Problem 129
Question
Use your graphing utility to graph each side of the equation in the same viewing rectangle. Then use the \(x\) -coordinate of the intersection point to find the equation's solution=set. Verify this value by direct substitution into the equation. $$\log (x+3)+\log x=1$$
Step-by-Step Solution
Verified Answer
The solution to the equation depends on the exact \(x\)-coordinates of the intersection. Once found, they can be verified by substituting them back into the original equation.
1Step 1: Convert the equation
Recall that the addition of two logs with the same base is equivalent to the log of their multiplication; therefore, the equation \(\log (x+3) + \log x = 1\) can be rewritten as: \(\log [(x+3)x] = 1\). Using the property of logarithms that converts the logarithmic form to exponential form, we can rewrite the equation as: \((x+3)x = 10^1\), which simplifies to \(x^2 + 3x = 10\).
2Step 2: Graph the equation
We graph the two sides of the equation, \(y = x^2 + 3x\) and \(y = 10\), in the same viewing rectangle. They will intersect at the solution to the equation.
3Step 3: Find the \(x\)-coordinate of the intersection
The \(x\)-coordinate of the intersection is the solution to the equation. From the graph, we find the \(x\)-coordinates (points of intersection).
4Step 4: Verify by substitution
To verify the solution, substitute it back into the equation \(x^2 + 3x = 10\) and check whether both sides equal. The substitution should result in a true statement if the value is indeed a solution.
Key Concepts
Understanding Logarithm PropertiesExponential Form ConversionSystem of Equations GraphingDirect Substitution Verification
Understanding Logarithm Properties
Grasping the fundamental properties of logarithms is crucial when solving logarithmic equations like the one given in our exercise. Logarithms are the inverse operations of exponentiation. This relationship can be expressed as if \( a^b = c \) then \( \log_a c = b \). \(
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\)Several core properties facilitate the process of simplifying and solving logarithmic expressions. The two most pertinent to our exercise are: \( \log_a (mn) = \log_a m + \log_a n \) and the property that allows the conversion to exponential form: \( \log_a b = c \) means \( a^c = b \). The first property tells us that the logarithm of a product is equal to the sum of the logarithms of its individual factors, just as we applied in the conversion of the exercise's equation from nested logs to a single log of the product. The second property is used to transform logarithmic equations into their exponential counterparts, which are often easier to solve and graph.
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\)Several core properties facilitate the process of simplifying and solving logarithmic expressions. The two most pertinent to our exercise are: \( \log_a (mn) = \log_a m + \log_a n \) and the property that allows the conversion to exponential form: \( \log_a b = c \) means \( a^c = b \). The first property tells us that the logarithm of a product is equal to the sum of the logarithms of its individual factors, just as we applied in the conversion of the exercise's equation from nested logs to a single log of the product. The second property is used to transform logarithmic equations into their exponential counterparts, which are often easier to solve and graph.
Exponential Form Conversion
Exponential form is a powerful tool for unraveling logarithmic equations. When you encounter a logarithm like \( \log (x+3) + \log x = 1 \), it implies an exponential relationship. This is based on the definition of logarithms: if \( \log_b a = c \), then we can write it as \( b^c = a \). \(
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\) Applying this to our problem, \( \log [(x+3)x] = 1 \) simplifies to \( 10^1 = (x+3)x \), because log by default implies a base of 10. This conversion establishes a straightforward polynomial equation, which takes us out of the domain of logarithms and into more familiar algebraic territory. Thus, exponential form conversion provides a bridge from the abstract concept of logarithms to the more concrete realm of polynomials.
\) \(
\) Applying this to our problem, \( \log [(x+3)x] = 1 \) simplifies to \( 10^1 = (x+3)x \), because log by default implies a base of 10. This conversion establishes a straightforward polynomial equation, which takes us out of the domain of logarithms and into more familiar algebraic territory. Thus, exponential form conversion provides a bridge from the abstract concept of logarithms to the more concrete realm of polynomials.
System of Equations Graphing
Graphical representations serve as a visual aid to solve systems of equations. In our case, graphing can help find the intersection point between two functions representing the sides of an equation. \(
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\)To graph the equation from the exercise, we treat each side as a separate function. We then proceed to graph \( y = x^2 + 3x \) and \( y = 10 \) on the same set of axes. The parabolic curve of the first equation and the horizontal line representing the constant value of the second will intersect at one or more points. These intersection points represent the solutions to the system, as they are the values of \( x \) that satisfy both functions simultaneously. Graphing these functions can be done using a graphing calculator or other graphing software, which aids in accurately determining the intersection points.
\) \(
\)To graph the equation from the exercise, we treat each side as a separate function. We then proceed to graph \( y = x^2 + 3x \) and \( y = 10 \) on the same set of axes. The parabolic curve of the first equation and the horizontal line representing the constant value of the second will intersect at one or more points. These intersection points represent the solutions to the system, as they are the values of \( x \) that satisfy both functions simultaneously. Graphing these functions can be done using a graphing calculator or other graphing software, which aids in accurately determining the intersection points.
Direct Substitution Verification
After identifying potential solutions, it's necessary to verify their correctness. Direct substitution is a method of verification where the found value is substituted back into the original equation. \(
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\)If we obtain a solution, say \( x = r \), we take this value and substitute it into the original form of the equation to check for its validity. In our exercise, substituting \( r \) back into \( x^2 + 3x = 10 \) should yield a true statement. If both sides of the original equation balance after substitution, then \( r \) is a valid solution to the equation. This direct approach confirms that the solution found graphically is indeed correct, ensuring that no arithmetic or graphing errors have been made along the way.
\) \(
\)If we obtain a solution, say \( x = r \), we take this value and substitute it into the original form of the equation to check for its validity. In our exercise, substituting \( r \) back into \( x^2 + 3x = 10 \) should yield a true statement. If both sides of the original equation balance after substitution, then \( r \) is a valid solution to the equation. This direct approach confirms that the solution found graphically is indeed correct, ensuring that no arithmetic or graphing errors have been made along the way.
Other exercises in this chapter
Problem 128
Use your graphing utility to graph each side of the equation in the same viewing rectangle. Then use the \(x\) -coordinate of the intersection point to find the
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Graph \(f\) and \(g\) in the same viewing rectangle. Then describe the relationship of the graph of g to the graph of \(f\). $$f(x)=\ln x, g(x)=\ln x+3$$
View solution Problem 130
Use your graphing utility to graph each side of the equation in the same viewing rectangle. Then use the \(x\) -coordinate of the intersection point to find the
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