Solving Logarithmic Equations Step By Step Log _5(5 X+2)-log _5(9 X-7)=2

Hey guys! Ever stumbled upon a logarithmic equation that just seems like a jumbled mess of numbers and logs? Don't worry, we've all been there! Logarithmic equations can seem intimidating at first, but with a systematic approach and a little bit of practice, you can totally conquer them. In this article, we're going to break down the solution to the equation $\log _5(5 x+2)-\log _5(9 x-7)=2$ step-by-step, making sure you understand the logic behind each move. So, grab your thinking caps, and let's dive into the fascinating world of logarithms!

Understanding Logarithms: The Foundation of Our Solution

Before we jump into solving the equation, let's quickly review the basics of logarithms. Think of a logarithm as the inverse operation of exponentiation. In simpler terms, if we have an equation like $a^b = c$, the logarithm answers the question: "To what power must we raise a to get c?" This is written as $\log_a(c) = b$. a is called the base of the logarithm, and it's crucial in determining the logarithm's value.

Why are logarithms important? Logarithms pop up in various fields, from calculating the magnitude of earthquakes (the Richter scale) to determining the pH of a solution in chemistry. They're also fundamental in computer science, particularly in analyzing the efficiency of algorithms. Understanding logarithms unlocks a powerful tool for solving problems in diverse areas.

In our equation, $\log _5(5 x+2)-\log _5(9 x-7)=2$, we're dealing with logarithms with a base of 5. This means we're asking, "To what power must we raise 5 to get $(5x + 2)$ or $(9x - 7)$?" The properties of logarithms, which we'll discuss next, will help us unravel this equation.

Key Properties of Logarithms

To effectively tackle logarithmic equations, you need to have a solid grasp of the properties of logarithms. These properties are like the secret weapons in your math arsenal!

1. The Quotient Rule: This rule is our primary weapon for this particular equation. It states that the logarithm of a quotient is equal to the difference of the logarithms. Mathematically, it's expressed as: $\log_b(x) - \log_b(y) = \log_b(\frac{x}{y})$. Notice that the bases of the logarithms must be the same for this rule to apply. This is perfect for our equation, where both logarithms have a base of 5!
2. The Power Rule: This rule says that the logarithm of a number raised to a power is equal to the power multiplied by the logarithm of the number. The formula is: $\log_b(x^p) = p \log_b(x)$. This rule isn't directly used in this specific problem, but it's a crucial tool in many other logarithmic equations.
3. The Product Rule: Similar to the quotient rule, this rule deals with the logarithm of a product. It states that the logarithm of a product is equal to the sum of the logarithms: $\log_b(xy) = \log_b(x) + \log_b(y)$.
4. Logarithm of the Base: This is a fundamental property: $\log_b(b) = 1$. Any number raised to the power of 1 equals itself.
5. Logarithm of 1: Another important property: $\log_b(1) = 0$. Any number raised to the power of 0 equals 1.

With these properties in our toolkit, we're well-equipped to solve our equation. The quotient rule is the key to simplifying the left side of the equation, allowing us to combine the two logarithmic terms into one.

Solving the Equation: A Step-by-Step Walkthrough

Okay, let's get our hands dirty and solve the equation $\log _5(5 x+2)-\log _5(9 x-7)=2$. We'll break it down into manageable steps, making sure you understand each action.

Step 1: Applying the Quotient Rule

The first thing we notice is that we have two logarithmic terms with the same base (5) being subtracted. This screams for the application of the quotient rule! Remember, $\log_b(x) - \log_b(y) = \log_b(\frac{x}{y})$.

Applying this rule to our equation, we get:

$\log _5(5 x+2)-\log _5(9 x-7) = \log_5(\frac{5x + 2}{9x - 7}) = 2$

Notice how we've combined the two logarithms into a single logarithm. This is a significant simplification, and it moves us closer to isolating x. The equation now looks much more manageable!

Step 2: Converting to Exponential Form

Now that we have a single logarithmic term, we can get rid of the logarithm altogether by converting the equation into its exponential form. Remember the relationship between logarithms and exponents: if $\log_a(c) = b$, then $a^b = c$.

In our case, we have $\log_5(\frac{5x + 2}{9x - 7}) = 2$. Applying the conversion, we get:

$5^2 = \frac{5x + 2}{9x - 7}$

We've successfully transformed the logarithmic equation into a regular algebraic equation. The logarithm is gone, and we're left with a fraction to deal with. This is something we can definitely handle!

Step 3: Simplifying and Solving for x

Let's simplify the equation further. We know that $5^2 = 25$, so we can rewrite the equation as:

$25 = \frac{5x + 2}{9x - 7}$

To get rid of the fraction, we can multiply both sides of the equation by $(9x - 7)$:

$25(9x - 7) = 5x + 2$

Now, let's distribute the 25 on the left side:

$225x - 175 = 5x + 2$

Next, we want to get all the x terms on one side and the constant terms on the other. Let's subtract $5x$ from both sides:

$220x - 175 = 2$

Now, add 175 to both sides:

$220x = 177$

Finally, divide both sides by 220 to isolate x:

$x = \frac{177}{220}$

We've found a potential solution for x! But hold on, we're not quite done yet. With logarithmic equations, it's crucial to check our solution to make sure it's valid.

Step 4: Checking for Extraneous Solutions

Why do we need to check for extraneous solutions? Logarithms are only defined for positive arguments. This means that the expressions inside the logarithms, $(5x + 2)$ and $(9x - 7)$ in our original equation, must be greater than zero. If our solution makes either of these expressions negative or zero, it's an extraneous solution and we must discard it.

Let's plug our solution, $x = \frac{177}{220}$, back into the original expressions:

• $5x + 2 = 5(\frac{177}{220}) + 2 = \frac{177}{44} + 2 = \frac{177 + 88}{44} = \frac{265}{44}$
• $9x - 7 = 9(\frac{177}{220}) - 7 = \frac{1593}{220} - 7 = \frac{1593 - 1540}{220} = \frac{53}{220}$

Both expressions, $\frac{265}{44}$ and $\frac{53}{220}$, are positive. Therefore, our solution, $x = \frac{177}{220}$, is valid!

The Final Answer

Phew! We made it through the maze of logarithms and algebra. The solution to the equation $\log _5(5 x+2)-\log _5(9 x-7)=2$ is:

$x = \frac{177}{220}$

Key Takeaways and Practice Tips

Solving logarithmic equations involves a few key steps:

1. Simplify using logarithm properties: The quotient, product, and power rules are your best friends.
2. Convert to exponential form: This eliminates the logarithm and turns the equation into an algebraic one.
3. Solve the algebraic equation: Use your algebra skills to isolate the variable.
4. Check for extraneous solutions: This is crucial! Make sure your solution doesn't result in taking the logarithm of a non-positive number.

Practice is key to mastering logarithmic equations. The more you practice, the more comfortable you'll become with the properties and the steps involved. Try working through different types of logarithmic equations, and don't be afraid to make mistakes – that's how you learn! Remember, every equation is a puzzle waiting to be solved. So, keep practicing, and you'll become a logarithm-solving pro in no time! Good luck, and happy solving!