Solving Logarithmic Equations Using The One-to-One Property

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In mathematics, logarithmic equations are equations that involve logarithms of variable expressions. Solving these equations often requires applying various properties of logarithms to isolate the variable. One such property is the one-to-one property of logarithms, which states that if log⁑bx=log⁑by{ \log_b{x} = \log_b{y} }, then x=y{ x = y }, provided that b{ b } is a positive number not equal to 1, and both x{ x } and y{ y } are positive. This property is particularly useful for solving equations where logarithms with the same base are set equal to each other.

In this article, we will delve into how to effectively use the one-to-one property to solve logarithmic equations. We'll break down the process into clear, manageable steps, illustrate with examples, and provide a detailed solution to the equation ln⁑(βˆ’15βˆ’19d)=ln⁑(18d){ \ln(-15 - 19d) = \ln(18d) }. Whether you're a student grappling with logarithmic equations or someone looking to refresh your math skills, this guide will equip you with the knowledge and techniques to tackle these problems confidently.

Understanding and applying the one-to-one property is crucial not only for solving equations but also for a deeper comprehension of logarithmic functions and their applications in various fields, including science, engineering, and finance. So, let's embark on this mathematical journey to master the art of solving logarithmic equations.

The one-to-one property of logarithms is a cornerstone in solving logarithmic equations. It simplifies the process by allowing us to eliminate logarithms when they have the same base on both sides of the equation. To fully grasp this concept, let’s break it down and understand its underlying principles.

The fundamental statement of this property is: if log⁑bx=log⁑by{ \log_b{x} = \log_b{y} }, then x=y{ x = y }, given that the base b{ b } is a positive number not equal to 1 (i.e., b>0{ b > 0 } and bβ‰ 1{ b \neq 1 }), and both x{ x } and y{ y } are positive (i.e., x>0{ x > 0 } and y>0{ y > 0 }).

Key Components of the Property

  1. Same Base: The logarithms on both sides of the equation must have the same base. This is crucial because the property relies on the logarithmic function being one-to-one for a given base. If the bases are different, this property cannot be directly applied.
  2. Positive Arguments: The arguments of the logarithms (x{ x } and y{ y } in this case) must be positive. This is because logarithms are only defined for positive numbers. If either argument is zero or negative, the logarithm is undefined.
  3. Base Conditions: The base b{ b } must be positive and not equal to 1. If b{ b } were 1, the logarithmic function would not be one-to-one, and if b{ b } were negative or zero, the logarithm would not be defined for all positive numbers.

How the Property Works

The one-to-one property stems from the fact that logarithmic functions are one-to-one functions. A function is one-to-one if each element of its range corresponds to exactly one element of its domain. In simpler terms, if the outputs (logarithms) are the same, the inputs (arguments) must also be the same.

Consider the equation log⁑28=log⁑28{ \log_2{8} = \log_2{8} }. Both sides are equal because 23=8{ 2^3 = 8 }. If we change the argument on one side, for example, log⁑28=log⁑216{ \log_2{8} = \log_2{16} }, the equality no longer holds because 23β‰ 24{ 2^3 \neq 2^4 }. This illustrates that for the same base, equal logarithms imply equal arguments.

Practical Application

To use the one-to-one property in solving equations, follow these steps:

  1. Verify the Conditions: Ensure that the logarithms on both sides of the equation have the same base and that the arguments are positive.
  2. Apply the Property: If the conditions are met, set the arguments equal to each other. That is, if log⁑bx=log⁑by{ \log_b{x} = \log_b{y} }, write x=y{ x = y }.
  3. Solve the Resulting Equation: Solve the algebraic equation obtained in the previous step for the variable.
  4. Check the Solutions: It is crucial to check the solutions in the original logarithmic equation to ensure that the arguments of the logarithms remain positive. Extraneous solutions may arise if this check is not performed.

Example

Let's consider a simple example: log⁑3(2x+1)=log⁑3(x+3){ \log_3{(2x + 1)} = \log_3{(x + 3)} }.

  1. Same Base: Both logarithms have the base 3.
  2. Apply the Property: Set the arguments equal: 2x+1=x+3{ 2x + 1 = x + 3 }.
  3. Solve: Subtract x{ x } from both sides to get x+1=3{ x + 1 = 3 }, then subtract 1 from both sides to find x=2{ x = 2 }.
  4. Check: Substitute x=2{ x = 2 } back into the original equation: log⁑3(2(2)+1)=log⁑35{ \log_3{(2(2) + 1)} = \log_3{5} } and log⁑3(2+3)=log⁑35{ \log_3{(2 + 3)} = \log_3{5} }. Both arguments are positive, so the solution is valid.

The one-to-one property is a powerful tool that simplifies the process of solving logarithmic equations. By understanding its conditions and applying it correctly, we can efficiently tackle a wide range of logarithmic problems.

Now, let's apply the one-to-one property to solve the given equation: ln⁑(βˆ’15βˆ’19d)=ln⁑(18d){ \ln(-15 - 19d) = \ln(18d) }. This equation involves natural logarithms (ln⁑{ \ln }), which are logarithms with base e{ e } (Euler's number, approximately 2.71828).

Step-by-Step Solution

  1. Verify the Conditions:

    • Same Base: Both logarithms are natural logarithms, meaning they have the same base e{ e }.
    • Positive Arguments: We need to ensure that both βˆ’15βˆ’19d{ -15 - 19d } and 18d{ 18d } are positive. This will be checked later after finding the solution for d{ d }.
  2. Apply the One-to-One Property:

    Since the bases are the same, we can set the arguments equal to each other: βˆ’15βˆ’19d=18d{ -15 - 19d = 18d }

  3. Solve the Resulting Equation:

    Now, we solve the linear equation for d{ d }:

    • Add 19d{ 19d } to both sides: βˆ’15=18d+19d{ -15 = 18d + 19d }
    • Combine like terms: βˆ’15=37d{ -15 = 37d }
    • Divide both sides by 37: d=βˆ’1537{ d = \frac{-15}{37} }

    So, the solution for d{ d } is βˆ’1537{ \frac{-15}{37} }.

  4. Check the Solutions:

    We must check if this solution makes the arguments of the original logarithms positive.

    • Check βˆ’15βˆ’19d{ -15 - 19d }: βˆ’15βˆ’19(βˆ’1537)=βˆ’15+28537{ -15 - 19\left(\frac{-15}{37}\right) = -15 + \frac{285}{37} } To combine these, we need a common denominator: βˆ’15β‹…3737+28537=βˆ’55537+28537=βˆ’27037{ \frac{-15 \cdot 37}{37} + \frac{285}{37} = \frac{-555}{37} + \frac{285}{37} = \frac{-270}{37} } Since βˆ’27037{ \frac{-270}{37} } is negative, the argument βˆ’15βˆ’19d{ -15 - 19d } is not positive.
    • Check 18d{ 18d }: 18(βˆ’1537)=βˆ’27037{ 18\left(\frac{-15}{37}\right) = \frac{-270}{37} } This is also negative, so the argument 18d{ 18d } is not positive.

    Since both arguments are negative when d=βˆ’1537{ d = \frac{-15}{37} }, this solution is extraneous and not valid. Therefore, the equation has no real solution.

Conclusion

After solving the equation ln⁑(βˆ’15βˆ’19d)=ln⁑(18d){ \ln(-15 - 19d) = \ln(18d) } using the one-to-one property of logarithms, we found a potential solution of d=βˆ’1537{ d = \frac{-15}{37} }. However, upon checking this solution in the original equation, we discovered that it makes the arguments of the logarithms negative, which is not allowed. Thus, there is no real solution for this equation.

When solving logarithmic equations, it’s easy to make mistakes if you're not careful. Understanding common pitfalls and how to avoid them can greatly improve your accuracy and efficiency. Here are some frequent errors and strategies to prevent them:

1. Forgetting to Check for Extraneous Solutions

Pitfall: The most common mistake is solving the equation and finding a value for the variable, but failing to check whether this value makes the arguments of the logarithms positive. Logarithms are only defined for positive arguments, so any solution that results in a non-positive argument is extraneous.

How to Avoid: Always substitute your solution(s) back into the original equation and check if the arguments of the logarithms are positive. If an argument is zero or negative, that solution is extraneous and must be discarded. This step is crucial for solving logarithmic equations accurately.

2. Incorrectly Applying Logarithmic Properties

Pitfall: Logarithmic properties, such as the product rule, quotient rule, and power rule, must be applied correctly. Misusing these properties can lead to incorrect simplification and, consequently, wrong solutions.

How to Avoid: Review the logarithmic properties thoroughly and ensure you understand when and how to apply them. For instance:

*   **Product Rule:** ${ \log_b{(MN)} = \log_b{M} + \log_b{N} }$
*   **Quotient Rule:** ${ \log_b{\left(\frac{M}{N}\right)} = \log_b{M} - \log_b{N} }$
*   **Power Rule:** ${ \log_b{M^k} = k \log_b{M} }$

Practice applying these properties in various contexts to build confidence and accuracy. Write out each step clearly to minimize errors.

3. Ignoring the Domain of Logarithmic Functions

Pitfall: The domain of a logarithmic function is restricted to positive real numbers. Ignoring this can lead to accepting solutions that are not valid within the domain.

How to Avoid: Be mindful of the domain of logarithmic functions. Before solving, identify any restrictions on the variable due to the arguments of the logarithms. For example, if you have log⁑b(xβˆ’2){ \log_b{(x - 2)} }, then xβˆ’2>0{ x - 2 > 0 }, so x>2{ x > 2 }. Keep these restrictions in mind when checking your solutions.

4. Making Algebraic Errors

Pitfall: Errors in basic algebra, such as incorrect distribution, combining like terms improperly, or sign errors, can derail the solution process.

How to Avoid: Practice algebraic manipulation regularly. Write out each step in detail, especially when dealing with complex equations. Double-check your work to catch any mistakes. Consider using a math tool or calculator to verify intermediate steps if necessary.

5. Assuming All Logarithms Have the Same Base

Pitfall: The one-to-one property and other logarithmic properties can only be applied directly when logarithms have the same base. If bases are different, you must first use the change of base formula or other methods to make them the same.

How to Avoid: Always verify that the logarithms have the same base before applying properties or the one-to-one principle. If the bases differ, use the change of base formula: log⁑bx=log⁑axlog⁑ab{ \log_b{x} = \frac{\log_a{x}}{\log_a{b}} } This allows you to convert logarithms to a common base before proceeding.

6. Not Simplifying Before Applying Properties

Pitfall: Applying logarithmic properties prematurely can sometimes complicate the equation unnecessarily. It's often best to simplify the equation as much as possible before using logarithmic rules.

How to Avoid: Look for opportunities to simplify the equation first. This might involve combining constants, distributing terms, or simplifying algebraic expressions within the arguments of the logarithms. Simplifying first can make the application of logarithmic properties more straightforward.

Example: Avoiding Extraneous Solutions

Consider the equation log⁑2(xβˆ’3)+log⁑2(xβˆ’1)=3{ \log_2{(x - 3)} + \log_2{(x - 1)} = 3 }.

  1. Combine Logarithms: Use the product rule to combine the logarithms: log⁑2((xβˆ’3)(xβˆ’1))=3{ \log_2{((x - 3)(x - 1))} = 3 }

  2. Convert to Exponential Form: (xβˆ’3)(xβˆ’1)=23{ (x - 3)(x - 1) = 2^3 }

  3. Simplify and Solve: x2βˆ’4x+3=8{ x^2 - 4x + 3 = 8 } x2βˆ’4xβˆ’5=0{ x^2 - 4x - 5 = 0 } (xβˆ’5)(x+1)=0{ (x - 5)(x + 1) = 0 } So, x=5{ x = 5 } or x=βˆ’1{ x = -1 }.

  4. Check for Extraneous Solutions:

    • For x=5{ x = 5 }: log⁑2(5βˆ’3)+log⁑2(5βˆ’1)=log⁑22+log⁑24=1+2=3{ \log_2{(5 - 3)} + \log_2{(5 - 1)} = \log_2{2} + \log_2{4} = 1 + 2 = 3 } This solution is valid.
    • For x=βˆ’1{ x = -1 }: log⁑2(βˆ’1βˆ’3)+log⁑2(βˆ’1βˆ’1)=log⁑2(βˆ’4)+log⁑2(βˆ’2){ \log_2{(-1 - 3)} + \log_2{(-1 - 1)} = \log_2{(-4)} + \log_2{(-2)} } Since the arguments are negative, x=βˆ’1{ x = -1 } is an extraneous solution.

By being aware of these common pitfalls and consistently applying strategies to avoid them, you can improve your accuracy and confidence in solving logarithmic equations.

In this comprehensive guide, we've explored the one-to-one property of logarithms and its application in solving logarithmic equations. We began by defining the property, emphasizing the conditions under which it can be applied: the logarithms must have the same base, and their arguments must be positive. We then demonstrated how to use this property to solve the equation ln⁑(βˆ’15βˆ’19d)=ln⁑(18d){ \ln(-15 - 19d) = \ln(18d) }, highlighting the importance of checking for extraneous solutions.

The step-by-step solution illustrated the process of setting the arguments equal to each other, solving the resulting algebraic equation, and verifying the solution by ensuring the arguments of the original logarithms remain positive. In this particular case, the initial solution obtained was extraneous, leading to the conclusion that the equation has no real solution. This underscores a critical aspect of solving logarithmic equations: the necessity of validating solutions within the domain of logarithmic functions.

Furthermore, we delved into common pitfalls encountered when solving logarithmic equations, such as forgetting to check for extraneous solutions, incorrectly applying logarithmic properties, ignoring the domain of logarithmic functions, making algebraic errors, assuming all logarithms have the same base, and not simplifying before applying properties. For each pitfall, we provided clear strategies to avoid these mistakes, ensuring a more accurate and efficient problem-solving process.

Mastering the one-to-one property, along with other logarithmic properties and solution verification techniques, is essential for anyone studying mathematics, particularly in areas such as algebra, calculus, and beyond. The ability to solve logarithmic equations is not only a valuable skill in academic settings but also in various real-world applications, including scientific research, engineering, finance, and computer science.

By understanding the concepts and practicing the techniques discussed in this article, you can confidently tackle logarithmic equations and enhance your mathematical proficiency. Remember, the key to success lies in a solid grasp of the fundamental principles, careful application of properties, and meticulous verification of solutions. With these tools at your disposal, you'll be well-equipped to navigate the complexities of logarithmic equations and excel in your mathematical endeavors.