Solving Logarithmic Equations Algebraically A Step-by-Step Guide

Logarithmic equations can seem daunting at first, but with a systematic approach and a solid understanding of logarithmic properties, they can be solved effectively. This guide will walk you through the process of solving the logarithmic equation $\log _5(x)+\log _5(x+1)=\log _5 6$ algebraically, providing a clear and detailed explanation of each step.

Understanding Logarithms

Before diving into the solution, let's briefly review what logarithms are and their fundamental properties. A logarithm is essentially the inverse operation of exponentiation. The expression $\log_b a = c$ means that $b^c = a$, where 'b' is the base, 'a' is the argument, and 'c' is the logarithm. Understanding this relationship is crucial for manipulating and solving logarithmic equations.

Key Properties of Logarithms

Several properties of logarithms are essential for solving equations:

1. Product Rule: $\log_b(mn) = \log_b m + \log_b n$
2. Quotient Rule: $\log_b(\frac{m}{n}) = \log_b m - \log_b n$
3. Power Rule: $\log_b(m^p) = p \log_b m$
4. Logarithm of 1: $\log_b 1 = 0$
5. Logarithm of the Base: $\log_b b = 1$
6. One-to-One Property: If $\log_b m = \log_b n$, then $m = n$

In this particular problem, the product rule and the one-to-one property will be particularly useful.

Problem Statement

Let's restate the problem clearly. We are tasked with solving the following logarithmic equation algebraically:

$\log _5(x)+\log _5(x+1)=\log _5 6$

Our goal is to find the value(s) of 'x' that satisfy this equation. Solving this equation requires a careful application of logarithmic properties and algebraic techniques.

Step-by-Step Solution

Step 1: Apply the Product Rule

The left side of the equation involves the sum of two logarithms with the same base. According to the product rule, we can combine these into a single logarithm:

$\log _5(x)+\log _5(x+1) = \log _5(x(x+1))$

So, the equation becomes:

$\log _5(x(x+1)) = \log _5 6$

By applying the product rule, we've simplified the equation into a more manageable form. This step is crucial for isolating the variable.

Step 2: Apply the One-to-One Property

Now we have a single logarithm on each side of the equation, both with the same base (5). The one-to-one property of logarithms states that if $\log_b m = \log_b n$, then $m = n$. Applying this property, we can eliminate the logarithms:

$x(x+1) = 6$

This step transforms the logarithmic equation into a quadratic equation. The one-to-one property is a powerful tool in solving logarithmic equations.

Step 3: Simplify and Rearrange

Next, we expand the left side of the equation and rearrange it into a standard quadratic form:

$x^2 + x = 6$

Subtract 6 from both sides:

$x^2 + x - 6 = 0$

Rearranging the equation into a standard form allows us to apply standard techniques for solving quadratic equations.

Step 4: Solve the Quadratic Equation

We now have a quadratic equation in the form $ax^2 + bx + c = 0$. We can solve this by factoring, completing the square, or using the quadratic formula. In this case, factoring is the most straightforward method.

We look for two numbers that multiply to -6 and add to 1. These numbers are 3 and -2. So, we can factor the quadratic as follows:

$(x + 3)(x - 2) = 0$

Setting each factor equal to zero gives us two potential solutions:

$x + 3 = 0$ or $x - 2 = 0$

Solving for x:

$x = -3$ or $x = 2$

Solving the quadratic equation provides us with potential solutions to the original logarithmic equation. However, it's crucial to check these solutions.

Step 5: Check for Extraneous Solutions

When solving logarithmic equations, it's essential to check for extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but do not satisfy the original logarithmic equation. This often happens because the domain of a logarithmic function is restricted to positive arguments.

We need to check both $x = -3$ and $x = 2$ in the original equation:

$\log _5(x)+\log _5(x+1)=\log _5 6$

Checking x = -3:

$\log _5(-3)+\log _5(-3+1)=\log _5 6$

$\log _5(-3)+\log _5(-2)=\log _5 6$

Since the logarithm of a negative number is undefined, $x = -3$ is an extraneous solution. Checking for extraneous solutions is a critical step in solving logarithmic equations.

Checking x = 2:

$\log _5(2)+\log _5(2+1)=\log _5 6$

$\log _5(2)+\log _5(3)=\log _5 6$

Using the product rule:

$\log _5(2 imes 3)=\log _5 6$

$\log _5(6)=\log _5 6$

This is true, so $x = 2$ is a valid solution.

Step 6: State the Solution

After checking for extraneous solutions, we find that only $x = 2$ satisfies the original equation. Therefore, the solution to the logarithmic equation is:

$x = 2$

Stating the solution clearly is the final step in the problem-solving process.

Conclusion

We have successfully solved the logarithmic equation $\log _5(x)+\log _5(x+1)=\log _5 6$ algebraically. The steps involved included applying the product rule, using the one-to-one property, solving a quadratic equation, and, most importantly, checking for extraneous solutions. Mastering these steps will enable you to solve a wide range of logarithmic equations.

By following this detailed guide, you can confidently approach and solve logarithmic equations. Remember to always check your solutions to avoid extraneous results and ensure the validity of your answers. Understanding the properties of logarithms and applying them correctly is the key to success in solving these types of equations. This comprehensive approach provides a strong foundation for tackling more complex logarithmic problems in the future.