Story problem

Question

A baseball card sold for $248 in 1975 and was sold again in 1985 for $460. Assume that the growth in the value V of the collector's item was exponential.

a)Find the value k of the exponential growth rate. Assume V=248

b)Find the exponential growth function interms of t, where t is the number of years since 1975.

c) Estimate the value of the baseball card in 2012.

d)What is the doubling time for the value of the baseball card to the nearest tenth of a year?

e)Find the amount of time after which the value of the baseball card will be $3095

Answer 1

OK. We start with V = V₀ e^kt. Here V₀ = 248.

That establishes the $248 price of the card in 1975 (year zero).

 

Ten years later, in 1985,

460 = 248 e^10k or k = 0.615/10 = 0.0615

 

In 2012, when t = 37,

V = 248 e^(0.0615)(37) or V = $2413 and change

 

To find the doubling time,

496 = 248e^0.0615t or t = 11.3 years

 

And when V = 3095,

3095 = 248e^0.0615t or t = 41 years (this year!)

Cheers!

 

 

 

Answer 2

Good evening, Nick.

 

I always tell my students that these are "two-pass" problems.  On the first pass, your goal is to find A.  On the second pass, your goal is to find k.  Once you have these, you have everything you need to answer all of the questions.

 

To start, let the first year, 1975, be t=0.

And when t=0 the baseball card sold for $248, this is your y-value or "V" as the problem states it.

 

Write down the "skeleton" equation:  V=Ae^kt  

 

Plug in V = 248 and t = 0.

 

248 = A e^(k)(0)

 

248 = A e⁰

 

248 = A(1)

 

So A = 248.  You have completed the first pass.

 

Start over and write the "skeleton" equation, but now you know A, so use it.

 

V=248e^kt

 

For the second pass, our goal is to find k.

 

Use the second pieces of information:  1985 it sold for $460.

 

If 1975 is t=0, then 1985 is t=10.

 

Plug in 460 for V and 10 for t

 

406 = 248 e^(k)(10)

 

Divide both sides by 248

 

1.6371 = e^10k

 

Take the "ln" of both sides to get the "10k" out of the exponent and down to where you can use it.

 

ln(1.6371) = ln(e^10k)

 

Since "ln" and "e^" are inverses of each other, they "undo" each other, leaving

 

0.4929244 = 10k

 

Now just divide both sides by 10, and you will have k.

 

k=0.04929

 

 

Knowing these values, you can write the equation that you will use to solve all of the rest of the questions.  It is

 

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

*                                                        *

*        V = 248e^0.04929t                                *

*                                                        *

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

 

We have already covered (a) and (b)

 

(c)  Find out how many years have passed (since 1975) in 2012,  (2012 - 1975 = 37 years) and use this for "t" in the equation surrounded by "*"s.

 

V = 248e^{(0.04929)(37)}

 

Put this into your calculator and get   V = $1,536.34

 

 

(d)  Doubling time means how long does it take to go from $248 to twice that, $496.  So plug $496 into V and solve for t.

 

 

496 = 248e^0.04929t

 

Divide both sides by 248

 

2 = e^0.04929t

 

"ln" of both sides

 

 

ln(2) = ln(e^0.04929t)

 

0.69315 = 0.04929t

 

 

Divide both sides by 0.04929

 

14.062633 = t

 

 

They say to round to the nearest tenth,

so the answer to (d) is 14.1 years

 

 

 

(e)  Plug 3095 in for V, and again, solve for t.

 

3095 = 248e^0.04929t)

 

Again, divide both sides by 248

 

12.47984 = e^0.04929t

 

"ln" both sides

 

ln(12.47984) = ln(e^0.04929t)

 

2.52411444 = 0.04929t

 

Divide both sides by 0.04929

 

51.2095 = t

 

So answer to (e) is 51.2 years

 

 

 

Hope that helped!  :-)