The problem describes a population of spotted woodpeckers that starts at 51 and quadruples every 19 years. We need to find the number of years, $t$, it takes for the population to reach 316.

1. Problem Description

The problem describes a population of spotted woodpeckers that starts at 51 and quadruples every 19 years. We need to find the number of years, $t$, it takes for the population to reach

3

1

6.

2. Solution Steps

We can model the population growth with the following exponential equation:

$P(t) = P_0 \cdot r^{\frac{t}{T}}$

where:

$P(t)$ is the population at time $t$,

$P_0$ is the initial population,

$r$ is the growth rate (quadrupling means $r = 4$),

$T$ is the time it takes for the population to quadruple (in this case, $T = 19$), and

$t$ is the time in years.

We are given:

$P_0 = 51$

$P(t) = 316$

$r = 4$

$T = 19$

We want to find $t$. Plugging in the values into the equation, we have:

$316 = 51 \cdot 4^{\frac{t}{19}}$

First, divide both sides by 51:

$\frac{316}{51} = 4^{\frac{t}{19}}$

Take the natural logarithm of both sides:

$ln(\frac{316}{51}) = ln(4^{\frac{t}{19}})$

Using the property of logarithms that $ln(a^b) = b \cdot ln(a)$, we get:

$ln(\frac{316}{51}) = \frac{t}{19} \cdot ln(4)$

Now, solve for $t$:

$t = 19 \cdot \frac{ln(\frac{316}{51})}{ln(4)}$

Calculate the value:

$t = 19 \cdot \frac{ln(6.196)}{ln(4)}$

$t = 19 \cdot \frac{1.824}{1.386}$

$t = 19 \cdot 1.316$

$t = 25.004$

Round to the nearest hundredth:

$t \approx 25.00$

3. Final Answer

25.00