Final answer:
To find out how long $70,000 at 2% interest compounded monthly takes to become $826,000, we use the compound interest formula. The formula is rearranged to solve for time (t) and involves logarithms to find t to the nearest tenth of a year after calculations.
Step-by-step explanation:
The question is asking how long it would take for an investment of $70,000 at an interest rate of 2% compounded monthly to grow to $826,000. To solve this, we use the formula for compound interest:
A = P(1 + \frac{r}{n})^{nt},
where:
- A is the amount of money accumulated after n years, including interest.
- P is the principal amount (the initial amount of money).
- r is the annual interest rate (decimal).
- n is the number of times that interest is compounded per year.
- t is the time the money is invested for, in years.
In this case, we're solving for t and our equation becomes:
826,000 = 70,000(1 + \frac{0.02}{12})^{12t}
First, we divide both sides by 70,000 to isolate the compound interest factor on the right-hand side:
11.8 = (1 + \frac{0.02}{12})^{12t}
Next, we take the natural logarithm of both sides:
ln(11.8) = ln((1 + \frac{0.02}{12})^{12t})
Since ln(a^b) = b*ln(a), we apply this property to simplify:
ln(11.8) = 12t*ln(1 + \frac{0.02}{12})
To solve for t, we divide both sides by 12*ln(1 + \frac{0.02}{12}):
t = \frac{ln(11.8)}{12*ln(1 + \frac{0.02}{12})}
After calculating the above expression using a calculator, we find the time t to the nearest tenth of a year.