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?
She came to me because she did not get the question. It's been a good five years since I've seen this stuff.
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.
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?
If it's the original case, to solve for k, we take the natural logarithm of each side of the equation to get:
$\displaystyle ln(23305) = ln(5000e^{20k})$
Using the properties of logarithms, when multiplication of two arguments is the same as adding the ln of each argument.
$\displaystyle ln(23305) = ln(5000) + ln(e^{20k})$
Now, we can simplify Ln(e^{20k}) by bring down the exponent and multiply it by ln (e). What is Ln(e) = ???Spoiler:
$\displaystyle ln(23305) = ln(5000) + (20k)ln(e)$
$\displaystyle ln(23305) = ln(5000) + (20k) \implies k = \frac{ ln(23305) - ln(5000)}{ 20} = 0.07696$
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....
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
$\displaystyle \begin{align*} V_{50} &= 5000 \left[ \left( \frac{4661}{1000} \right) ^{\frac{1}{20}} \right] ^{50} \\ &= 5000 \left( \frac{4661}{1000} \right) ^{\frac{5}{2}} \\ &\approx 234\,513.45 \end{align*}$