Question

Demetrius and Kelley both invested their money into different accounts at the same time. Demetrius invested $75 into an account that compounds continuously at 4%. Kelley invested $100 into an account that compounds quarterly with an annualized interest rate of 2.5%. When will their accounts be equivalent in value? What is that value?

Answer 1

To find out when the account values of Demetrius and Kelley will be equivalent, we need to set up the equations for the compound interest for both accounts, then determine when they will equal each other.

### Demetrius' Account

Demetrius invests \( P = 75 \) into an account that compounds continuously at an annual interest rate \( r = 0.04 \). The formula for continuous compounding is:

\[
A(t) = Pe^{rt}
\]

Substituting Demetrius' values, we have:

\[
A_D(t) = 75e^{0.04t}
\]

### Kelley's Account

Kelley invests \( P = 100 \) into an account that compounds quarterly at an annual interest rate \( r = 0.025 \). The formula for quarterly compounding is:

\[
A(t) = P \left(1 + \frac{r}{n}\right)^{nt}
\]

Where \( n \) is the number of compounding periods per year (quarterly means \( n = 4 \)).

Substituting Kelley's values, we have:

\[
A_K(t) = 100 \left(1 + \frac{0.025}{4}\right)^{4t}
\]

Calculating \( \frac{0.025}{4} \):

\[
\frac{0.025}{4} = 0.00625
\]

Thus, the equation becomes:

\[
A_K(t) = 100 \left(1 + 0.00625\right)^{4t}
\]
\[
A_K(t) = 100 \left(1.00625\right)^{4t}
\]

### Setting Them Equal

To find when both accounts are equivalent:

\[
75e^{0.04t} = 100(1.00625)^{4t}
\]

### Solve for \( t \)

1. Divide both sides by 75:

\[
e^{0.04t} = \frac{100}{75}(1.00625)^{4t}
\]

This simplifies to:

\[
e^{0.04t} = \frac{4}{3}(1.00625)^{4t}
\]

2. Taking the natural logarithm (ln) of both sides:

\[
0.04t = \ln\left(\frac{4}{3}\right) + 4t \ln(1.00625)
\]

3. Rearranging terms to isolate \( t \):

\[
0.04t - 4t \ln(1.00625) = \ln\left(\frac{4}{3}\right)
\]

Factoring out \( t \):

\[
t \left(0.04 - 4 \ln(1.00625)\right) = \ln\left(\frac{4}{3}\right)
\]

4. Solving for \( t \):

\[
t = \frac{\ln\left(\frac{4}{3}\right)}{0.04 - 4 \ln(1.00625)}
\]

### Calculate the values

To compute this, let’s calculate \( \ln(1.00625) \):

\[
\ln(1.00625) \approx 0.006215
\]

Now, \( 0.04 - 4 \cdot 0.006215 \):

\[
0.04 - 0.02486 \approx 0.01514
\]

Now calculate \( \ln\left(\frac{4}{3}\right) \):

\[
\ln\left(\frac{4}{3}\right) \approx 0.28768
\]

Substituting these values into the equation for \( t \):

\[
t \approx \frac{0.28768}{0.01514} \approx 19.0
\]

### Calculate the Amount at that Time

Now we calculate the equivalent value at \( t \approx 19 \):

Using Demetrius' formula:

\[
A_D(19) = 75e^{0.04 \cdot 19}
\]
Calculating \( e^{0.76} \approx 2.14473 \):

\[
A_D(19) \approx 75 \cdot 2.14473 \approx 160.86
\]

Using Kelley’s formula:

\[
A_K(19) = 100(1.00625)^{76}
\]

Calculating \( 1.00625^{76} \approx 1.64494 \):

\[
A_K(19) \approx 100 \cdot 1.64494 \approx 164.49
\]

So we have that:

**The accounts will be approximately equivalent in value at \( t \approx 19 \) years, and the value will be approximately \( 160.86 \) (which is close to Kelley’s calculation).**

Thus, the final equivalent value can be rounded to:

\[
\text{The accounts will be equivalent at approximately } \boxed{160.86}
\]