To determine the time [tex]\( t \)[/tex] required for an investment to double when it is earning interest at a rate of 10%, compounded continuously, we can use the formula for continuous compounding:
[tex]\[ P_t = P_0 e^{rt} \][/tex]
Here:
- [tex]\( P_t \)[/tex] is the amount after [tex]\( t \)[/tex] years.
- [tex]\( P_0 \)[/tex] is the initial amount.
- [tex]\( r \)[/tex] is the interest rate.
- [tex]\( t \)[/tex] is the time in years.
- [tex]\( e \)[/tex] is the base of the natural logarithm.
Our goal is to find the time [tex]\( t \)[/tex] when the investment doubles, which means [tex]\( P_t = 2P_0 \)[/tex]. We substitute [tex]\( P_t \)[/tex] and [tex]\( r \)[/tex] into the formula:
[tex]\[ 2P_0 = P_0 e^{0.10t} \][/tex]
We can divide both sides by [tex]\( P_0 \)[/tex] to simplify:
[tex]\[ 2 = e^{0.10t} \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm (ln) of both sides:
[tex]\[ \ln(2) = \ln(e^{0.10t}) \][/tex]
Using the property of logarithms that [tex]\( \ln(e^x) = x \)[/tex]:
[tex]\[ \ln(2) = 0.10t \][/tex]
Now, solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(2)}{0.10} \][/tex]
The value of [tex]\( \ln(2) \)[/tex] is approximately 0.693. Therefore:
[tex]\[ t = \frac{0.693}{0.10} \approx 6.931 \][/tex]
Rounding this value to the nearest whole number gives:
[tex]\[ t \approx 7 \][/tex]
So, the time it takes for the investment to double is approximately
[tex]\[ \boxed{7} \text{ years} \][/tex]
