Logarithms/ Pe^(rt): note pays $50K in 5.5 yrs; find price

[Gretta99]

My book gave me this problem.

A promissory note will pay $50,000 at maturity 5.5 years from now. How much should you pay for the note now if the note gains value at a rate of 10% compounded continuously?

I put it into the Pe^(rt) formula.
50000=Pe^(.1*5.5)
50000=Pe^(5.5)
ln(50000)=ln (Pe^5.5)
ln(50000)=5.5 +ln (P)
10.8198=5.5 +ln (P)
10.2698=ln(P)

I'm not sure where to go from here, or if I even did it right. Please help soon!!!

Michelle

 

[Deleted member 4993]

My book gave me this problem.

A promissory note will pay $50,000 at maturity 5.5 years from now. How much should you pay for the note now if the note gains value at a rate of 10% compounded continuously?

I put it into the Pe^(rt) formula.
50000=Pe^(.1*5.5)
50000=Pe^(5.5)
ln(50000)=ln (Pe^5.5)
ln(50000)=5.5 +ln (P)
10.8198=5.5 +ln (P)
10.2698=ln(P)

I'm not sure where to go from here, or if I even did it right. Please help soon!!!

Michelle

What did you need to find?

What is "P"?

 

[stapel]

A promissory note will pay $50,000 at maturity 5.5 years from now. How much should you pay for the note now if the note gains value at a rate of 10% compounded continuously?

You are assuming that you can get 10%, continuously compounded, on some investment. You have the option to buy a note with a certain fixed value at a certain future time. You are being asked the "value" (the price above which you should not pay) of the note, under the stated assumptions.

You are given that the note will have a value of $50,000 in 5.5 years. Viewing this note as an investment, and using the given interest rate, what must the current "value" be?

I put it into the Pe^(rt) formula.

I will guess that you actually mean the formula "A = Pe[sup:rtt8576d]rt[/sup:rtt8576d]", where "A" is the ending balance, "P" is the initial investment, "r" is the interest rate (usually "annual"), and "t" is the time (usually in years). If so, then plug in the given values, and solve. :idea:

Hint: The investment is worth more than $10.27. :wink:

Eliz.

 

[Denis]

> 50000=Pe^(.1*5.5)
> 50000=Pe^(5.5)
NO; .1 * 5.5 = .55

Easier if next step is: P = 50000 / e^(.55) ; you should get in vicinity of 30000