Question

an ecologist began studying a certain type of plant species in a wetlands area in 2013. in 2015 (t = 2), there were 50 plants. in 2021 (t = 8), there were 118 plants. the number of plants of this species can be modeled by the function p given by p(t) = ab^t, where p(t) is the number of plants during year t, and t is the number of years since 2013. part a (i) use the given data to write two equations that can be used to find the values for constants a and b in the expression for p(t). (ii) find the values for a and b as decimal approximations.

Explanation:

Step1: Identify the values of t and P(t)

In 2013, \( t = 0 \) (since \( t \) is years since 2013), and \( P(0)=50 \). In 2015, \( t = 2 \), and \( P(2)=? \) Wait, the problem says in 2015 (\( t = 2 \)) there were 50 plants? Wait, no, let's re - read. The ecologist began studying in 2013. In 2015 (\( t = 2 \)), there were 50 plants? Wait, no, the text: "In 2015 (\( t = 2 \)), there were 50 plants. In 2021 (\( t = 8 \)), there were 118 plants." And the function is \( P(t)=ab^{t} \).

For \( t = 0 \) (2013), \( P(0)=ab^{0}=a\times1=a \). Wait, but the problem says in 2015 (\( t = 2 \)) there were 50 plants? Wait, maybe I misread. Let's check again.

Wait, the problem: "An ecologist began studying a certain type of plant species in a wetlands area in 2013. In 2015 (\( t = 2 \)), there were 50 plants. In 2021 (\( t = 8 \)), there were 118 plants. The number of plants of this species can be modeled by the function \( P(t)=ab^{t} \), where \( P(t) \) is the number of plants during year \( t \), and \( t \) is the number of years since 2013."

So when \( t = 2 \) (2015), \( P(2)=50 \); when \( t = 8 \) (2021), \( P(8)=118 \). And also, when \( t = 0 \) (2013), what's \( P(0) \)? Wait, the function is \( P(t)=ab^{t} \), so when \( t = 0 \), \( P(0)=a\times b^{0}=a \). But the problem doesn't give \( P(0) \) directly. Wait, maybe the first data point is \( t = 2 \), \( P(2)=50 \) and \( t = 8 \), \( P(8)=118 \), and we also know that when \( t = 0 \), maybe \( P(0) \) is the initial number? Wait, no, the problem says "began studying in 2013", so \( t = 0 \) is 2013. But the problem states "In 2015 (\( t = 2 \)), there were 50 plants. In 2021 (\( t = 8 \)), there were 118 plants." So we have two points: \( (t_1,P(t_1))=(2,50) \) and \( (t_2,P(t_2))=(8,118) \), and the function \( P(t)=ab^{t} \).

So for \( t = 2 \): \( P(2)=ab^{2}=50 \)

For \( t = 8 \): \( P(8)=ab^{8}=118 \)

These are the two equations.

Step2: Solve for a and b

First, from the equation when \( t = 2 \): \( a=\frac{50}{b^{2}} \)

Substitute this into the equation when \( t = 8 \):

\( \frac{50}{b^{2}}\times b^{8}=118 \)

Simplify: \( 50\times b^{6}=118 \)

Then \( b^{6}=\frac{118}{50}=2.36 \)

Take the sixth root of both sides: \( b = 2.36^{\frac{1}{6}} \)

Calculate \( 2.36^{\frac{1}{6}} \). Let's compute:

\( \ln(2.36)\approx0.8589 \), then \( \frac{\ln(2.36)}{6}\approx\frac{0.8589}{6}\approx0.1432 \), then \( e^{0.1432}\approx1.153 \)

Alternatively, using a calculator to find the sixth root of 2.36: \( 2.36^{\frac{1}{6}}=\sqrt[6]{2.36}\approx1.15 \) (approximate)

Then, substitute \( b \) back into \( a=\frac{50}{b^{2}} \)

If \( b\approx1.153 \), then \( b^{2}\approx(1.153)^{2}\approx1.329 \), so \( a=\frac{50}{1.329}\approx37.6 \)

Wait, but let's do it more accurately.

First, solve for \( b \):

\( b^{6}=\frac{118}{50}=2.36 \)

\( b = 2.36^{\frac{1}{6}} \)

Using a calculator: \( 2.36^{\frac{1}{6}}\approx1.152 \)

Then \( b^{2}=(1.152)^{2}\approx1.327 \)

\( a=\frac{50}{1.327}\approx37.7 \)

Wait, but maybe we made a mistake in the initial data. Wait, the problem says "In 2015 (\( t = 2 \)), there were 50 plants. In 2021 (\( t = 8 \)), there were 118 plants."

Alternatively, maybe the first data point is \( t = 0 \), \( P(0)=a \), but the problem doesn't say. Wait, no, the function is \( P(t)=ab^{t} \), so when \( t = 0 \), \( P(0)=a \). But the problem states "began studying in 2013", so maybe \( t = 0 \) is 2013, and the number of plants in 2013 is \( a \). But the problem doesn't give that. Wait, maybe the user made a typo, but according to the given data, we have \( t = 2 \), \( P(2)=50 \) and \( t = 8 \), \( P(8)=118 \).

So for part (i), the two equations are:

When \( t = 2 \): \( 50=ab^{2} \)

When \( t = 8 \): \( 118=ab^{8} \)

For part (ii), we can solve these two equations.

From the first equation: \( a=\frac{50}{b^{2}} \)

Substitute into the second equation:

\( 118=\frac{50}{b^{2}}\times b^{8}=50b^{6} \)

So \( b^{6}=\frac{118}{50}=2.36 \)

\( b = \sqrt[6]{2.36}\approx1.152 \)

Then \( a=\frac{50}{b^{2}}=\frac{50}{(\sqrt[6]{2.36})^{2}}=\frac{50}{\sqrt[3]{2.36}} \) (since \( (b^{6})^{\frac{1}{3}}=b^{2} \))

\( \sqrt[3]{2.36}\approx1.33 \)

\( a=\frac{50}{1.33}\approx37.6 \)

Wait, let's calculate \( b \) more accurately.

\( \ln(2.36)=0.85893 \)

\( \frac{\ln(2.36)}{6}=0.143155 \)

\( e^{0.143155}\approx1.153 \)

Then \( b^{2}=(1.153)^{2}=1.329 \)

\( a = 50\div1.329\approx37.6 \)

So \( a\approx37.6 \), \( b\approx1.15 \)

Answer:

(i) The two equations are \( 50 = ab^{2} \) and \( 118=ab^{8} \)

(ii) \( a\approx37.6 \), \( b\approx1.15 \)