Solve Log((x-9)/(4x)) = 0: A Step-by-Step Guide

Hey guys! Today, we're going to embark on a mathematical adventure to solve a fascinating logarithmic equation: $\log \frac{x-9}{4 x}=0$. Logarithmic equations might seem daunting at first, but trust me, with a bit of understanding and the right approach, they become quite manageable. We'll break down each step, ensuring you grasp the underlying concepts and feel confident tackling similar problems in the future. So, grab your thinking caps, and let's dive in!

Understanding Logarithms: The Foundation of Our Journey

Before we jump into solving the equation, let's take a moment to solidify our understanding of logarithms. Logarithms are essentially the inverse operation of exponentiation. Think of it this way: if we have an exponential expression like $b^y = x$, the logarithm answers the question, "What power ($y$) do we need to raise the base ($b$) to in order to get $x$?" This is represented as $\log_b x = y$.

In our equation, $\log \frac{x-9}{4 x}=0$, we're dealing with a common logarithm, which means the base is 10 (even though it's not explicitly written). So, the equation is essentially asking, "To what power do we need to raise 10 to get $\frac{x-9}{4 x}$?" The answer, according to the equation, is 0.

Now, let's delve deeper into the properties of logarithms that will be crucial for solving our equation:

• The fundamental property: This is the key to unlocking the equation! The definition of a logarithm tells us that if $\log_b x = y$, then $b^y = x$. This allows us to convert a logarithmic equation into its equivalent exponential form.
• The logarithm of 1: An extremely important property is that the logarithm of 1 to any base is always 0. Mathematically, $\log_b 1 = 0$. This is because any number raised to the power of 0 equals 1.
• The domain of logarithms: This is a critical consideration. Logarithms are only defined for positive arguments. In other words, we can only take the logarithm of a positive number. This means that in our equation, the expression $\frac{x-9}{4 x}$ must be greater than 0. We'll need to keep this in mind when we find our solutions and check for extraneous solutions.

Understanding these fundamental concepts and properties is paramount to successfully navigating logarithmic equations. They provide the tools and the framework for us to manipulate and solve these types of problems. So, with this knowledge in our arsenal, let's move on to the next step: transforming our logarithmic equation into a more manageable form.

Transforming the Equation: From Logarithmic to Exponential

Alright, now that we have a solid grasp of logarithms, let's get our hands dirty and start solving the equation $\log \frac{x-9}{4 x}=0$. The first crucial step is to transform this logarithmic equation into its equivalent exponential form. Remember that fundamental property we discussed? If $\log_b x = y$, then $b^y = x$. We're going to use that principle right now.

In our case, the base b is 10 (since it's a common logarithm), x is the expression $\frac{x-9}{4 x}$, and y is 0. Applying the property, we can rewrite the equation as:

$$10^0 = \frac{x-9}{4 x}$$

This transformation is a game-changer! We've successfully eliminated the logarithm, and now we have a simple algebraic equation to solve. But before we rush into solving, let's simplify this a bit further. We know that any non-zero number raised to the power of 0 is 1. So, $10^0$ simplifies to 1. Our equation now looks like this:

$$1 = \frac{x-9}{4 x}$$

See how much cleaner and easier to work with this is? This is the power of understanding logarithmic properties. By converting the equation from logarithmic to exponential form, we've essentially removed the complexity and made it accessible to standard algebraic techniques.

Now, to solve for x, we need to get rid of the fraction. We can do this by multiplying both sides of the equation by $4x$. This is a valid algebraic manipulation as long as $x$ is not equal to 0 (we'll need to keep this in mind when we check our solutions later). Multiplying both sides by $4x$ gives us:

$$4x * 1 = 4x * \frac{x-9}{4 x}$$

Simplifying, we get:

$$4x = x - 9$$

We're getting closer! Now we have a linear equation that's very straightforward to solve. The next step involves isolating x on one side of the equation. This transformation from a logarithmic equation to a simple linear equation highlights the beauty and power of mathematical principles. By applying the fundamental property of logarithms, we've effectively "unlocked" the equation and made it solvable. In the next section, we'll tackle the remaining algebraic steps to find the value(s) of x that satisfy the equation.

Solving for x: The Algebraic Journey

We've successfully transformed our logarithmic equation into a linear equation: $4x = x - 9$. Now, it's time to put on our algebraic hats and solve for x. The goal here is to isolate x on one side of the equation, revealing its value.

To begin, let's subtract x from both sides of the equation. This will consolidate the x terms on the left side:

$$4x - x = x - 9 - x$$

Simplifying, we get:

$$3x = -9$$

We're almost there! Now, to completely isolate x, we need to divide both sides of the equation by 3:

$$\frac{3x}{3} = \frac{-9}{3}$$

This gives us our solution:

$$x = -3$$

Fantastic! We've found a potential solution for x. But hold on a second! We're not quite done yet. Remember that crucial point we discussed earlier about the domain of logarithms? We need to check if this solution is valid within the context of the original logarithmic equation.

Checking for Extraneous Solutions: The Crucial Verification Step

We've arrived at a potential solution: $x = -3$. However, in the world of logarithmic equations (and many other types of equations, for that matter), it's absolutely crucial to check for extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but do not satisfy the original equation. They often arise due to the restrictions on the domain of the functions involved.

In our case, the original equation is $\log \frac{x-9}{4 x}=0$. Remember that the argument of a logarithm (the expression inside the logarithm) must be strictly positive. This means that $\frac{x-9}{4 x}$ must be greater than 0. Let's plug our solution, $x = -3$, into this expression and see what happens:

$$\frac{-3 - 9}{4 * -3} = \frac{-12}{-12} = 1$$

Okay, so the argument of the logarithm is 1 when $x = -3$. This seems fine, since 1 is a positive number. But we need to dig a little deeper. We also need to ensure that both the numerator ($x - 9$) and the denominator ($4x$) have the same sign (either both positive or both negative) for the fraction to be positive.

Let's analyze the numerator and denominator separately when $x = -3$:

• Numerator: $x - 9 = -3 - 9 = -12$ (Negative)
• Denominator: $4x = 4 * -3 = -12$ (Negative)

Since both the numerator and the denominator are negative, their quotient is indeed positive. So far, so good. However, there's another potential issue we need to consider. We also need to make sure that the expressions inside the original equation are defined. In this case, we need to make sure that $x - 9$ and $4x$ are not equal to zero.

Since $x=-3$, we have:

• $x-9 = -3-9 = -12 \neq 0$
• $4x = 4(-3) = -12 \neq 0$

Therefore, $x = -3$ is a valid solution.

The Final Verdict: Our Solution to the Logarithmic Puzzle

After a thorough journey through the world of logarithms and algebraic manipulations, we've arrived at our solution. We started with the equation $\log \frac{x-9}{4 x}=0$, transformed it into an exponential equation, solved for x, and meticulously checked for extraneous solutions.

Our hard work has paid off, and we can confidently state that the solution to the equation is:

$$x = -3$$

This entire process highlights the importance of understanding the fundamental properties of logarithms and the necessity of checking for extraneous solutions. Logarithmic equations can be tricky, but by breaking them down into manageable steps and applying the correct principles, we can conquer them with confidence. So, keep practicing, keep exploring, and keep unlocking the secrets of mathematics!