Question

6700 dollars is placed in an account with an annual interest rate of 8%. How much will be in the account after 24 years, to the nearest cent?
Example: Exponential Functions:
y, equals, a, b, to the power x
y=ab
x

a, equals, starting value = , 4800
a=starting value = 4800
r, equals, rate = , 6, point, 5, percent, equals, 0, point, 0, 6, 5
r=rate = 6.5%=0.065
Exponential Growth:
Exponential Growth:
b, equals, 1, plus, r, equals, 1, plus, 0, point, 0, 6, 5, equals, 1, point, 0, 6, 5
b=1+r=1+0.065=1.065
Write Exponential Function:
Write Exponential Function:
y, equals, 4800, left parenthesis, 1, point, 0, 6, 5, right parenthesis, to the power x
y=4800(1.065)
x

Put it all together
Plug in time for x:
Plug in time for x:
y, equals, 4800, left parenthesis, 1, point, 0, 6, 5, right parenthesis, start superscript, 14, end superscript
y=4800(1.065)
14

y, equals, 11591, point, 3, 9, 6, 0, 8, 5, 9
y=11591.3960859
Evaluate
y, approximately equals, 11591, point, 4
y≈11591.4
round

Answer 1

42485.91

Answer 2

To calculate the future value of an investment using the formula for exponential growth, we can follow a similar format as you have shown in your example.

Given:
- Starting amount (a) = \$6700
- Annual interest rate (r) = 8% = 0.08
- Time (x) = 24 years

### Step 1: Calculate \( b \)
The growth factor \( b \) is calculated as:
\[
b = 1 + r = 1 + 0.08 = 1.08
\]

### Step 2: Write the Exponential Function
The exponential function can be written as:
\[
y = a(1 + r)^x = 6700(1.08)^x
\]

### Step 3: Plug in Time for \( x \)
Substituting \( x = 24 \):
\[
y = 6700(1.08)^{24}
\]

### Step 4: Calculate \( (1.08)^{24} \)
Calculating \( (1.08)^{24} \):
\[
(1.08)^{24} \approx 5.278
\]

### Step 5: Calculate the Future Value \( y \)
Now, plug this back into the equation:
\[
y = 6700 \times 5.278 \approx 35365.6
\]

### Step 6: Round to the Nearest Cent
Rounding to the nearest cent:
\[
y \approx 35365.60
\]

So, the amount in the account after 24 years will be approximately **$35,365.60**.