To determine how long it will take for a [tex]$2000 investment to grow to $[/tex]3424 at an annual interest rate of 7.8%, compounded monthly, follow these steps:
### Step 1: Identify the given values
- Initial investment ([tex]\(P\)[/tex]): \[tex]$2000
- Final amount (\(A\)): \$[/tex]3424
- Annual interest rate ([tex]\(r\)[/tex]): 7.8% or 0.078 in decimal form
- Number of times interest is compounded per year ([tex]\(n\)[/tex]): 12 (since it is compounded monthly)
### Step 2: Write the formula for compound interest
The compound interest formula is
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\(A\)[/tex] is the amount of money accumulated after n years, including interest.
- [tex]\(P\)[/tex] is the principal amount (the initial sum of money).
- [tex]\(r\)[/tex] is the annual interest rate (decimal).
- [tex]\(n\)[/tex] is the number of times that interest is compounded per year.
- [tex]\(t\)[/tex] is the number of years the money is invested for.
### Step 3: Rearrange the formula to solve for [tex]\(t\)[/tex]
We need to solve the equation for [tex]\(t\)[/tex]:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
First, divide both sides by [tex]\(P\)[/tex]:
[tex]\[ \frac{A}{P} = \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Next, take the natural logarithm (ln) of both sides to get:
[tex]\[ \ln \left(\frac{A}{P}\right) = \ln \left[\left(1 + \frac{r}{n}\right)^{nt}\right] \][/tex]
By the properties of logarithms, this is equivalent to:
[tex]\[ \ln \left(\frac{A}{P}\right) = nt \cdot \ln \left(1 + \frac{r}{n}\right) \][/tex]
Now, solve for [tex]\(t\)[/tex]:
[tex]\[ t = \frac{\ln \left(\frac{A}{P}\right)}{n \cdot \ln \left(1 + \frac{r}{n}\right)} \][/tex]
### Step 4: Plug in the given values and calculate [tex]\(t\)[/tex]
- [tex]\(P = 2000\)[/tex]
- [tex]\(A = 3424\)[/tex]
- [tex]\(r = 0.078\)[/tex]
- [tex]\(n = 12\)[/tex]
[tex]\[ t = \frac{\ln \left(\frac{3424}{2000}\right)}{12 \cdot \ln \left(1 + \frac{0.078}{12}\right)} \][/tex]
[tex]\[ t = \frac{\ln (1.712)}{12 \cdot \ln \left(1 + 0.0065\right)} \][/tex]
[tex]\[ t = \frac{\ln (1.712)}{12 \cdot \ln (1.0065)} \][/tex]
Using logarithm values:
[tex]\[ \ln (1.712) \approx 0.537 \][/tex]
[tex]\[ \ln (1.0065) \approx 0.00649 \][/tex]
So:
[tex]\[ t = \frac{0.537}{12 \cdot 0.00649} \][/tex]
[tex]\[ t = \frac{0.537}{0.07788} \][/tex]
[tex]\[ t \approx 6.915484528589429 \][/tex]
### Step 5: Round the final answer to the nearest hundredth
[tex]\[ t \approx 6.92 \][/tex]
Therefore, it will take approximately 6.92 years for the [tex]$2000 investment to grow to \( \$[/tex]3424 \) at an annual interest rate of 7.8%, compounded monthly.
