Question

Every year, a marine biologist measures the sea turtle population in Sunny Harbor. When he first measured the population one year ago, there were 640 sea turtles. Now, the sea turtle population has increased to 800. The biologist expects the population to continue increasing each year.
Write an exponential equation in the form y=a(b)x that can model the sea turtle population, y, x years after the initial measurement.
Use whole numbers, decimals, or simplified fractions for the values of a and b.
y =

How many years after the marine biologist first measured the population does he expect it will reach 1,250?

Answer 1

To write the exponential equation in the form \( y = a(b)^x \), we first identify the initial population and the growth rate.

1. **Determine values for \( a \) and \( b \)**:
- The initial population, when \( x = 0 \) (one year ago), is \( a = 640 \).
- The current population \( (y) = 800 \) corresponds to \( x = 1 \) (one year later).

We can use these points to find \( b \):

\[
y = a(b)^x
\]

Substituting the values we have for \( x = 1 \):

\[
800 = 640(b)^1
\]

Now we can solve for \( b \):

\[
b = \frac{800}{640} = \frac{5}{4} = 1.25
\]

Now, we can write the exponential equation:

\[
y = 640(1.25)^x
\]

2. **Finding when the population reaches 1,250**:
We need to determine \( x \) when \( y = 1250 \):

\[
1250 = 640(1.25)^x
\]

Now, divide both sides by 640:

\[
\frac{1250}{640} = (1.25)^x
\]

Calculating \( \frac{1250}{640} \):

\[
\frac{1250}{640} = \frac{125}{64} \approx 1.953125
\]

So we have:

\[
(1.25)^x = \frac{125}{64}
\]

To solve for \( x \), we can take the logarithm of both sides:

\[
x \cdot \log(1.25) = \log\left(\frac{125}{64}\right)
\]

Now we calculate \( \log(1.25) \) and \( \log\left(\frac{125}{64}\right) \):

Using a calculator:

1. \( \log(1.25) \approx 0.09691 \)
2. \( \log(125) = \log(5^3) = 3 \cdot \log(5) \approx 3 \cdot 0.6990 \approx 2.0970 \)
3. \( \log(64) = \log(2^6) = 6 \cdot \log(2) \approx 6 \cdot 0.3010 \approx 1.8060 \)

Now we calculate \( \log\left(\frac{125}{64}\right) \):

\[
\log\left(\frac{125}{64}\right) = \log(125) - \log(64) \approx 2.0970 - 1.8060 = 0.2910
\]

Now, we substitute back into the equation:

\[
x \cdot 0.09691 = 0.2910
\]

Solving for \( x \):

\[
x = \frac{0.2910}{0.09691} \approx 3.00
\]

Thus, the marine biologist expects the population will reach 1,250 in approximately 3 years after the first measurement.

In summary:

- The exponential equation is:
\[
y = 640(1.25)^x
\]
- The time it will take to reach a population of 1,250 is about **3 years**.