Problem 84
Question
Determine whether each statement makes sense or does not make sense, and explain your reasoning. I'm using a photocopier to reduce an image over and over by \(50 \%,\) so the exponential function \(f(x)=\left(\frac{1}{2}\right)^{x}\) models the new image size, where \(x\) is the number of reductions.
Step-by-Step Solution
Verified Answer
The statement makes sense because the function \(f(x)=\left(\frac{1}{2}\right)^{x}\) correctly models the described scenario where the image size is consistently reduced by half with each copy, thus representing an exponential decay.
1Step 1: Understand the Exponential Function
In the exponential function \(f(x)=\left(\frac{1}{2}\right)^{x}\), the base is \(0.5\) (which is less than 1, indicating an exponential decay), and \(x\) is the exponent which represents the number of operations(reductions in this case).
2Step 2: Relate the function to scenario
Each reduction is shrinking the size by 50% or half. This aligns with the base of the function (\(\frac{1}{2}\)). The exponent \(x\) indicates the number of reductions which also fits the scenario.
3Step 3: Confirm the logic
To confirm, consider first two reductions. After one reduction, size of image will be \(\frac{1}{2}\) of original and after two reduction it will be \(\frac{1}{2} * \frac{1}{2}\) = \(\frac{1}{4}\) which is in line with \(f(2) = \frac{1}{2}^{2}\). This reasoning applies to any number of reductions.
Key Concepts
Exponential DecayAlgebraFunction Modeling
Exponential Decay
Exponential decay is a fundamental concept in mathematics that describes a process that decreases at a consistent percentage rate over time. This is often seen in phenomena like radioactive decay, population decreases, and naturally, in our scenario of reducing an image size using a photocopier. In these situations, after each instance of reduction or elapsed time, the quantity being measured decreases by a fixed percentage. For an exponential decay function, like in the exercise, the formula takes the form \( f(x) = a \times b^x \).
Here, \(a\) is the initial quantity (starting size), \(b\) is the base representing the decay factor (in this case, 0.5), and \(x\) is the exponent which shows how many times this process has happened. With a base less than 1, the function models decay over time. Each increment in \(x\) by 1 results in multiplying the current value by the base of 0.5, effectively halving the size at each step. This explains why continuously reducing the image size by 50% reflects exponential decay behavior.
Here, \(a\) is the initial quantity (starting size), \(b\) is the base representing the decay factor (in this case, 0.5), and \(x\) is the exponent which shows how many times this process has happened. With a base less than 1, the function models decay over time. Each increment in \(x\) by 1 results in multiplying the current value by the base of 0.5, effectively halving the size at each step. This explains why continuously reducing the image size by 50% reflects exponential decay behavior.
Algebra
Algebra is a branch of mathematics dealing with symbols and the rules for manipulating these symbols. It's the language of modeling situations using mathematical representations. In the given problem, algebra helps formulate the relationship between the number of reductions and the size of the image mathematically.
We used an algebraic expression \( f(x) = \left( \frac{1}{2} \right)^x \) which is concise and captures the essence of repeated halving of the image's size. Algebra allows us to understand this pattern clearly: every time you operate the photocopier, the next step is essentially calculator work following this expression. This helps us predict the final size of the image after any number of reductions, e.g., after three reductions, the size is \( \left( \frac{1}{2} \right)^3 = \frac{1}{8} \) of the original.
Algebraic models are powerful tools in describing and predicting real-world situations succinctly.
We used an algebraic expression \( f(x) = \left( \frac{1}{2} \right)^x \) which is concise and captures the essence of repeated halving of the image's size. Algebra allows us to understand this pattern clearly: every time you operate the photocopier, the next step is essentially calculator work following this expression. This helps us predict the final size of the image after any number of reductions, e.g., after three reductions, the size is \( \left( \frac{1}{2} \right)^3 = \frac{1}{8} \) of the original.
Algebraic models are powerful tools in describing and predicting real-world situations succinctly.
Function Modeling
Function modeling is the mathematical practice of representing real-world phenomena with functions. It helps in understanding how quantities change relative to each other. In our exercise, the function \( f(x) = \left( \frac{1}{2} \right)^x \) is used to model the image shrinking process.
This model is effective as it can predict the resulting size of the image after a given number of operations. Function modeling is not limited to just mathematical exercises; it is widely applied by engineers, scientists, and businesses to make informed predictions and decisions.
Using this exponential decay function, you can forecast outcomes without needing to perform the practical experiment. With each pass \(x\), the model shows consistent reductions in size. Function models like these illustrate the power of equations in simplifying the complexities of real-life processes into predictable patterns.
This model is effective as it can predict the resulting size of the image after a given number of operations. Function modeling is not limited to just mathematical exercises; it is widely applied by engineers, scientists, and businesses to make informed predictions and decisions.
Using this exponential decay function, you can forecast outcomes without needing to perform the practical experiment. With each pass \(x\), the model shows consistent reductions in size. Function models like these illustrate the power of equations in simplifying the complexities of real-life processes into predictable patterns.
Other exercises in this chapter
Problem 84
Exercises \(83-85\) will help you prepare for the material covered in the first section of the next chapter. Graph \(x+2 y=2\) and \(x-2 y=6\) in the same recta
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Evaluate or simplify each expression without using a calculator. $$\log 10^{8}$$
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Let \(\log _{b} 2=A\) and \(\log _{b} 3=C .\) Write each expression in terms of \(A\) and \(C\). \(\log _{b} 6\)
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Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer.
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