Thread: Word Problem Exponential Equations /Value

1. Word Problem Exponential Equations /Value

Painting that was purchased 20 years ago for $5000 is now worth$23,305. Calculate how much the painting will be worth 30 years from now?

2. Re: Word Problem Exponential Equations /Value

Originally Posted by tdotodot
Painting that was purchased 20 years ago for \$5000 is now worth \$23,305. Calculate how much the painting will be worth 30 years from now?
$V = V_0e^{kt}$

$23305 = 5000e^{20k}$

solve for $k$, then use that value for $k$ to determine the value when $t = 50$

3. Re: Word Problem Exponential Equations /Value

sorry this is my sisters high school question I'm trying to help her with.
how do i solve for k, without e?

4. Re: Word Problem Exponential Equations /Value

e is Euler's constant , e=2.718. The symbol 'e' should be somewhere on your calculator

5. Re: Word Problem Exponential Equations /Value

Originally Posted by tdotodot
sorry this is my sisters high school question I'm trying to help her with.
how do i solve for k, without e?
if that is the case, your sister should know something about using logarithms to solve for $k$ in an exponential equation

6. Re: Word Problem Exponential Equations /Value

k=9.62698x10^-9 ?

7. Re: Word Problem Exponential Equations /Value

Originally Posted by tdotodot
k=9.62698x10^-9 ?
not even close

8. Re: Word Problem Exponential Equations /Value

why don't you have your sister show how she solved for k?

can't fix it if I don't know where the mistake(s) are ...

9. Re: Word Problem Exponential Equations /Value

Originally Posted by tdotodot
sorry this is my sisters high school question I'm trying to help her with.
how do i solve for k, without e?
Well, if e doesn't enter the picture, then it's regular compounding.
Tried annual : 5000(1 + i)^20 = 23305
Solve for i : you'll get .080000496512...
So "seems" teacher set that up so you'll get 8% annual compounding.

If Sis can't wrap up with that, well....

10. Re: Word Problem Exponential Equations /Value

Originally Posted by skeeter
why don't you have your sister show how she solved for k?

can't fix it if I don't know where the mistake(s) are ...
She came to me because she did not get the question. It's been a good five years since I've seen this stuff.

Originally Posted by DenisB
Well, if e doesn't enter the picture, then it's regular compounding.
Tried annual : 5000(1 + i)^20 = 23305
Solve for i : you'll get .080000496512...
So "seems" teacher set that up so you'll get 8% annual compounding.

If Sis can't wrap up with that, well....
That would be getting into finance, i asked her and thats her next subject. Is there not a generic formula which can be used to directly solve this type of problem?

I know people don't like to give answers out and like to let the posters think for themselves... but I'm asking for someone to explain this to me, so i can explain it to her.

11. Re: Word Problem Exponential Equations /Value

I provided the general formula for continuous exponential growth in post #2. If this is the formula she is to use, then she should be familiar with it. So, the question is, what has she learned about exponential growth? Kind of difficult to proceed without some additional background information. Does she recall using any formulas in class?

12. Re: Word Problem Exponential Equations /Value

If it's the original case, to solve for k, we take the natural logarithm of each side of the equation to get:

$ln(23305) = ln(5000e^{20k})$

Using the properties of logarithms, when multiplication of two arguments is the same as adding the ln of each argument.

$ln(23305) = ln(5000) + ln(e^{20k})$

Now, we can simplify Ln(e^{20k}) by bring down the exponent and multiply it by ln (e). What is Ln(e) = ???
Spoiler:
ln(e)=1

$ln(23305) = ln(5000) + (20k)ln(e)$

$ln(23305) = ln(5000) + (20k) \implies k = \frac{ ln(23305) - ln(5000)}{ 20} = 0.07696$

13. Re: Word Problem Exponential Equations /Value

Well, because of the "nice" 8% result from assuming annual compounding,
I'm bettin' ya'll a Canadian Toonie (thassa 2 dollar coin worth ~1.52US)
that this is what the teacher expects.

And looks like the "sister" routine is a ploy to convince us to give you full solution.
If not, then your sister was looking at the muscular boy in the seat next to hers
when the teacher was teaching....

14. Re: Word Problem Exponential Equations /Value

Originally Posted by tdotodot
Painting that was purchased 20 years ago for $5000 is now worth$23,305. Calculate how much the painting will be worth 30 years from now?
If every year the value of the painting increases by a certain rate "r", then every year:

\displaystyle \begin{align*} V_1 &= V_0 + r\,V_0 \\ &= V_0 \left( 1 + r \right) \\ \\ V_2 &= V_1 + r\,V_1 \\ &= V_1 \left( 1 + r \right) \\ &= V_0 \left( 1 + r \right) \left( 1 + r \right) \\ &= V_0 \left( 1 + r \right) ^2 \\ \\ V_3 &= V_2 + r\,V_2 \\ &= V_2 \left( 1 + r \right) \\ &= V_0 \left( 1 + r \right) ^2 \left( 1 + r \right) \\ &= V_0 \left( 1 + r \right) ^3 \\ \\ V_n &= V_0 \left( 1 + r \right) ^n \end{align*}

So from this we can find

\displaystyle \begin{align*} 23\,305 &= 5000 \left( 1 + r \right) ^{20} \\ \frac{23\,305}{5000} &= \left( 1 + r \right) ^{20} \\ \frac{4661}{1000} &= \left( 1 + r \right) ^{20} \\ \left( \frac{4661}{1000} \right) ^{\frac{1}{20}} &= 1 + r \end{align*}

In 30 years it will be \displaystyle \begin{align*} V_{50} \end{align*}, (30 years after the 20th year) so that means

\displaystyle \begin{align*} V_{50} &= 5000 \left[ \left( \frac{4661}{1000} \right) ^{\frac{1}{20}} \right] ^{50} \\ &= 5000 \left( \frac{4661}{1000} \right) ^{\frac{5}{2}} \\ &\approx 234\,513.45 \end{align*}

15. Re: Word Problem Exponential Equations /Value

Originally Posted by Prove It
In 30 years it will be \displaystyle \begin{align*} V_{50} \end{align*}, (30 years after the 20th year) so that means

\displaystyle \begin{align*} V_{50} &= 5000 \left[ \left( \frac{4661}{1000} \right) ^{\frac{1}{20}} \right] ^{50} \\ &= 5000 \left( \frac{4661}{1000} \right) ^{\frac{5}{2}} \\ &\approx 234\,513.45 \end{align*}
...which is 8% cpd annually....to be precise: 8.00004965121881505....