Problem: You begin with $250.  At 7% compounded continuously, how long will it take you to reach $1000?

To solve this problem, we will use the formula for continuous compound interest:

\[
A = P e^{rt}
\]

Where:
- $A$ is the final amount,
- $P$ is the initial principal,
- $r$ is the rate of interest (as a decimal),
- $t$ is the time (in years),
- $e$ is the mathematical constant approximately equal to 2.718.

We are given:
- $A = 1000$,
- $P = 250$,
- $r = 0.07$.

We need to find $t$, the time it takes to grow from $250 to $1000.

### Step 1: Set up the equation
\[
1000 = 250 e^{0.07t}
\]

### Step 2: Isolate $e^{0.07t}$
\[
\frac{1000}{250} = e^{0.07t}
\]
\[
4 = e^{0.07t}
\]

### Step 3: Take the natural logarithm of both sides
\[
\ln(4) = 0.07t
\]

### Step 4: Solve for $t$
\[
t = \frac{\ln(4)}{0.07}
\]

Now I'll compute the value of $t$.It will take approximately 19.80 years to grow the initial $250 to $1000 at a 7% interest rate compounded continuously.

Would you like more details or have any questions? Here are 5 related questions to explore:

1. What is the formula for compound interest when not compounded continuously?
2. How would the time change if the interest rate were increased to 8%?
3. What is the meaning of the natural logarithm in this context?
4. How would the result differ if the compounding were done monthly instead of continuously?
5. What is the future value of $250 after 10 years at a 7% continuously compounded rate?

**Tip:** Continuous compounding uses the exponential function $e^{rt}$, which leads to faster growth than other compounding methods (like yearly or monthly).

You begin with $250. At 7% compounded continuously, how long will it take you to reach $1000?

Discover how to calculate the time it takes for $250 to grow to $1000 at a 7% continuously compounded interest rate. This problem involves exponential growth, logarithms, and the continuous compound interest formula. Suitable for students in Grades 10-12.

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