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a. Show that the values in the table have the multiply-add property. b. Use the first and last points to find algebraically the particular equation of the natural logarithmic function that fits the points. c. Show that the equation in part b gives the other points in the table. $$\begin{array}{rr}x & y \\\\\hline 3.6 & 1 \\\14.4 & 2 \\\57.6 & 3 \\\230.4 & 4 \\\921.6 & 5\end{array}$$

Short Answer

Expert verified
The multiply-add property holds with a multiplying factor of 4 for x and an addition of 1 for y. The particular natural logarithmic function that fits the points is \(y = \ln(3.6a \times x)\) with \(a \approx \frac{e}{3.6}\). This equation gives the correct y-values for the other points in the table.

Step by step solution

01

Checking Multiply-add Property

The multiply-add property means that if you multiply the input by a certain factor 'a', and the output adds a constant 'b'. To verify this, calculate the ratios of consecutive x-values and differences of consecutive y-values. The ratio for x-values should be constant, and the difference for y-values should also be constant.
02

Calculate Ratios and Differences

For x-values: 14.4/3.6 = 4, 57.6/14.4 = 4, 230.4/57.6 = 4, and 921.6/230.4 = 4. For y-values: 2 - 1 = 1, 3 - 2 = 1, 4 - 3 = 1, and 5 - 4 = 1. Since the ratios of x are always 4 and the differences of y are always 1, the multiply-add property holds.
03

Determine the Particular Equation

Using the form of the natural logarithmic function \(y = \ln(ax)\), substitute the first point (3.6, 1) and solve for \(a\). Then, use the last point (921.6, 5) to verify or adjust the value of \(a\).
04

Find 'a' Using the First Point

Substitute (3.6, 1) into the equation, giving \(1 = \ln(3.6a)\). By exponentiating both sides, we find \(a = e^{1}/3.6\).
05

Verify 'a' Using the Last Point

Substitute (921.6, 5) and the found \(a\) value into the equation: \(5 = \ln(921.6 \times \frac{e}{3.6})\). If the equation holds true, then we have our particular function.
06

Verify Other Points with the Found Equation

Using the equation \(y = \ln(3.6a \times x)\) with the confirmed value of \(a\), calculate \(y\) for the remaining x-values in the table to see if they match the given y-values.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Properties
Logarithmic functions, such as the natural logarithm, have a unique set of properties that allow us to manipulate and solve equations involving them. One key property is the logarithm's ability to turn multiplication into addition. This property is based on the fundamental logarithmic identity, \( \log(b^m) = m \log(b) \) for any positive base \( b \) and real number \( m \). In the context of our exercise, we see this property at work when examining the 'multiply-add' characteristic of the given table.

In simpler terms, when we have two numbers, say \(A \) and \(B \) where \(B = aA \) (\(\ a \) is a constant multiplier), their natural logarithms relate in a straightforward way: \( \ln(B) = \ln(a) + \ln(A) \). This is a direct result of the logarithmic property \( \ln(xy) = \ln(x) + \ln(y) \). It's essential for students to understand this property to grasp how logarithmic functions behave, both in theoretical mathematics and real-world exponential growth scenarios, such as population dynamics and interest calculations.

Another important logarithmic property is the power rule, where \( \ln(x^n) = n \ln(x) \) for a natural number \( n \). This applies to the 'multiply-add' property we've confirmed through the table because we are effectively applying the power rule in reverse, turning our constant multiples of \( x \) into linear additions of \( y \) by logarithmic transformation.
Exponential Functions
Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. These functions are closely related to logarithms, as natural logarithms are the inverse of exponential functions with base \( e \) (Euler's number, approximately 2.71828). The function \( f(x) = e^x \) is the most common example of an exponential function and the basis for the natural logarithmic function \( y = \ln(x) \).

In the given exercise, we use an exponential function when we convert the logarithmic equation back into its exponential form to solve for the constant \( a \) in Step 4. This move from logarithmic to exponential form is achieved using one of the defining properties of logarithms: \( b = e^{\ln(b)} \), meaning any number \( b \) can be expressed as an exponent of \( e \) using the natural logarithm of \( b \) itself.

Understanding how exponential functions work is crucial for students. These functions model a wide range of phenomena like radioactive decay, population growth, and even the compounding of interest in finance. The ability to switch between exponential and logarithmic forms expands the toolset students have available for solving real-world problems.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying mathematical expressions and equations in order to solve for variables. It is a fundamental skill set for working with all types of functions, including logarithmic and exponential functions.

In the given problem, algebraic manipulation enters into play when we are asked to find the particular equation that fits the given set of points. By substituting the points into the natural logarithmic function, we are setting up equations to be solved for the unknown constant \( a \) in Steps 4 and 5. This process involves exponentiation - to 'undo' a logarithm, simplifying expressions, and isolating variables to arrive at a solution.

Effective algebraic techniques include understanding how to properly move terms from one side of the equation to the other, applying inverse operations, and factoring when necessary. These skills are not only essential in this specific exercise but are widely applicable across all of mathematics. Being proficient in algebraic manipulation allows students to unlock the potential of equations to describe patterns, make predictions, and solve problems across various scientific disciplines.

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Most popular questions from this chapter

Test your knowledge of the definition of logarithm Write in logarithmic forme \(d^{5}=k\)

Suppose that \(y\) varies directly with the square of \(x\) and that \(z\) is a quadratic function of \(x\) Explain why the direct-square-variation function is a quadratic function but the quadratic function is not necessarily a direct- square-variation function.

By considering second differences, show that a quadratic function does not fit the points in this table. $$\begin{array}{lr} x & y \\ \hline 4 & 5 \\ 5 & 7 \\ 6 & 11 \\ 7 & 17 \\ 8 & 27 \end{array}$$ What would the last \(y\) -value have to be in order for a quadratic function to fit exactly?

Let \(f(x)=a x^{2}+b x+c .\) Let \(d\) be the constant difference between successive \(x\) -values. Find \(f(x+d), f(x+2 d),\) and \(f(x+3 d) .\) Simplify. By subtracting consecutive \(y\) -values, find the three first differences. By subtracting consecutive first differences, show that the two second differences equal the constant \(2 a d^{2}\)

In 1896 Samuel Langley successfully flew a model of an airplane he was designing. In \(1903,\) he tried unsuccessfully to fly the full-sized airplane. Assume that the full-sized plane was 4 times the length of the model (Figure \(7-3 \mathrm{h}\) ). a. The wing area, and thus the lift, of similarly shaped airplanes is directly proportional to the square of the length of each plane. How many times more wing area did the full-sized plane have than the model? b. The volume, and thus the weight, of similarly shaped airplanes is directly proportional to the cube of the length. How many times heavier was the full- sized plane than the model? c. Why do you think the model was able to fly but the full-sized plane was not?

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