Solving Logarithmic Equations: A Step-by-Step Guide

Are you grappling with logarithmic equations? Do logarithmic equations seem like a puzzle you can't quite solve? Fear not! This comprehensive guide will walk you through the process of solving the equation $\log _6(x^2+8)=1+\log _6(x)$, providing a clear, step-by-step solution and offering valuable insights into the world of logarithms. By the end of this article, you'll not only be able to solve this specific equation but also gain a solid understanding of the techniques involved in tackling similar logarithmic problems.

Understanding Logarithms

Before diving into the solution, let's quickly recap what logarithms are. A logarithm is essentially the inverse operation of exponentiation. The logarithm of a number to a given base is the exponent to which the base must be raised to produce that number. In simpler terms, if we have $b^y = x$, then the logarithm of $x$ to the base $b$ is $y$, written as $\log_b(x) = y$. Understanding this fundamental relationship is crucial for manipulating and solving logarithmic equations.

Logarithms have several important properties that we'll use in solving our equation. Some key properties include:

• Product Rule: $\log_b(mn) = \log_b(m) + \log_b(n)$
• Quotient Rule: $\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n)$
• Power Rule: $\log_b(m^p) = p \log_b(m)$
• Change of Base Formula: $\log_b(a) = \frac{\log_c(a)}{\log_c(b)}$
• Logarithm of the Base: $\log_b(b) = 1$

These properties allow us to simplify complex logarithmic expressions and transform equations into more manageable forms. In the context of our problem, we'll primarily use the properties related to combining logarithms and the definition of a logarithm to eliminate the logarithmic terms.

Step-by-Step Solution

Now, let's tackle the equation $\log _6(x^2+8)=1+\log _6(x)$. We'll break down the solution into manageable steps, explaining the reasoning behind each operation.

Step 1: Combine Logarithmic Terms

Our first goal is to bring all the logarithmic terms to one side of the equation. We can do this by subtracting $\log_6(x)$ from both sides:

$\log _6(x^2+8) - \log _6(x) = 1$

Now, we can use the quotient rule of logarithms, which states that $\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n)$. Applying this rule, we get:

$\log _6(\frac{x^2+8}{x}) = 1$

This step is crucial because it condenses the two logarithmic terms into a single term, making it easier to eliminate the logarithm in the next step.

Step 2: Convert to Exponential Form

To eliminate the logarithm, we need to convert the equation from logarithmic form to exponential form. Recall the definition of a logarithm: if $\log_b(x) = y$, then $b^y = x$. Applying this definition to our equation, where the base is 6, the argument is $\frac{x^2+8}{x}$, and the result is 1, we get:

$6^1 = \frac{x^2+8}{x}$

This simplifies to:

$6 = \frac{x^2+8}{x}$

By converting to exponential form, we've transformed the logarithmic equation into a standard algebraic equation, which we can solve using familiar techniques.

Step 3: Solve the Quadratic Equation

To solve for $x$, we first multiply both sides of the equation by $x$ to get rid of the fraction:

$6x = x^2 + 8$

Now, rearrange the equation to form a quadratic equation by subtracting $6x$ from both sides:

$0 = x^2 - 6x + 8$

This is a quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = -6$, and $c = 8$. We can solve this equation by factoring. We need to find two numbers that multiply to 8 and add up to -6. These numbers are -2 and -4. Thus, we can factor the quadratic as:

$(x - 2)(x - 4) = 0$

Setting each factor equal to zero gives us the possible solutions:

$x - 2 = 0$ or $x - 4 = 0$

Solving for $x$, we find:

$x = 2$ or $x = 4$

These are our potential solutions for the equation. However, we need to check these solutions in the original equation to ensure they are valid.

Step 4: Check for Extraneous Solutions

It's crucial to check our solutions in the original logarithmic equation because logarithms are only defined for positive arguments. This means we need to make sure that the values we found for $x$ don't result in taking the logarithm of a negative number or zero in the original equation. The original equation is:

$\log _6(x^2+8)=1+\log _6(x)$

Let's check $x = 2$:

$\log _6(2^2+8) = \log _6(4+8) = \log _6(12)$

$1 + \log _6(2)$

Using the logarithm property to combine the right side $ \log _6(6) + \log _6(2) = \log _6(6*2) = \log _6(12)$.

Since both sides are equal, $x = 2$ is a valid solution.

Now, let's check $x = 4$:

$\log _6(4^2+8) = \log _6(16+8) = \log _6(24)$

$1 + \log _6(4)$

Using the logarithm property to combine the right side $ \log _6(6) + \log _6(4) = \log _6(6*4) = \log _6(24)$.

Since both sides are equal, $x = 4$ is also a valid solution.

Therefore, both $x = 2$ and $x = 4$ are solutions to the equation.

Conclusion

In this guide, we've walked through a detailed solution to the logarithmic equation $\log _6(x^2+8)=1+\log _6(x)$. We started by understanding the fundamental properties of logarithms, then applied these properties to simplify the equation. We converted the logarithmic equation to an algebraic equation, solved the resulting quadratic equation, and finally, checked for extraneous solutions. The solutions to the equation are $x = 2$ and $x = 4$. By following these steps, you can confidently solve a wide range of logarithmic equations. Remember to always check your solutions in the original equation to avoid extraneous solutions, which are values that satisfy the transformed equation but not the original one. Keep practicing, and you'll master the art of solving logarithmic equations!

For further exploration and practice, you might find helpful resources on websites like Khan Academy's Logarithm Section.