Solve Logarithmic Equations: A Step-by-Step Guide

Ever stared at a logarithmic equation and felt a pang of dread? You're not alone! Logarithms can seem a bit intimidating at first, but with the right approach, they become quite manageable. Today, we're going to tackle a specific type of logarithmic equation: $\\log _4(x-3)+\\log _4(x+3)=2$. This isn't just about finding an answer; it's about understanding the process that leads us to the correct solution set. We'll break down each step, explain the reasoning behind it, and ensure you feel confident in solving similar problems. Get ready to demystify logarithmic equations and discover the elegance of their solutions!

Understanding the Properties of Logarithms

Before we dive into solving our specific equation, it's crucial to have a firm grasp on the fundamental properties of logarithms. These properties are our toolkit for simplifying and manipulating logarithmic expressions. The one we'll heavily rely on today is the product rule for logarithms. This rule states that the sum of logarithms with the same base is equal to the logarithm of the product of their arguments. Mathematically, this is expressed as: $\\log_b(M) + \\log_b(N) = \\log_b(M imes N)$. Think of it as combining two separate logarithmic terms into a single, more manageable one. It's like merging two smaller ingredients into a single, more potent mixture. This property is indispensable when dealing with equations where you have multiple logarithmic terms added together. Without it, isolating the variable would be a much more convoluted process. Another property that often comes into play is the definition of a logarithm itself: if $\\log_b(y) = x$, then $b^x = y$. This allows us to convert a logarithmic equation into an exponential one, which is often easier to solve directly. We'll see how these two properties work in tandem to unravel our problem.

Applying the Product Rule

Now, let's apply the product rule to our equation: $\\log _4(x-3)+\\log _4(x+3)=2$. Notice that both logarithms have the same base, which is 4. This is the key that allows us to use the product rule. We can combine the two logarithmic terms on the left side of the equation into a single logarithm: $\\log _4((x-3)(x+3))=2$. This step is critical because it simplifies the equation significantly. Instead of dealing with two separate logarithmic expressions, we now have just one. This is where the power of logarithmic properties truly shines. The expression $(x-3)(x+3)$ might look familiar to some of you. It's a classic example of the difference of squares factorization, which states that $(a-b)(a+b) = a^2 - b^2$. Applying this to our expression, we get $(x-3)(x+3) = x^2 - 3^2 = x^2 - 9$. So, our equation transforms into a much simpler form: $\\log _4(x^2-9)=2$. This transformation is a testament to the algebraic manipulation that accompanies logarithmic solving. We're not just blindly applying rules; we're using algebraic identities to make the equation even more streamlined. This iterative simplification is what makes complex problems solvable.

Converting to Exponential Form

We've successfully simplified the equation to $\\log _4(x^2-9)=2$. Now, to solve for 'x', we need to get rid of the logarithm. This is where the definition of a logarithm comes into play. Remember, if $\\log_b(y) = x$, then $b^x = y$. In our equation, the base 'b' is 4, the result 'x' (the exponent) is 2, and the argument 'y' is $(x^2-9)$. Applying the definition, we can rewrite the equation in its exponential form: $4^2 = x^2-9$. This conversion is a pivotal moment in solving the equation because it transforms a logarithmic equation into a standard algebraic equation that we know how to solve using basic arithmetic and algebra. The exponential form is often more intuitive for isolating variables. Calculating $4^2$ gives us 16. So, our equation becomes $16 = x^2-9$. This is a significant simplification, moving us closer to finding the value(s) of x. The transition from logarithmic to exponential form is a fundamental technique that opens the door to solving a wide array of logarithmic equations. It's like switching from a coded message to plain text – suddenly, everything is much clearer.

Solving the Quadratic Equation

We've arrived at the equation $16 = x^2-9$. This is a quadratic equation, and we can solve it by isolating the $x^2$ term and then taking the square root. First, let's add 9 to both sides of the equation to isolate $x^2$: $16 + 9 = x^2$. This gives us $25 = x^2$. Now, to find the value(s) of x, we need to take the square root of both sides: $\\\sqrt{25} = \\sqrt{x^2}$. Remember that when taking the square root of both sides of an equation, there are generally two possible solutions: a positive and a negative one. Therefore, $x = \\pm 5$. This means our potential solutions are $x = 5$ and $x = -5$. This is where careful consideration comes into play, as we're not done yet. We need to check these potential solutions against the original equation to ensure they are valid.

The Importance of Domain and Extraneous Solutions

It's crucial to understand that not all potential solutions derived from algebraic manipulation are valid for logarithmic equations. This is due to the domain restrictions of logarithms. The argument of a logarithm (the expression inside the parentheses) must always be positive. In our original equation, $\\log _4(x-3)+\\log _4(x+3)=2$, we have two arguments: $(x-3)$ and $(x+3)$. For the logarithms to be defined, both of these must be greater than zero:

• $x-3 > 0 \\implies x > 3$
• $x+3 > 0 \\implies x > -3$

For both conditions to be true, 'x' must be greater than 3. This combined condition, $x > 3$, represents the valid domain for our equation. Now, let's check our potential solutions, $x=5$ and $x=-5$, against this domain:

• For $x = 5$: Is $5 > 3$? Yes, it is. This solution satisfies the domain restriction.
• For $x = -5$: Is $-5 > 3$? No, it is not. This solution does not satisfy the domain restriction.

Therefore, $x = -5$ is an extraneous solution. It's a solution that arises during the solving process but doesn't work in the original equation. This highlights the importance of always checking your solutions. The only valid solution is $x = 5$. This process of verifying solutions is a non-negotiable step in solving logarithmic equations to ensure accuracy and avoid errors.

Conclusion: The Solution Set

After carefully following the steps – applying logarithmic properties to simplify the equation, converting to exponential form, solving the resulting algebraic equation, and critically checking for extraneous solutions against the domain – we have arrived at our final answer. The equation $\\log _4(x-3)+\\log _4(x+3)=2$ has one valid solution. We found that $x=5$ satisfies the domain requirements ($x>3$) and the original equation, while $x=-5$ does not. Thus, the solution set for this equation is simply x = 5. Understanding this entire process, from the basic rules of logarithms to the final domain check, empowers you to confidently tackle a wide range of similar problems. Remember, practice and attention to detail are your greatest allies in mathematics!

For further exploration and resources on solving logarithmic equations and understanding their properties, you can visit Khan Academy's mathematics section or Brilliant.org.