Problem 75
Question
The following table is based on a functional relationship between \(x\) and \(y\) that is either an exponential or a power function: \begin{tabular}{cc} \hline \(\boldsymbol{x}\) & \(\boldsymbol{y}\) \\ \hline\(-1\) & \(0.398\) \\ \(-0.5\) & \(1.26\) \\ 0 & 4 \\ \(0.5\) & \(12.68\) \\ 1 & \(40.18\) \\ \hline \end{tabular} Use an appropriate logarithmic transformation and a graph to decide whether the table comes from a power function or an exponential function, and find the functional relationship between \(x\) and \(y\).
Step-by-Step Solution
Verified Answer
The functional relationship is exponential: \(y = 4 \times 10^x\).
1Step 1: Identify Possible Models
We need to decide whether the data fits an exponential model, of the form \(y = ab^x\), or a power model, of the form \(y = ax^b\). Depending on the model, we will apply different logarithmic transformations.
2Step 2: Logarithmic Transformations
For the exponential model \(y = ab^x\), taking the natural log gives \(\ln y = \ln a + x\ln b\). For the power model \(y = ax^b\), taking the log base 10 gives \(\log y = \log a + b \log x\). We apply the transformations suitable for each model to the data.
3Step 3: Transform Data for Exponential Model
Compute \(\ln y\) for each \(y\) value in the table: \[\begin{align*}\ln(0.398) & \approx -0.921, \\ln(1.26) & \approx 0.232, \\ln(4) & \approx 1.386, \\ln(12.68) & \approx 2.541, \\ln(40.18) & \approx 3.692.\end{align*}\] Then plot \(x\) against \(\ln y\). If the points form a straight line, the relationship is exponential.
4Step 4: Transform Data for Power Model
Compute \(\log y\) and \(\log x\) for each pair \((x, y)\): \[\begin{align*}\log(0.398) & \approx -0.400, & \log(-1) & \text{not defined}, \\log(1.26) & \approx 0.100, & \log(-0.5) & \text{not defined}, \\log(4) & \approx 0.602, & \log(0) & \text{not defined}, \\log(12.68) & \approx 1.103, & \log(0.5) & \approx -0.301, \\log(40.18) & \approx 1.604, & \log(1) & = 0.\end{align*}\]Recall: The logarithm of negative or zero values is undefined, which eliminates the power model since some \(x\) values are negative.
5Step 5: Conclusion from Graph
Since the logarithms for negative and zero \(x\) values are undefined, making graphing for the power function impossible, we focus on the graph of \(x\) versus \(\ln y\). In this case, plotting these points should reveal a straight line, confirming an exponential relationship.
6Step 6: Determine Exponential Relationship
Once a linear relationship in the graph confirms an exponential model, use two points to find the slope \(\ln b\) and \ln a using \(\ln y = \ln a + x \ln b\). For example, using points \((0, 1.386)\) and \((1, 3.692)\):\[\begin{align*}1.306 & = 0 + \ln b, \3.692 & = 1.386 + 1\ln b,\end{align*}\]Re-solve equations for \(a\) and \(b\):\(b \approx 10\) and \(a \approx 4\). The exponential relationship is \(y = 4 \times 10^x\).
Key Concepts
Exponential FunctionsPower FunctionsLogarithmic TransformationsGraphing Techniques
Exponential Functions
Exponential functions are mathematical expressions that model situations where a quantity grows or decays at a constant percentage rate over time. The classic form of an exponential function is given by the equation \(y = ab^x\), where:
- \(a\) is the initial value of \(y\) when \(x=0\).
- \(b\) is the base, which determines the rate of growth or decay; if \(b > 1\) the function represents growth, if \(0 < b < 1\) it represents decay.
- \(x\) represents the independent variable or time.
Power Functions
Power functions are another type of mathematical relationship, often written as \(y = ax^b\), where:
- \(a\) is a constant that scales the function.
- \(b\) is the exponent which determines the function's degree, affecting the steepness of the curve.
- \(x\) is the independent variable raised to the power \(b\).
Logarithmic Transformations
Logarithmic transformations are crucial techniques used to linearize relationships, especially when dealing with exponential and power functions. These transformations change a multiplicative relationship into an additive one, making patterns clearer. For exponential functions, the transformation is by taking the natural logarithm of \(y\), converting \(y = ab^x\) into \(\ln y = \ln a + x \ln b\). This formula’s linearity is beneficial for graphing and analysis.
Similarly, for power functions, applying the transformation \(\log y = \log a + b \log x\) simplifies \(y = ax^b\) so we can analyze the linearity between \(\log x\) and \(\log y\). However, in our exercise with some \(x\) values negative or zero, this approach isn't feasible due to logarithmic limitations preventing accurate transformation and graphing for power functions. Logarithmic transformations become an analytical tool that assists in determining whether a relationship follows an exponential pattern by changing the nature of the graph based on the type of function we are examining.
Similarly, for power functions, applying the transformation \(\log y = \log a + b \log x\) simplifies \(y = ax^b\) so we can analyze the linearity between \(\log x\) and \(\log y\). However, in our exercise with some \(x\) values negative or zero, this approach isn't feasible due to logarithmic limitations preventing accurate transformation and graphing for power functions. Logarithmic transformations become an analytical tool that assists in determining whether a relationship follows an exponential pattern by changing the nature of the graph based on the type of function we are examining.
Graphing Techniques
Graphing techniques help visually determine the nature of a relationship between variables. A powerful graphing method involves transforming data to assess if a linear relationship is present. In this case, by plotting \(x\) against \(\ln y\), we establish whether the data adheres to an exponential model. A graph that yields a straight line after transformation strongly signals an exponential connection.
- If the plots of \(x\) against \(\ln y\) reveal a clear straight line, then we're observing evidence of \(y = ab^x\).
- If instead, the graph were plotted as \(\log y\) against \(\log x\) and showed linearity (in an appropriate scenario where \(\log x\) is defined), the data would suggest a power function.
Other exercises in this chapter
Problem 74
The reciprocal of a function \(f(x)\) can be written as either \(1 / f(x)\) or \([f(x)]^{-1} .\) The point of this problem is to make clear that a reciprocal of
View solution Problem 74
Evaluate the following exponential expressions: (a) \(\left(2^{4} 2^{-3 / 2}\right)^{2}\) (b) \(\left(\frac{6^{5 / 2} 6^{2 / 3}}{6^{1 / 3}}\right)^{3}\) (c) \(\
View solution Problem 75
Find the inverse of \(f(x)=3^{x}, x \in \mathbf{R}\), together with its domain, and graph both functions in the same coordinate system.
View solution Problem 75
Which real number \(x\) satisfies (a) \(\log _{4} x=-2\) ? (b) \(\log _{1 / 3} x=-3 ?\) (c) \(\log _{10} x=-2\) ?
View solution