To determine how long the money has been invested, we start with the given formula for continuous compounding:
[tex]\[ 200 e^{0.08 t} = 350 \][/tex]
Here, [tex]\(200\)[/tex] is the principal (initial amount), [tex]\(350\)[/tex] is the future value, and [tex]\(0.08\)[/tex] is the annual interest rate. We need to solve for [tex]\( t \)[/tex], the number of years the money has been invested.
Step-by-Step Solution:
1. Divide both sides of the equation by 200 to isolate the exponential term:
[tex]\[ e^{0.08 t} = \frac{350}{200} \][/tex]
2. Simplify the fraction on the right side:
[tex]\[ e^{0.08 t} = 1.75 \][/tex]
3. Take the natural logarithm (ln) on both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(e^{0.08 t}) = \ln(1.75) \][/tex]
4. Use the properties of logarithms to move the exponent down:
[tex]\[ 0.08 t \cdot \ln(e) = \ln(1.75) \][/tex]
Since [tex]\(\ln(e) = 1\)[/tex]:
[tex]\[ 0.08 t = \ln(1.75) \][/tex]
5. Solve for [tex]\( t \)[/tex] by dividing both sides by 0.08:
[tex]\[ t = \frac{\ln(1.75)}{0.08} \][/tex]
6. Use a calculator to find the natural logarithm of 1.75 and perform the division:
[tex]\[ \ln(1.75) \approx 0.5596 \][/tex]
[tex]\[ t = \frac{0.5596}{0.08} \approx 6.995 \][/tex]
7. Round the answer to the nearest whole number:
[tex]\[ t \approx 7 \][/tex]
Therefore, the money has been invested for about [tex]\( \boxed{7} \)[/tex] years.
