Solving Ln(x) = Ln(5) + Ln(x-3) A Step-by-Step Guide

July 12, 2025 by Scholario Team 53 views

Solving the Logarithmic Equation: ln(x) = ln(5) + ln(x-3)

In the realm of mathematics, solving equations is a fundamental skill. Among the various types of equations, logarithmic equations hold a significant place, particularly in fields like calculus, physics, and engineering. Logarithmic equations involve logarithmic functions, which are the inverse of exponential functions. This article delves into a step-by-step approach to solving a specific logarithmic equation, $\ln x = \ln 5 + \ln (x-3)$ , while providing a broader understanding of the techniques and principles involved in solving logarithmic equations in general.

Understanding Logarithms

Before we dive into solving the equation, it's crucial to have a solid grasp of what logarithms are. A logarithm answers the question: "To what power must we raise a base to get a certain number?" The expression $\log_b a = c$ means that $b^c = a$ , where 'b' is the base, 'a' is the argument (the number we're taking the logarithm of), and 'c' is the exponent (the logarithm itself). In our equation, we encounter the natural logarithm, denoted as $\ln$ . The natural logarithm is simply the logarithm to the base 'e', where 'e' is an irrational number approximately equal to 2.71828. Thus, $\ln x$ is equivalent to $\log_e x$ .

Properties of Logarithms

Several key properties of logarithms are essential for solving logarithmic equations. These include:

Product Rule: $\log_b (mn) = \log_b m + \log_b n$
Quotient Rule: $\log_b (m/n) = \log_b m - \log_b n$
Power Rule: $\log_b (m^p) = p \log_b m$
Logarithm of the Base: $\log_b b = 1$
Logarithm of 1: $\log_b 1 = 0$
One-to-One Property: If $\log_b m = \log_b n$ , then $m = n$

These properties allow us to manipulate logarithmic expressions, combine or separate logarithms, and ultimately simplify equations to solve for the unknown variable. In the context of our problem, the product rule will be particularly useful.

Solving the Equation: ln(x) = ln(5) + ln(x-3)

Now, let's tackle the equation $\ln x = \ln 5 + \ln (x-3)$ . We'll break down the solution into a series of steps:

Step 1: Apply the Product Rule

Our main objective is to simplify the equation. The right side of the equation involves the sum of two logarithms. We can use the product rule of logarithms, which states that the sum of logarithms is equal to the logarithm of the product. Applying this rule, we get:

$\ln x = \ln [5(x-3)]$

This step combines the two logarithmic terms on the right side into a single term, making the equation easier to work with.

Step 2: Use the One-to-One Property

Now we have a logarithm on both sides of the equation. The one-to-one property of logarithms states that if $\log_b m = \log_b n$ , then $m = n$ . In our case, this means that if $\ln x = \ln [5(x-3)]$ , then:

$x = 5(x-3)$

This step eliminates the logarithms, transforming the logarithmic equation into a simple algebraic equation.

Step 3: Solve the Algebraic Equation

We now have a linear equation in terms of 'x'. Let's solve for 'x':

$x = 5x - 15$

Subtract 5x from both sides:

$-4x = -15$

Divide both sides by -4:

$x = \frac{15}{4}$

So, we have found a potential solution: $x = \frac{15}{4}$ .

Step 4: Check for Extraneous Solutions

An essential step in solving logarithmic equations is checking for extraneous solutions. Extraneous solutions are values that satisfy the transformed algebraic equation but do not satisfy the original logarithmic equation. This can occur because the domain of a logarithmic function is restricted to positive values. We must ensure that the arguments of all logarithms in the original equation are positive when we substitute our solution.

In our original equation, $\ln x = \ln 5 + \ln (x-3)$ , we have two logarithmic terms: $\ln x$ and $\ln (x-3)$ . For these to be defined, we must have:

$x > 0$
$x - 3 > 0$ which implies $x > 3$

Our solution, $x = \frac{15}{4} = 3.75$ , satisfies both of these conditions ( $3.75 > 0$ and $3.75 > 3$ ). Therefore, it is a valid solution.

Step 5: State the Solution

Having checked for extraneous solutions, we can confidently state that the solution to the equation $\ln x = \ln 5 + \ln (x-3)$ is:

$x = \frac{15}{4}$

General Strategies for Solving Logarithmic Equations

The process we followed for this specific equation highlights the general strategies for solving logarithmic equations. Here's a summary of these strategies:

Isolate Logarithmic Terms: Use algebraic manipulations to isolate the logarithmic terms on one side of the equation.
Apply Logarithmic Properties: Use the properties of logarithms (product rule, quotient rule, power rule) to combine multiple logarithmic terms into a single term.
Convert to Exponential Form: If you have a single logarithmic term, convert the equation to exponential form. For example, if $\log_b a = c$ , rewrite it as $b^c = a$ .
Use the One-to-One Property: If you have logarithms with the same base on both sides of the equation, use the one-to-one property to equate the arguments.
Solve the Resulting Equation: After eliminating the logarithms, you'll be left with an algebraic equation. Solve this equation using standard techniques.
Check for Extraneous Solutions: Always substitute your potential solutions back into the original logarithmic equation to check for extraneous solutions. This is crucial because the domain of logarithmic functions is restricted.

Common Mistakes to Avoid

When solving logarithmic equations, it's easy to make mistakes. Here are some common pitfalls to watch out for:

Forgetting to Check for Extraneous Solutions: This is the most common mistake. Always check your solutions to ensure they satisfy the original equation.
Incorrectly Applying Logarithmic Properties: Make sure you understand and apply the logarithmic properties correctly. For example, $\log (a + b)$ is not equal to $\log a + \log b$ .
Ignoring the Domain of Logarithms: Remember that the argument of a logarithm must be positive. Pay attention to this when checking for extraneous solutions and when determining the domain of the equation.
Dividing by Zero: Be careful not to divide by zero when solving the resulting algebraic equation. Check if any of your solutions make the denominator zero.

Examples of Other Logarithmic Equations

To further solidify your understanding, let's look at a few more examples of logarithmic equations and how to solve them.

Example 1: Solving an Equation with Multiple Logarithmic Terms

Solve for x: $\log_2 (x + 2) + \log_2 (x - 1) = 2$

Apply the product rule: $\log_2 [(x + 2)(x - 1)] = 2$
Convert to exponential form: $(x + 2)(x - 1) = 2^2$
Simplify and solve the quadratic equation: $x^2 + x - 2 = 4$ $x^2 + x - 6 = 0$ $(x + 3)(x - 2) = 0$ $x = -3$ or $x = 2$
Check for extraneous solutions: For $x = -3$ : $\log_2 (-3 + 2)$ and $\log_2 (-3 - 1)$ are undefined, so $x = -3$ is extraneous. For $x = 2$ : $\log_2 (2 + 2) = \log_2 4 = 2$ and $\log_2 (2 - 1) = \log_2 1 = 0$ , so the equation holds.
Solution: $x = 2$

Example 2: Solving an Equation with a Single Logarithmic Term

Solve for x: $2 \ln x = \ln 9$

Apply the power rule: $\ln (x^2) = \ln 9$
Use the one-to-one property: $x^2 = 9$
Solve for x: $x = \pm 3$
Check for extraneous solutions: For $x = -3$ : $\ln (-3)$ is undefined, so $x = -3$ is extraneous. For $x = 3$ : $2 \ln 3 = \ln (3^2) = \ln 9$ , so the equation holds.
Solution: $x = 3$

Applications of Logarithmic Equations

Logarithmic equations aren't just abstract mathematical exercises; they have practical applications in various fields. Here are a few examples:

Chemistry: The pH scale, which measures the acidity or alkalinity of a solution, is based on logarithms. The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration.
Physics: The decibel scale, used to measure sound intensity, is logarithmic. The intensity of a sound is often expressed in decibels relative to a reference intensity.
Finance: Compound interest calculations often involve logarithms. For example, determining the time it takes for an investment to double at a given interest rate can be solved using logarithmic equations.
Computer Science: Logarithms are used in the analysis of algorithms, particularly in determining the efficiency of search and sorting algorithms.
Earthquakes: The Richter scale, used to measure the magnitude of earthquakes, is logarithmic. Each whole number increase on the Richter scale represents a tenfold increase in the amplitude of the seismic waves.

Conclusion

Solving logarithmic equations requires a solid understanding of logarithms, their properties, and the techniques for manipulating logarithmic expressions. By following the steps outlined in this article and practicing with various examples, you can master the art of solving logarithmic equations. Remember to always check for extraneous solutions and be mindful of the domain of logarithmic functions. Whether you're a student, a scientist, or an engineer, the ability to solve logarithmic equations is a valuable skill that will serve you well in many contexts. The equation $\ln x = \ln 5 + \ln (x-3)$ , which we solved in detail, serves as a prime example of the process involved. By mastering the general strategies and avoiding common mistakes, you'll be well-equipped to tackle a wide range of logarithmic problems. The applications of logarithmic equations are vast and varied, highlighting their importance in diverse fields of study and real-world scenarios. As you continue your mathematical journey, remember that practice makes perfect, and the more you work with logarithmic equations, the more confident and proficient you'll become.

This comprehensive guide provides a strong foundation for understanding and solving logarithmic equations, empowering you to approach these problems with clarity and precision. Remember, the key is to break down the problem into manageable steps, apply the appropriate properties, and always verify your solutions. With dedication and practice, you'll unlock the power of logarithms and their applications in various domains. The journey of learning mathematics is a continuous one, and logarithmic equations are just one piece of the puzzle. Embrace the challenge, explore the concepts, and enjoy the process of discovery. By doing so, you'll not only enhance your mathematical skills but also gain a deeper appreciation for the beauty and elegance of mathematics itself.