What Was the Rate of Return on the Picasso Painting Sold in 1995?

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Homework Help Overview

The discussion centers around calculating the rate of return on a Picasso painting purchased in 1903 for $600 and sold in 1995 for $29,152,000. The subject area involves financial mathematics, specifically the application of compound interest formulas.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of the single payment compound interest formula and discuss the implications of logarithmic calculations in isolating the interest rate. Questions arise regarding the base of the logarithm used and the interpretation of the variables involved.

Discussion Status

Some participants have suggested alternative methods for isolating the interest rate, such as taking the n-th root instead of using logarithms. There is an ongoing examination of the assumptions regarding the compounding period, with some participants questioning whether the painting's value was compounded annually or only once at the time of sale.

Contextual Notes

There is uncertainty regarding the correct interpretation of the compounding period (n), with discussions indicating that it may not be as straightforward as initially assumed. Additionally, confusion exists around the conversion of percentage rates and their application in the original formula.

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Homework Statement



In 1903, a Picasso painting was purchased for $600. The family of the original owner sold the painting in 1995 for $29,152,000. What rate of return (interest) did the family receive on the investment?

Homework Equations



Single Payment Compound Interest Formula:

F = P(1+i)n

where,

F= a future some of money (future value)
P= a present sum of money
n= number of interest periods
i = interest rate per interest period.

The Attempt at a Solution



solve for i:

F = P(1+i)n
F/P = (1+i)n
log(F/P) = n log (1+i)

This is as far as I get. I know that n log (1+i) does not equal n log 1 + n log i, but I don't know what to do to isolate the i. I can't just divide both sides by n log because there is no such thing as n log. I thought about moving n log (1+i) to the left side and setting the equation to zero, but I didn't get anywhere with that either.

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
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What is the base of "log"? When I look at it, I assume common log (base 10), but IIRC in other places the natural log (base e) is assumed. Whatever it is, you can raise a number to both sides. In other words,
[tex]x = y \rightarrow a^x = a^y[/tex]
The number you choose for a should probably be the base of the logarithm you were using earlier. :wink:
 
Last edited by a moderator:
eumyang said:
What is the base of "log"? When I look at it, I assume common log (base 10), but IIRC in other places the natural log (base e) is assumed. Whatever it is, you can raise a number to both sides. In other words,
[tex]x = y \rightarrow a^x = a^y[/tex]
The number you choose for a should probably be the base of the logarithm you were using earlier. :wink:


69

The problem says nothing about the base, so I can only assume it is base 10. So then, if I understand you correctly, I would end up with:

10log(F/P) = 10n log(1+i)

Please excuse my ignorance, but I've always had a problem grasping logs.

so, if the above is correct, then I should be able to use the alog u = u property and get:

F/P = (1+i)

i = (F/P) - 1

Is this correct?
 
Last edited:
nope. That's not correct. That gives me an answer of 48,585.67. Seems a bit high for an interest rate. Plus it doesn't work when I plug the numbers into the original formula.
 
When you get to this step,
F/P = (1+i)n,
instead of taking the log of both sides, take the n-th root of both sides (n is known). That will isolate 1 + i.
 
Mark44 said:
When you get to this step,
F/P = (1+i)n,
instead of taking the log of both sides, take the n-th root of both sides (n is known). That will isolate 1 + i.

You are right, n is known. Even so, if I am doing this right I still end up with i = F/P - 1

(F/P)92 = (1+i)92

i = (F/P) - 1

= ($29,152,000/$600) - 1

= 48,586.67%

Ok, this works out if I take $600 x 48,586.67%, however, when I try to plug $48586.67% into the original F=P(1+i)n formula, I get an overflow error when I should get $29,152,000.

F = P(1+i)n

= $600(1+48,586.67%)92 = overflow error
 
I just thought of something. I assumed that n = 1995 - 103 = 92 years. This might be wrong. The painting was not compounded yearly. It was only sold once. Therefore, n=1.

Now it works

F = $600(1+48,586.67%)1
= $29,152,600.00

Thank you everyone for your help.
 
Jim01 said:
You are right, n is known. Even so, if I am doing this right I still end up with i = F/P - 1

(F/P)92 = (1+i)92
No, this isn't right.
Assuming that n = 92 is correct, if you start with F/P = (1 + i)92, and then take the 92nd root (not power) of each side, what do you get?
Jim01 said:
i = (F/P) - 1

= ($29,152,000/$600) - 1

= 48,586.67%

Ok, this works out if I take $600 x 48,586.67%, however, when I try to plug $48586.67% into the original F=P(1+i)n formula,
$48586.67% doesn't make any sense. Is it a dollar amount or a percentage? It can't be both. Note that 48,586.67% = 485.8667 as a decimal value.
Jim01 said:
I get an overflow error when I should get $29,152,000.

F = P(1+i)n

= $600(1+48,586.67%)92 = overflow error
 

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