Math Problem Statement

Determine the time necessary for dollars to double when it is invested at an interest rate of 8.5%, compounded annually, monthly, daily, and continuously.

Solution

It looks like the question involves determining the time required for an investment to double at an interest rate of 8.5%, compounded in different ways (annually, monthly, daily, and continuously). The correct answer for continuous compounding is already provided as 8.15 years, but the other values for daily, monthly, and annual compounding appear to be incorrect.

To solve these, we can use the following formulas:

  1. Continuous Compounding:
    The formula is A=PertA = P e^{rt}, where AA is the final amount, PP is the principal, rr is the annual interest rate, and tt is the time in years. Since the money doubles, A=2PA = 2P, so: 2=e0.085t2 = e^{0.085t} Taking the natural logarithm on both sides: ln(2)=0.085t\ln(2) = 0.085t Solving for tt: t=ln(2)0.0858.15 yearst = \frac{\ln(2)}{0.085} \approx 8.15 \text{ years}

  2. Compounding nn times per year (for annually, monthly, daily compounding):
    The formula is A=P(1+rn)ntA = P \left( 1 + \frac{r}{n} \right)^{nt}, where nn is the number of times the interest is compounded per year. Again, A=2PA = 2P, so: 2=(1+rn)nt2 = \left( 1 + \frac{r}{n} \right)^{nt} Taking the natural logarithm on both sides and solving for tt, we get: t=ln(2)nln(1+rn)t = \frac{\ln(2)}{n \ln\left( 1 + \frac{r}{n} \right)} We need to apply this for n=1n = 1 (annually), n=12n = 12 (monthly), and n=365n = 365 (daily).

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Math Problem Analysis

Mathematical Concepts

Exponential Growth
Compound Interest

Formulas

Continuous Compounding: A = P e^(rt)
Compounded n times per year: A = P (1 + r/n)^(nt)

Theorems

Natural Logarithm
Compounding Interest Theorem

Suitable Grade Level

Grade 11-12 (Advanced High School or College Level)