The population of a certain species of wildflower in a national park increases by 300\% ev… — SAT

The population of a certain species of wildflower in a national park increases by $300\%$ every $8$ years. The population can be modeled by the function $P(t) = P_0(1 + r)^t$ , where $P_0$ is the initial population, $r$ is the constant annual growth rate, and $t$ is the time in years. Which of the following is closest to the value of $r$ ?

A
0.15
0.19Answer
C
0.38
D
1.19

Answer

The correct value of the annual growth rate is

0.19

An increase of

300\%

means the population becomes

100\% + 300\% = 400\%

of its initial value, which corresponds to a growth factor of

4

over an

8

-year period. The annual growth rate

r

satisfies the equation

(1 + r)^8 = 4

. Solving for

1 + r

gives

1 + r = 4^{1/8} = (2^2)^{1/8} = 2^{1/4} \approx 1.1892

. Subtracting

1

from both sides yields the annual growth rate

r \approx 0.1892

, which is closest to

0.19

Step-by-Step Solution

Determine the total growth factor over the 8-year period from the percentage increase.

The population increases by

300\%

, meaning the new population is

100\% + 300\% = 400\%

of the initial population. This corresponds to a growth factor of

4

every

8

years.

An increase of

300\%

adds

3

times the original value to the original value, resulting in a total of

4

times the original value.

Set up the equation relating the annual growth rate

r

to the 8-year growth factor.

(1+r)^8 = 4

Since the population is multiplied by

1+r

each year, after

8

years it is multiplied by

(1+r)^8

, which must equal the total growth factor of

4

Solve the equation for

r

and round to the nearest hundredth.

1+r = 4^{1/8} \approx 1.1892 \implies r \approx 0.1892 \approx 0.19

Taking the eighth root of both sides gives the annual growth factor

1+r \approx 1.19

. Subtracting

1

isolates the annual growth rate

r

Key Concept

Relating exponential growth rates and growth factors across different time periods

Alternative Method

Instead of solving

(1+r)^8 = 4

, we can write the function directly using the 8-year growth factor as

P(t) = P_0 \cdot 4^{t/8}

. Rewriting this in the form

P(t) = P_0(4^{1/8})^t

shows that the annual growth factor is

4^{1/8}

. We can approximate

4^{1/8} \approx 1.1892

, which corresponds to an annual growth rate

r = 1.1892 - 1 = 0.1892

, or approximately

0.19

Estimated Time:1m 30s

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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