Chapter 8: Problem 17
The growth of a new social networking site can be modelled by the exponential function \(N(t)=1.1^{t},\) where \(N\) is the number of users after \(t\) days. a) Write the equation of the inverse. b) How long will it take, to the nearest day, for the number of users to exceed \(1 000 000 ?\)
Short Answer
Expert verified
a) The inverse equation is: \(N^{-1}(t) = \frac{\log(t)}{\log(1.1)}\). b) It will take approximately 145 days for the number of users to exceed 1,000,000.
Step by step solution
01
Understand the Exponential Function
The given exponential function is \(N(t) = 1.1^t\), where \(N(t)\) represents the number of users after \(t\) days.
02
Write the Inverse Equation
To find the inverse of the function \(N(t) = 1.1^t\), exchange \(N(t)\) and \(t\). Thus, we get: \(t = 1.1^{N(t)}\). Solve for \(N(t)\) by taking the logarithm of both sides: \[ t = 1.1^{N(t)} \] \[ \log(t) = N(t) \cdot \log(1.1) \] Therefore, the inverse equation is: \[ N^{-1}(t) = \frac{\log(t)}{\log(1.1)} \]
03
Set Up the Inequality for Users
We need to find the time \(t\) when the number of users exceeds 1,000,000. So, we set up the inequality: \[ 1.1^t > 1,000,000 \]
04
Solve the Inequality Using Logarithms
To solve the inequality \(1.1^t > 1,000,000\), take the log of both sides: \[ \log(1.1^t) > \log(1,000,000) \] Using the power rule of logarithms, this becomes: \[ t \cdot \log(1.1) > \log(1,000,000) \] Now solve for \(t\): \[ t > \frac{\log(1,000,000)}{\log(1.1)} \]
05
Calculate the Value of t
Using a calculator, determine the values: \[ \log(1,000,000) = 6 \] and \[ \log(1.1) \approx 0.0414 \] Therefore, \[ t > \frac{6}{0.0414} \] \[ t > 144.93 \] Since \(t\) must be an integer, round up to the nearest day: \[ t \approx 145 \] days.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inverse Function
An inverse function is a function that reverses the effect of the original function. Essentially, if the original function takes input and produces output, the inverse function takes that output and brings you back to the original input. For the given exponential function, finding the inverse involves a few steps:
First, switch the roles of the dependent and independent variables.
Then, you solve for the new independent variable by using logarithms, a tool that is key when dealing with exponential functions.
For the function given by the exercise, which is an exponential function, the inverse is found by swapping the variables and then isolating the new input variable using logarithms. The calculation shows:
\( t = 1.1^{N(t)} \)
Taking the logarithm of both sides, it becomes:
\( \log(t) = N(t) \cdot \log(1.1) \)
Finally isolating the variable gives us the inverse equation:
\( N^{-1}(t) = \frac{\log(t)}{\log(1.1)} \).
Inverse functions allow you to understand how long it takes for the input to reach a particular output.
First, switch the roles of the dependent and independent variables.
Then, you solve for the new independent variable by using logarithms, a tool that is key when dealing with exponential functions.
For the function given by the exercise, which is an exponential function, the inverse is found by swapping the variables and then isolating the new input variable using logarithms. The calculation shows:
\( t = 1.1^{N(t)} \)
Taking the logarithm of both sides, it becomes:
\( \log(t) = N(t) \cdot \log(1.1) \)
Finally isolating the variable gives us the inverse equation:
\( N^{-1}(t) = \frac{\log(t)}{\log(1.1)} \).
Inverse functions allow you to understand how long it takes for the input to reach a particular output.
Logarithms
Logarithms are the inverse operations of exponentiation. They answer the question: 'To what exponent must we raise a specific number (the base) to obtain another number?' When dealing with exponential functions like \(N(t) = 1.1^t\), logarithms are essential for solving for the unknown exponent.
For example, if you have an equation \( 1.1^t = 1,000,000 \), using logarithms allows you to solve for \(t\). Taking the log of both sides returns:
\( \log(1.1^t) = \log(1,000,000) \)
Applying the power rule (\( \log(a^b) = b\log(a) \)), this becomes:
\( t \cdot \log(1.1) = \log(1,000,000) \)
Then solving for \(t\):
\( t = \frac{\log(1,000,000)}{\log(1.1)} \)
Logarithms simplify calculations involving exponential growth, making them vital in fields such as science, economics, and many areas of everyday life.
For example, if you have an equation \( 1.1^t = 1,000,000 \), using logarithms allows you to solve for \(t\). Taking the log of both sides returns:
\( \log(1.1^t) = \log(1,000,000) \)
Applying the power rule (\( \log(a^b) = b\log(a) \)), this becomes:
\( t \cdot \log(1.1) = \log(1,000,000) \)
Then solving for \(t\):
\( t = \frac{\log(1,000,000)}{\log(1.1)} \)
Logarithms simplify calculations involving exponential growth, making them vital in fields such as science, economics, and many areas of everyday life.
Inequalities
Inequalities are mathematical expressions involving the symbols \( >, <, \geq, \leq \). They are used to show the relative size of two values. In this exercise, the inequality \( 1.1^t > 1,000,000 \) is set up to find the time \(t\) when the number of users exceeds 1,000,000. To solve this inequality, we:
Take the logarithm of both sides:
\( \log(1.1^t) > \log(1,000,000) \)
Using the property of logarithms \( \log(a^b) = b\log(a) \), the inequality becomes:
\( t \cdot \log(1.1) > \log(1,000,000) \)
Solving for \(t\):
\( t > \frac{\log(1,000,000)}{\log(1.1)} \)
Calculating these logarithm values allows us to find the minimum number of days needed. In mathematical modeling and real-life scenarios, inequalities help set bounds and establish thresholds, ensuring conditions are met.
Take the logarithm of both sides:
\( \log(1.1^t) > \log(1,000,000) \)
Using the property of logarithms \( \log(a^b) = b\log(a) \), the inequality becomes:
\( t \cdot \log(1.1) > \log(1,000,000) \)
Solving for \(t\):
\( t > \frac{\log(1,000,000)}{\log(1.1)} \)
Calculating these logarithm values allows us to find the minimum number of days needed. In mathematical modeling and real-life scenarios, inequalities help set bounds and establish thresholds, ensuring conditions are met.
Exponential Equations
Exponential equations are equations in which variables appear as exponents. They often model real-world situations like population growth, radioactive decay, and, as seen in the exercise, the growth of users on a social networking site. The general form of an exponential function is \( N(t) = a^t \), where \(a\) is a constant base, and \(t\) is the exponent.
Solving exponential equations usually involves logarithms. For example, given an exponential equation \( 1.1^t = y \), taking the logarithm of both sides results in:
\( \log(1.1^t) = \log(y) \)
By applying logarithm rules, it simplifies to:
\( t \cdot \log(1.1) = \log(y) \)
The value of \(t\) can then be isolated:
\( t = \frac{\log(y)}{\log(1.1)} \)
Exponential functions describe rapid increases or decreases over time and are prominent in many scientific and financial disciplines. Understanding how to manipulate and solve these equations is fundamental for analyzing exponential growth patterns effectively.
Solving exponential equations usually involves logarithms. For example, given an exponential equation \( 1.1^t = y \), taking the logarithm of both sides results in:
\( \log(1.1^t) = \log(y) \)
By applying logarithm rules, it simplifies to:
\( t \cdot \log(1.1) = \log(y) \)
The value of \(t\) can then be isolated:
\( t = \frac{\log(y)}{\log(1.1)} \)
Exponential functions describe rapid increases or decreases over time and are prominent in many scientific and financial disciplines. Understanding how to manipulate and solve these equations is fundamental for analyzing exponential growth patterns effectively.