A culture of bacteria has a population of 150150150 cells when it is first observed. The population doubles every 12 hr12~\text{hr}12 hr, which means its population is governed by the function p(t)=150⋅2t12p\left(t\right)=150\cdot{$$2^{\frac{t}$${12}}}p(t)=150⋅212t, where ttt is the number of hours after the first observation.How long does it take the population to triple in size?

Ch. 1 - Functions

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 1 - Functions

Problem 1.82d

Chapter 1, Problem 1.82d

A culture of bacteria has a population of $150$ cells when it is first observed. The population doubles every $12~$\text{hr}$$ , which means its population is governed by the function $p$\left$(t$\right$)=150$\cdot{2^{\frac{t}{12}$}}$ , where $t$ is the number of hours after the first observation.
How long does it take the population to triple in size?

Verified step by step guidance

Identify the initial population and the function governing the population growth. The initial population is 150 cells, and the function is given by $ p(t) = 150 \cdot 2^{\frac{t}{12}} $.

Determine the target population size for the bacteria to triple. Since the initial population is 150, the target population is $ 3 \times 150 = 450 $ cells.

Set up the equation to find the time $ t $ when the population reaches 450 cells: $ 450 = 150 \cdot 2^{\frac{t}{12}} $.

Divide both sides of the equation by 150 to isolate the exponential term: $ 3 = 2^{\frac{t}{12}} $.

Take the logarithm of both sides to solve for $ t $. Use the property of logarithms that $ \log(a^b) = b \cdot \log(a) $: $ \log(3) = \frac{t}{12} \cdot \log(2) $. Solve for $ t $ by multiplying both sides by 12 and dividing by $ \log(2) $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Exponential Growth

Exponential growth occurs when a quantity increases at a rate proportional to its current value. In the context of the bacteria population, the population doubles every 12 hours, which can be modeled by the function p(t) = 150 * 2^(t/12). This type of growth leads to rapid increases over time, making it essential to understand when analyzing population dynamics.

Doubling Time

Doubling time is the period it takes for a quantity to double in size. For the bacteria population, the doubling time is given as 12 hours. This concept is crucial for calculating future population sizes and understanding the rate of growth, as it provides a consistent measure of how quickly the population expands.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions and are used to solve for time in growth problems. To determine how long it takes for the population to triple, one can set up the equation 3 * 150 = 150 * 2^(t/12) and solve for t using logarithms. This concept is vital for transitioning from exponential growth equations to practical time calculations.

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Textbook Question

A culture of bacteria has a population of $150$ cells when it is first observed. The population doubles every $12~$\text{hr}$$ , which means its population is governed by the function $p$\left$(t$\right$)=150$\cdot{2^{\frac{t}{12}$}}$ , where $t$ is the number of hours after the first observation.

What is the population $4~$\text{days}$$ after the first observation?

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Textbook Question

Identify the symmetry (if any) in the graphs of the following equations.

$y^2-4x^2=4$

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Taxicab fees A taxicab ride costs \$3.50 plus \$2.50 per mile. Let m be the distance (in miles) from the airport to a hotel. Find and graph the function c(m) that represents the cost of taking a taxi from the airport to the hotel. Also determine how much it will cost if the hotel is 9 miles from the airport.

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Textbook Question

Decide whether $f$ , $g$ , or both represent one-to-one functions. <IMAGE>

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Textbook Question

Inverse of composite functions

c. Explain why if g and h are one-to-one, the inverse of ƒ(x) = g(h(x)) exists.

Textbook Question

Composite functions

Let ƒ(x) = x³, g (x) = sin x and h(x) = √x .

Find the domain of ƒ o g.