Peo needs R30,835.42 to buy a second-hand trailer. Approximately how long will it take him to save towards this amount if he deposits R25,000 now into a savings account earning 10.5% interest per year compounded weekly?



Answer :

To determine how long it will take for Peo to save R30,835.42 by depositing R25,000 into a savings account with an annual interest rate of 10.5% compounded weekly, we use the formula for compound interest:

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

where:
- [tex]\( A \)[/tex] is the future value of the investment (R30,835.42),
- [tex]\( P \)[/tex] is the principal amount (R25,000),
- [tex]\( r \)[/tex] is the annual interest rate (10.5%, or 0.105 as a decimal),
- [tex]\( n \)[/tex] is the number of times the interest is compounded per year (52 for weekly),
- [tex]\( t \)[/tex] is the number of years the money is invested.

We need to solve for [tex]\( t \)[/tex]. Starting with the compound interest formula:

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

We rearrange the formula to solve for [tex]\( t \)[/tex]:

1. Divide both sides by [tex]\( P \)[/tex]:

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

2. Take the natural logarithm (logarithm base e) of both sides:

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

3. Apply the logarithm power rule [tex]\( \ln(x^y) = y \ln(x) \)[/tex]:

[tex]\[ \ln\left(\frac{A}{P}\right) = nt \cdot \ln\left(1 + \frac{r}{n}\right) \][/tex]

4. Solve for [tex]\( t \)[/tex]:

[tex]\[ t = \frac{\ln\left(\frac{A}{P}\right)}{n \cdot \ln\left(1 + \frac{r}{n}\right)} \][/tex]

Now we plug in the values:
- [tex]\( A = 30,835.42 \)[/tex]
- [tex]\( P = 25,000 \)[/tex]
- [tex]\( r = 0.105 \)[/tex]
- [tex]\( n = 52 \)[/tex]

[tex]\[ t = \frac{\ln\left(\frac{30,835.42}{25,000}\right)}{52 \cdot \ln\left(1 + \frac{0.105}{52}\right)} \][/tex]

Using the values provided, we calculate:

[tex]\[ \frac{30,835.42}{25,000} = 1.2334168 \][/tex]

[tex]\[ \ln(1.2334168) \approx 0.210721 \][/tex]

[tex]\[ 1 + \frac{0.105}{52} = 1.002019 \][/tex]

[tex]\[ \ln(1.002019) \approx 0.002016 \][/tex]

[tex]\[ 52 \cdot 0.002016 = 0.104832 \][/tex]

Finally:

[tex]\[ t = \frac{0.210721}{0.104832} \approx 2.009603 \][/tex]

Rounding to a more reasonable precision, Peo will need approximately 2 years to reach his goal of R30,835.42 with an initial deposit of R25,000 in a savings account earning 10.5% annual interest compounded weekly.