Example Question - exponential growth
Here are examples of questions we've helped users solve.
Bacterial Colony Doubling Problem
Let the initial number of bacteria be \( P_0 \). Given that the colony doubles every 3 hours, the number of bacteria after 3 hours (\( P_3 \)) is \( 2P_0 \). The number of bacteria after 6 hours (\( P_6 \)) is \( 2P_3 = 2 \times 2P_0 = 2^2P_0 \). The number of bacteria after 9 hours (\( P_9 \)) is \( 2P_6 = 2 \times 2^2P_0 = 2^3P_0 \). The number of bacteria after 12 hours (\( P_{12} \)) is \( 2P_9 = 2 \times 2^3P_0 = 2^4P_0 \). Thus, after 12 hours, the number of bacteria is \( P_{12} = 2^4P_0 \). Since \( P_0 = 10 \): \( P_{12} = 2^4 \times 10 \) \( P_{12} = 16 \times 10 \) \( P_{12} = 160 \) Therefore, the number of bacteria in the colony after 12 hours is 160. The answer is option C.
Exponential Growth of Rabbit Population
Given that the size of a rabbit population doubles every year, we use the formula for exponential growth: \( r(t) = r_0 \cdot 2^{t} \), where \( t \) is time in years and \( r_0 \) is the initial population size. In this scenario, \( r_0 = 1,000 \), hence the correct formula becomes: \[ r(t) = 1,000 \cdot 2^{t} \] The answer that matches this formula is: \[ r = 1,000(2^{t}) \]
Calculating Population Growth Using Exponential Growth Formula
Um diese Aufgabe zu lösen, können wir die Formel für exponentielles Wachstum verwenden: \[ P(t) = P_0 \cdot e^{rt} \] Hierbei ist: - \( P(t) \) die Bevölkerung nach Zeit \( t \) - \( P_0 \) die anfängliche Bevölkerung (6,1 Mrd. Menschen) - \( r \) die jährliche Wachstumsrate (1,5% oder 0,015 als Dezimalzahl) - \( e \) die Basis des natürlichen Logarithmus (etwa 2,71828) - \( t \) die Zeit in Jahren Wir wollen herausfinden, nach wie vielen Jahren \( t \) die Bevölkerung auf 12 Milliarden ansteigt, also setzen wir \( P(t) = 12 \). Zuerst formen wir die Gleichung um, um \( t \) zu isolieren: \[ 12 = 6,1 \cdot e^{0,015t} \] Dann teilen wir beide Seiten durch 6,1: \[ \frac{12}{6,1} = e^{0,015t} \] Anschließend nehmen wir den natürlichen Logarithmus (ln) beider Seiten der Gleichung: \[ \ln\left(\frac{12}{6,1}\right) = \ln\left(e^{0,015t}\right) \] Wir verwenden die Eigenschaft des Logarithmus, dass \( \ln(e^x) = x \): \[ \ln\left(\frac{12}{6,1}\right) = 0,015t \] Nun teilen wir beide Seiten der Gleichung durch 0,015, um \( t \) zu finden: \[ t = \frac{\ln\left(\frac{12}{6,1}\right)}{0,015} \] Jetzt berechnen wir \( t \) mit einem Taschenrechner oder einer Rechensoftware: \[ t = \frac{\ln\left(\frac{12}{6,1}\right)}{0,015} \approx \frac{\ln(1,967213115)}{0,015} \approx \frac{0,678034403}{0,015} \approx 45,20229353 \] Die Bevölkerung würde also ungefähr nach 45 Jahren die 12-Milliarden-Grenze überschreiten, wenn sie jährlich um 1,5% zunimmt.
Exponential Growth in Fruit Fly Population
The image contains a math problem regarding exponential growth. The problem is as follows: "15,000 fruit flies were released into the Western River for a scientific study. The function N(t) = 15,000(3^t) represents the number of blue-tailed fruit flies after t years. What would happen to this population?" The options are: A) It will decrease by about 2,000. B) It will decrease by about 3,000. C) It will increase by about 2,000. D) It will increase by about 3,000. The function provided, N(t) = 15,000(3^t), models exponential growth. As t increases, N(t) also increases since 3^t is always greater than 1 for any positive t. Therefore, the population will increase over time, not decrease. Since no specific time frame is given in the question or the options (like t = 1 year or 5 years), we cannot calculate the exact amount of increase. Nonetheless, we can definitively say that the population will increase, not decrease. The correct answer to what will happen to this population is between either option C or D (since options A and B are about a decrease), and without additional context or a specific time frame (value of t), we cannot determine the exact amount of increase from the options provided. However, given that the question implies a short period (probably a year), the increase would be substantial due to the nature of exponential growth, which would suggest option D is more likely. But to reiterate, without a specific time frame mentioned in the problem, the exact numerical increase cannot be determined from the options given.
Population Growth Calculation with Doubling Time
The question in the image pertains to exponential growth, more specifically the doubling time of a population. The doubling time of a population is the time required for that population to double in size at its current growth rate. The image states: "The doubling time of a population of flies is 5 hours. By what factor does the population increase in 30 hours? By what factor does the population increase in 3 weeks?" Let's solve this in two parts: 1. Calculation of the population increase factor in 30 hours: Given that the population doubles every 5 hours, we can find out how many doubling periods there are in 30 hours by dividing 30 by the doubling time of 5 hours: 30 hours / 5 hours per doubling = 6 doublings To determine the factor by which the population has increased after 6 doublings, we calculate 2 to the power of 6 (since each doubling multiplies the population by 2): 2^6 = 64 Thus, the population increases by a factor of 64 in 30 hours. 2. Calculation of the population increase factor in 3 weeks: First, let's convert 3 weeks into hours. There are 24 hours in a day and 7 days in a week, so: 3 weeks * 7 days/week * 24 hours/day = 504 hours Next, we'll calculate how many doublings occur in 504 hours: 504 hours / 5 hours per doubling = 100.8 doublings However, part of a doubling period does not result in a full doubling, so we must consider only the complete doublings. Therefore, we will consider 100 doublings. Now let's calculate the factor for 100 doublings: 2^100 This number is extremely large (roughly a 30-digit number), so it is impractical to write out in its entirety. For simplicity, it suffices to say that the population increases by a factor of 2^100 after 3 weeks. These calculations assume that the growth can continue without any limitations, which in real-world scenarios may not be possible due to factors like resources and space constraints.
Analyzing Data for Optimal Function Modeling
The image contains a question asking which type of function would best model the given data, providing options A. Linear, B. Quadratic, C. Exponential, and D. Absolute value. The data provided in the table shows the population of a city in thousands (y) for time in years (x) after the beginning of a decade. The data is as follows: - Time (years), x: 0, 1, 2, 6, 8 - Population (thousands), y: 52, 57, 137, 152, 227 To determine the best model for this data, we must examine how the population changes as time increases. A linear function is represented by a straight line and involves a constant rate of change. A quadratic function describes a parabolic relationship and involves acceleration or deceleration of the rate of change. An exponential function involves growth or decay that increases or decreases at a nonlinear rate, typically becoming more pronounced as time goes on. An absolute value function would typically create a V-shaped graph, which does not seem to be indicative of the data points. Looking at the intervals: - From x=0 to x=1, the population grows from 52 to 57 (a difference of 5). - From x=1 to x=2, the population grows from 57 to 137 (a difference of 80). - From x=2 to x=6, the population grows from 137 to 152 (a difference of 15 over 4 years). - From x=6 to x=8, the population grows from 152 to 227 (a difference of 75 over 2 years). Considering the significant jump from 57 to 137 and then the more gradual changes followed by another large jump, this suggests that an exponential growth trend might be more suitable. A linear or quadratic model would not typically show the sharp increases seen between certain intervals (like from year 1 to year 2), and absolute value is not reflective of the increasing growth pattern. Therefore, the type of function that would best model this data is C. Exponential.
Exponential Growth in Voter Convincing
This is a problem involving exponential growth. Simon starts by convincing 10 people to vote. On each subsequent day, each person he has convinced convinces 10 more people to vote. Here's how it works for five days: - On day one, Simon convinces 10 people. - On day two, those 10 people each convince 10 more people, so \(10 \times 10 = 100\) people are convinced. - On day three, each of these 100 people convinces 10 more people, so \(100 \times 10 = 1,000\) people are convinced. - On day four, each of these 1,000 people convinces 10 more people, so \(1,000 \times 10 = 10,000\) people are convinced. - On day five, each of these 10,000 people convinces 10 more people, so \(10,000 \times 10 = 100,000\) people are convinced. So the total number of people convinced to vote by the fifth day is 100,000. The correct answer to the question, "How many people will be convinced to vote on the fifth day?" is: D. 100,000
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