To solve this problem, let's break it down step by step:

1. Writing the Exponential Equation:

The general form of an exponential growth equation is: $y = a(b)^x$ where:

• $a$ is the initial amount,
• $b$ is the growth factor (rate of increase),
• $x$ is the number of years,
• $y$ is the amount of money in the account after $x$ years.

Given:

• Patrick's initial deposit is $a = 894$ dollars.
• The account grows by 2% per year. This means the growth factor $b = 1 + 0.02 = 1.02$.

Thus, the equation becomes: $y = 894(1.02)^x$ This is the exponential equation that models the amount of money in Patrick's account.

2. Finding When the Account Exceeds $1,000:

We need to find out after how many years $x$ the account will have more than $1,000. This means we want to solve the equation: $1000 = 894(1.02)^x$

To solve for $x$: $\frac{1000}{894} = (1.02)^x$ $1.1189 = (1.02)^x$

Now, take the natural logarithm (ln) of both sides: $\ln(1.1189) = x \ln(1.02)$ $x = \frac{\ln(1.1189)}{\ln(1.02)}$ Using a calculator: $x \approx \frac{0.11241}{0.01980} \approx 5.68$

So, Patrick's account will have more than $1,000 after about 6 years.

Do you have any questions or would you like more details? Here are some related questions to extend your understanding:

1. What would the balance be after 10 years?
2. How would the equation change if the interest rate was 3% instead of 2%?
3. How does exponential growth differ from simple interest growth?
4. Can you graph the function $y = 894(1.02)^x$ to visualize the growth?
5. What happens if Patrick adds more deposits each year? How would that affect the equation?

Tip: When solving exponential equations, using logarithms is a powerful tool for isolating the exponent.