The population of a certain species of plankton in a lake is modeled by the function P(t) … — SAT

The population of a certain species of plankton in a lake is modeled by the function

P(t) = P_0 \cdot (1.69)^{\frac{t}{d}}

where

P_0

is the initial population,

t

is the time, in days, since the population was first measured, and

d

is a positive constant. The population of the plankton increases by

30\%

every

h

hours, where

h

is a positive constant. If the model is accurate, what is the value of the ratio

\frac{h}{d}

A
0.5
B
2
12Answer
D
48

Answer

The correct value of the ratio is 12.

The correct answer shows the value of 12. Since

t

is measured in days, the time interval of

h

hours must be converted to

\frac{h}{24}

days. A

30\%

increase corresponds to a multiplier of

1.3

. Substituting

t = \frac{h}{24}

into the function gives

(1.69)^{\frac{h}{24d}} = 1.3

. Rewriting

1.69

1.3^2

yields

\left(1.3^2\right)^{\frac{h}{24d}} = 1.3

, which simplifies to

1.3^{\frac{h}{12d}} = 1.3^1

. Equating the exponents gives

\frac{h}{12d} = 1

, which solves to

\frac{h}{d} = 12

Step-by-Step Solution

Express the growth interval in days to match the unit of

t

in the model.

Since there are 24 hours in a day, an interval of

h

hours is equivalent to

\frac{h}{24}

days.

The variable

t

in the model

P(t) = P_0 \cdot (1.69)^{\frac{t}{d}}

represents time in days, so all time variables must be in days.

Relate the percentage increase to the population multiplier.

An increase of

30\%

corresponds to a multiplier of

1 + 0.30 = 1.3

30\%

increase means the new population is

130\%

of the previous population.

Set up the equation using the given model at

t = \frac{h}{24}

P_0 \cdot (1.69)^{\frac{h}{24d}} = 1.3 \cdot P_0

, which simplifies to

(1.69)^{\frac{h}{24d}} = 1.3

t = \frac{h}{24}

days, the population must be

1.3

times the initial population

P_0

Rewrite the base of the exponential term and solve for the ratio.

Since

1.69 = 1.3^2

, the equation becomes

\left(1.3^2\right)^{\frac{h}{24d}} = 1.3^1

, which simplifies to

1.3^{\frac{2h}{24d}} = 1.3^1

, leading to

\frac{h}{12d} = 1

Expressing both sides with the same base allows for equating their exponents to solve for the target ratio.

Solve for the final ratio

\frac{h}{d}

\frac{h}{d} = 12

Multiplying both sides of the equation

\frac{h}{12d} = 1

by 12 isolates the ratio

\frac{h}{d}

Key Concept

Exponential Equations with Base Manipulation and Unit Conversion

Alternative Method

Instead of substituting a specific value for

t

, we can equate the daily growth factors. The model's daily growth factor is

(1.69)^{1/d}

. A

30\%

increase every

h

hours corresponds to a factor of

1.3

every

h

hours. Since there are

24

hours in a day, there are

\frac{24}{h}

such intervals in a day, yielding a daily growth factor of

(1.3)^{24/h}

. Setting the factors equal:

(1.69)^{1/d} = (1.3)^{24/h} \implies (1.3^2)^{1/d} = (1.3)^{24/h} \implies \frac{2}{d} = \frac{24}{h} \implies \frac{h}{d} = 12

Estimated Time:3m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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