Problem 62
Question
Use a logarithmic transformation to find a linear relationship between the given quantities and graph the resulting linear relationship on a log-log plot. $$ y=5 x^{3} $$
Step-by-Step Solution
Verified Answer
Transformation gives \( \ln(y) = \ln(5) + 3\ln(x) \), a linear relationship on log-log plot.
1Step 1: Identify the Equation
The given equation is \( y = 5x^3 \). We need to transform this into a linear form using logarithmic transformation.
2Step 2: Apply Logarithmic Transformation
Take the natural logarithm of both sides of the equation to linearize it. \[ \ln(y) = \ln(5x^3) \] Using the properties of logarithms, this becomes \[ \ln(y) = \ln(5) + \ln(x^3) \] Further applying the power rule \( \ln(a^b) = b\ln(a) \), this simplifies to \[ \ln(y) = \ln(5) + 3\ln(x) \]. This equation is in the form \( Y = B + mX \) where \( Y = \ln(y) \), \( B = \ln(5) \), \( m = 3 \), and \( X = \ln(x) \).
3Step 3: Understand the Linear Relationship
Through the logarithmic transformation, the relationship between \( \ln(y) \) and \( \ln(x) \) is linear, with a slope of 3 and y-intercept of \( \ln(5) \). This indicates that on a log-log plot, you expect a straight line.
4Step 4: Graph the Linear Relationship
To graph this function, plot \( \ln(y) \) on the y-axis against \( \ln(x) \) on the x-axis. The plot will be a straight line with a slope of 3 and a y-intercept of approximately \( \ln(5) \approx 1.609 \).
Key Concepts
Linear relationshipLog-log plotProperties of logarithms
Linear relationship
Establishing a linear relationship using logarithmic transformation involves converting a nonlinear equation into a linear form. This transformation is particularly useful if you want to analyze exponential relationships in data.
In the exercise provided, the original equation is \( y = 5x^3 \). It's a power function, which represents a curved relationship in a regular Cartesian coordinate system. To convert it to a linear relationship that is easier to analyze, take the natural logarithm of both sides.
The transformed equation, \( \ln(y) = \ln(5) + 3\ln(x) \), is in a format corresponding to the linear equation \( Y = mX + B \). Here,
In the exercise provided, the original equation is \( y = 5x^3 \). It's a power function, which represents a curved relationship in a regular Cartesian coordinate system. To convert it to a linear relationship that is easier to analyze, take the natural logarithm of both sides.
The transformed equation, \( \ln(y) = \ln(5) + 3\ln(x) \), is in a format corresponding to the linear equation \( Y = mX + B \). Here,
- \( Y = \ln(y) \)
- \( X = \ln(x) \)
- \( m = 3 \), representing the slope
- \( B = \ln(5) \), which is the y-intercept
Log-log plot
A log-log plot is a graphical representation where both the axes are on a logarithmic scale. This type of plot is particularly useful for illustrating power-law relationships, such as the one in the exercise.
In the example of \( \ln(y) = 3\ln(x) + \ln(5) \), plotting \( \ln(y) \) against \( \ln(x) \) on a log-log plot will yield a straight line. This is because the plot shrinks the scale, compressing large values of \( x \) and \( y \) more aggressively, thus linearizing the relationship visually. The slope of the line on this plot represents the power in the original equation (3 in this example), while the y-intercept represents \( \ln(5) \).
Using a log-log plot can help reveal patterns in data that are not immediately apparent in standard plots, especially when dealing with exponential or polynomial relationships. It has the advantage of extending the length over which a data set can be fitted onto a graph within a given size.
In the example of \( \ln(y) = 3\ln(x) + \ln(5) \), plotting \( \ln(y) \) against \( \ln(x) \) on a log-log plot will yield a straight line. This is because the plot shrinks the scale, compressing large values of \( x \) and \( y \) more aggressively, thus linearizing the relationship visually. The slope of the line on this plot represents the power in the original equation (3 in this example), while the y-intercept represents \( \ln(5) \).
Using a log-log plot can help reveal patterns in data that are not immediately apparent in standard plots, especially when dealing with exponential or polynomial relationships. It has the advantage of extending the length over which a data set can be fitted onto a graph within a given size.
Properties of logarithms
Logarithms have several key properties that are instrumental in transforming complex mathematical relationships into simpler, linear ones. These properties make working with exponential and power functions much more manageable.
The primary properties of logarithms that were used in this exercise include:
By understanding and using these logarithmic rules, any exponential form can be simplified into a linear relationship, which is easier to handle in algebraic calculations and graphical analysis, such as in log-log plots. These properties help in deciphering complex relationships in data and are foundational in fields such as physics and economics, where such transformations are frequently necessary.
The primary properties of logarithms that were used in this exercise include:
- **Product Rule:** \( \ln(ab) = \ln(a) + \ln(b) \)
- **Power Rule:** \( \ln(a^b) = b\ln(a) \)
By understanding and using these logarithmic rules, any exponential form can be simplified into a linear relationship, which is easier to handle in algebraic calculations and graphical analysis, such as in log-log plots. These properties help in deciphering complex relationships in data and are foundational in fields such as physics and economics, where such transformations are frequently necessary.
Other exercises in this chapter
Problem 61
After 7 days, a particular radioactive substance decays to half of its original amount. Find the decay rate of this substance.
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Find the center and the radius of the circle given by the equation $$ 0=x^{2}+y^{2}-4 x+2 y-11 $$
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After 5 days, a particular radioactive substance decays to \(37 \%\) of its original amount. Find the half-life of this substance.
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Find the center and the radius of the circle given by the equation $$ x^{2}+y^{2}+2 x-4 y+1=0 $$
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