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Magazine Chapter 12: Problem 62
Solve. Where appropriate, include approximations to three decimal places. $$ \ln (x-6)+\ln (x+3)=\ln 22 $$
Short Answer
Expert verified
The solution is \( x = 8 \).
Step by step solution
01
Use Logarithm Property
Combine the logarithmic expressions on the left side using the property \( \ln a + \ln b = \ln (a \cdot b) \), so the equation becomes: \[ \ln ((x-6)(x+3)) = \ln 22 \]
02
Exponentiate Both Sides
To remove the natural logarithm, exponentiate both sides using the base e: \[ e^{\ln ((x-6)(x+3))} = e^{\ln 22} \] Since \( e^{\ln y} = y \), this simplifies to: \[ (x-6)(x+3) = 22 \]
03
Expand and Simplify
Expand the left side: \[ x^2 + 3x - 6x - 18 = 22 \] Simplify the equation: \[ x^2 - 3x - 18 = 22 \]
04
Bring to Standard Form
Subtract 22 from both sides to bring the equation to standard quadratic form: \[ x^2 - 3x - 40 = 0 \]
05
Solve the Quadratic Equation
Solve the quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -3 \), and \( c = -40 \): \[ x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-40)}}{2(1)} = \frac{3 \pm \sqrt{9 + 160}}{2} = \frac{3 \pm \sqrt{169}}{2} = \frac{3 \pm 13}{2} \]
06
Find the Roots
Calculate the two potential solutions: \[ x_1 = \frac{3 + 13}{2} = 8 \] \[ x_2 = \frac{3 - 13}{2} = -5 \]
07
Verify Validity of Solutions
Check both solutions in the context of the logarithmic function. \( x = -5 \) is invalid because it makes both terms inside the logarithms negative (logarithms of non-positive numbers are undefined). Thus, \( x = 8 \) is the only valid solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
natural logarithm
The natural logarithm is a special type of logarithm that uses the base \(e\), which is an irrational constant approximately equal to 2.71828. It is denoted as \(\text{ln}(x)\). The natural logarithm has several unique properties that make it useful in calculus and solving equations involving exponential growth and decay.
One such property is \(\text{ln}(e) = 1\) because \(e\) raised to the power of 1 is \(e\). Another important property is that the natural logarithm of a product is the sum of the natural logarithms of the factors: \text{ln}(ab) = \text{ln}(a) + \text{ln}(b)\$.
This property is particularly useful in simplifying and solving logarithmic equations by condensing multiple natural logarithm terms into a single term.
One such property is \(\text{ln}(e) = 1\) because \(e\) raised to the power of 1 is \(e\). Another important property is that the natural logarithm of a product is the sum of the natural logarithms of the factors: \text{ln}(ab) = \text{ln}(a) + \text{ln}(b)\$.
This property is particularly useful in simplifying and solving logarithmic equations by condensing multiple natural logarithm terms into a single term.
exponentiation in equations
Exponentiation is a mathematical operation involving two numbers, the base and the exponent. When solving logarithmic equations, exponentiation is a crucial step that helps us eliminate the logarithms. In the given exercise, the equation \text{ln}((x-6)(x+3)) = \text{ln}(22)\$ is exponentiated using base \(e\) to remove the logarithms.
Exponentiating both sides, we get: \(e^{\text{ln}((x-6)(x+3))} = e^{\text{ln}(22)}\). Since the exponential function and the natural logarithm are inverse operations, this simplifies to \text{(x-6)}\text{(x+3)} = 22\$.
This transformation turns the logarithmic equation into a polynomial equation, which is easier to solve.
Exponentiating both sides, we get: \(e^{\text{ln}((x-6)(x+3))} = e^{\text{ln}(22)}\). Since the exponential function and the natural logarithm are inverse operations, this simplifies to \text{(x-6)}\text{(x+3)} = 22\$.
This transformation turns the logarithmic equation into a polynomial equation, which is easier to solve.
quadratic formula
The quadratic formula is used to find the solutions to a quadratic equation of the form \(ax^2 + bx + c = 0\). The formula states that the solutions for \(x\) are: \frac{-b \text{卤} \text{ 鈭歿b^2 - 4ac}}}{2a}\$.
In the exercise, after simplification, we obtained the quadratic equation \(x^2 - 3x - 40 = 0\). Here, \(a = 1\), \(b = -3\), and \(c = -40\). Plugging these values into the quadratic formula, we get: \text{x} = \frac{-(-3) \text{卤} \text{ 鈭歿(-3)^2 - 4(1)(-40)}}}{2(1)}\$ = \text{x} = \frac{3 \text { 卤 } \text{ 鈭歿9 + 160}}}{2}\$ = \text{x} = \frac{3 \text { 卤 } \text{ 鈭歿169}}}{2}\$ = \text{x} = \frac{3 \text { 卤 } 13}{2}\$.
This yields the solutions \(\text{x_1} = 8\) and \(\text{x_2} = -5\).
In the exercise, after simplification, we obtained the quadratic equation \(x^2 - 3x - 40 = 0\). Here, \(a = 1\), \(b = -3\), and \(c = -40\). Plugging these values into the quadratic formula, we get: \text{x} = \frac{-(-3) \text{卤} \text{ 鈭歿(-3)^2 - 4(1)(-40)}}}{2(1)}\$ = \text{x} = \frac{3 \text { 卤 } \text{ 鈭歿9 + 160}}}{2}\$ = \text{x} = \frac{3 \text { 卤 } \text{ 鈭歿169}}}{2}\$ = \text{x} = \frac{3 \text { 卤 } 13}{2}\$.
This yields the solutions \(\text{x_1} = 8\) and \(\text{x_2} = -5\).
logarithmic properties
Logarithmic properties are essential tools for simplifying and solving logarithmic equations. Here are some key properties:
Using these properties helps us break down complex logarithmic expressions into more manageable parts. In the given exercise, we used the product property to combine \(\text{ln} (x-6)\) and \(\text{ln} (x+3)\) into \(\text{ln} ((x-6)(x+3))\).
This simplification is crucial for transforming the logarithmic equation into a form that allows us to solve for \(x\).
- Product Property:\text{ln}(ab) = \text{ln}(a) + \text{ln}(b)\$.
- Quotient Property:\text{ln}\text(\frac{a}{b}) = \text{ln}(a) - \text{ln}(b)\$.
- Power Property:\text{ln}(a^b) = b \text{ln}(a)\$.
Using these properties helps us break down complex logarithmic expressions into more manageable parts. In the given exercise, we used the product property to combine \(\text{ln} (x-6)\) and \(\text{ln} (x+3)\) into \(\text{ln} ((x-6)(x+3))\).
This simplification is crucial for transforming the logarithmic equation into a form that allows us to solve for \(x\).
