteamKing]

While I have no doubt that is correct, the given equation is the extent to this curriculum (it's a special word problem in the logarithm chapter). Is there anyway to do it using the standard formula?

Not that I am aware of.

You should also remember that logarithms have a special relationship with exponential functions, namely

log (e^x) = x, where "log" stands for the "natural logarithm", and is sometimes written ln (e^x) = x.

 

[certainly]

While I have no doubt that is correct, the given equation is the extent to this curriculum (it's a special word problem in the logarithm chapter). Is there anyway to do it using the standard formula?

You will be able to solve the problem even without logarithms, but since the question wants you to do it exponentially, it is recommended you do so.

My issue is I don't know how to use the 6 years and the 4 years. I've done it with semi-annually, annually, quarterly, etc, but this one deals with years > 1 . For instance, if this was semi, it would be (75% / 2) /100 , to get i.

$i$ would be $\frac{\frac{75}{2}}{100}$ if the rate was 75% annually and compounded semi-anually. What is the time period for the rate in this case ?

 

[certainly]

A vehicle purchase for $32,000 depreciates at a rate of 75% every 6 years

Also note that this is a clear indication that the period of compounding is 6 years, not 1. Therefore you must count $n$ with multiplicity.
[EDIT:- and accordingly for the other case.]

 

[Ray Vickson]

Homework Statement

A vehicle purchase for $32,000 depreciates at a rate of 75% every 6 years. Another vehicle purchased for $16,000 depreciates at a rate of 50% every 4 years. Create an exponential function for each situation, and use the functions to algebraically determine the amount of time it would take for the vehicles to be equal in value.2. Homework Equations

A = P(1 \pm i )^{n}

A is the future amount
P is the present/principal amount
i = interest rate per compounding period in decimal form
n = number of compounding periods

The Attempt at a Solution

Algebraically:

A_{1} = A_{2}

P_{1}(1-i_{1})^{n} = P_{2}(1-i_{2})^{n}

\log P_{1} + n\log (1-i_{1}) = \log P_{2} + n\log (1-i_{2})

n\log (1-i_{1}) - n\log (1-i_{2}) = \log P_{2} - \log P_{1}

n = \frac{ \log P_{2} - \log P_{1}}{\log (1-i_{1}) - \log (1-i_{2}) }

My issue is I don't know how to use the 6 years and the 4 years. I've done it with semi-annually, annually, quarterly, etc, but this one deals with years > 1 . For instance, if this was semi, it would be (75% / 2) /100 , to get i.

One approach is to use the 1-year depreciation rate for each car. For Car 1 (the $32,000 one), say the depreciation is $100\, r_1$ percent per year, while for Car 2 it is $100\, r_2$ per year. Then you need to solve
\frac{32000}{(1+r_1)^n} = \frac{16000}{(1+r_2)^n}.
Do you see how to get $r_1$ and $r_2$ from the information given in the problem?

 

[SteamKing]

One approach is to use the 1-year depreciation rate for each car. For Car 1 (the $32,000 one), say the depreciation is $100\, r_1$ percent per year, while for Car 2 it is $100\, r_2$ per year. Then you need to solve
\frac{32000}{(1+r_1)^n} = \frac{16000}{(1+r_2)^n}.
Do you see how to get $r_1$ and $r_2$ from the information given in the problem?

IMO, I don't think that's the way to solve this problem.

Based on the problem statement, this is an example of continuous compounding with a twist: instead of the principal increasing in value over time, an asset is depreciating.

http://cs.selu.edu/~rbyrd/math/continuous/

The formula for continuous compounding involves the exponential function (e^x) as described in the problem statement as the method of solution. By using a negative exponent, the continuous compounding formula is converted into a continuous depreciation formula.

 

[certainly]

One approach is to use the 1-year depreciation rate for each car.

Hi Ray,
Since this is compound interest, and the period of compounding is 6 years and 4 years respectively, I don't think you will be able to find the depreciation rate for 1 year for compound interest.

The formula for continuous compounding involves the exponential function (ex) as described in the problem statement as the method of solution. By using a negative exponent, the continuous compounding formula is converted into a continuous depreciation formula.

But the problem can still be solved by using simply $A=P(1\pm i)^n$.
Cheers.

 

[Ray Vickson]

IMO, I don't think that's the way to solve this problem.

Based on the problem statement, this is an example of continuous compounding with a twist: instead of the principal increasing in value over time, an asset is depreciating.

http://cs.selu.edu/~rbyrd/math/continuous/

The formula for continuous compounding involves the exponential function (e^x) as described in the problem statement as the method of solution. By using a negative exponent, the continuous compounding formula is converted into a continuous depreciation formula.

In post #3 the OP (sort-of) indicated that the exponential formula was unfamiliar and (more-or-less) outside the coverage of the course up to then. So, while I personally agree with you, I tried to put the method into a form that would be familiar to the OP, hence wrote $1/(1+r)^n$ instead of $e^{-vn}$. But, of course, $e^{-v} = 1/(1+r)$. The remaining issue is whether $n$ turns out to be integer, or not.

 

[Ray Vickson]

Hi Ray,
Since this is compound interest, and the period of compounding is 6 years and 4 years respectively, I don't think you will be able to find the depreciation rate for 1 year for compound interest.

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Finding the 1-year rate is easy if you ASSUME a steady compounding/discounting rate, such as $100 r\%$ per year. For an asset worth $P$ initially, the discounted value after 1 year is $P/(1+r)$. After 2 years it is $P/(1+r)^2$, etc. After $m$ years it is ____? If you fill in the blank you will see how to find $r$ from the given problem information.

***************************

But the problem can still be solved by using simply $A=P(1\pm i)^n$.

Cheers.

 

[Ray Vickson]

Hi Ray,
Since this is compound interest, and the period of compounding is 6 years and 4 years respectively, I don't think you will be able to find the depreciation rate for 1 year for compound interest.

But the problem can still be solved by using simply $A=P(1\pm i)^n$.
Cheers.

Well, if we write $1-i = 1/(1+r)$ they are exactly the same. The issue of finding $r$ is the same as the issue of finding $i$.

 

[SteamKing]

This problem, as stated, does not necessarily require the use of the compound interest formula or even writing an exponential function to solve. Given the present values of the two cars, and knowing by how much each depreciates over a fixed term, one can simply set up a calculation using the stated amount of depreciation and the time involved in order to determine when the two cars reach equal value.

I suspect the purpose of this exercise was not so much for the student to calculate how long it takes for equal value to be obtained, but in how well the student follows directions, especially since the problem statement was quite specific in the manner in which it desired the solution to be obtained.