Solving Logarithmic Equations Log_5(x+20) - Log_5(x-4) = 2

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In the realm of mathematics, logarithmic equations often present a unique challenge. These equations, involving logarithms, require a methodical approach to solve accurately. This article delves into the process of solving a specific logarithmic equation, $\log _5(x+20)-\log _5(x-4)=2$, while emphasizing the crucial step of verifying solutions against the domain of the original logarithmic expressions. We will explore the properties of logarithms, the steps involved in solving the equation, and the importance of checking for extraneous solutions. Understanding these concepts is fundamental for anyone studying algebra, calculus, or related fields.

Understanding Logarithmic Equations

Before diving into the solution, it's essential to grasp the fundamentals of logarithmic equations. A logarithm is essentially the inverse operation to exponentiation. The logarithmic expression $\log_b(a) = c$ can be rewritten in exponential form as $b^c = a$, where 'b' is the base, 'a' is the argument, and 'c' is the logarithm. In our given equation, the base is 5, and we have two logarithmic terms with different arguments. Solving logarithmic equations often involves using the properties of logarithms to simplify the equation into a manageable form. This may include combining logarithmic terms, eliminating logarithms, and solving the resulting algebraic equation. However, a critical step is to always check the solutions against the domain of the original logarithmic expressions to avoid extraneous solutions, which are solutions that satisfy the transformed equation but not the original one.

Properties of Logarithms

The properties of logarithms are the cornerstone of solving logarithmic equations. These properties allow us to manipulate and simplify logarithmic expressions, making it easier to isolate the variable and find the solution. Here are some key properties that are particularly relevant to solving equations like $\log _5(x+20)-\log _5(x-4)=2$:

  1. Quotient Rule: $\log_b(m) - \log_b(n) = \log_b(\frac{m}{n})$ This property allows us to combine two logarithms with the same base that are being subtracted. In our equation, we can use this property to combine the two logarithmic terms on the left side.
  2. Logarithmic to Exponential Form: If $\log_b(a) = c$, then $b^c = a$. This property is crucial for eliminating logarithms from the equation and converting it into a more familiar algebraic form.
  3. Domain of Logarithms: The argument of a logarithm must be positive. That is, for $\log_b(x)$, $x > 0$. This is a critical consideration when solving logarithmic equations, as we must ensure that our solutions do not result in taking the logarithm of a negative number or zero in the original equation.

Understanding and applying these properties correctly is essential for solving logarithmic equations efficiently and accurately. In the next section, we will apply these properties to solve the given equation step-by-step.

Solving the Equation Step-by-Step

Now, let's apply these principles to solve the equation $\log _5(x+20)-\log _5(x-4)=2$. This process involves several key steps, each building upon the previous one to lead us to the solution.

Step 1: Apply the Quotient Rule

The first step is to use the quotient rule of logarithms to combine the two logarithmic terms on the left side of the equation. The quotient rule states that $\log_b(m) - \log_b(n) = \log_b(\frac{m}{n})$. Applying this rule to our equation, we get:

log5(x+20)−log5(x−4)=log5(x+20x−4)\\\\log _5(x+20)-\\\\log _5(x-4) = \\\\log _5(\frac{x+20}{x-4})

So, our equation now becomes:

log5(x+20x−4)=2\\\\log _5(\frac{x+20}{x-4}) = 2

This step simplifies the equation by reducing two logarithmic terms into one, making it easier to proceed with solving.

Step 2: Convert to Exponential Form

The next step is to convert the logarithmic equation into its equivalent exponential form. This is done using the definition of logarithms: if $\log_b(a) = c$, then $b^c = a$. In our case, the base is 5, the argument is $\frac{x+20}{x-4}$, and the logarithm is 2. Therefore, we can rewrite the equation as:

52=fracx+20x−45^2 = \\frac{x+20}{x-4}

Simplifying the left side, we have:

25=fracx+20x−425 = \\frac{x+20}{x-4}

This conversion eliminates the logarithm, transforming the equation into a rational equation that we can solve using algebraic techniques.

Step 3: Solve the Algebraic Equation

Now that we have a rational equation, we can solve for 'x'. To do this, we first multiply both sides of the equation by (x-4) to eliminate the fraction:

25(x−4)=x+2025(x-4) = x+20

Next, we distribute the 25 on the left side:

25x−100=x+2025x - 100 = x + 20

Now, we move all the terms involving 'x' to one side and the constants to the other side. Subtract 'x' from both sides:

24x−100=2024x - 100 = 20

Add 100 to both sides:

24x=12024x = 120

Finally, divide both sides by 24 to solve for 'x':

x=frac12024x = \\frac{120}{24}

x=5x = 5

So, we have found a potential solution: x = 5. However, we must verify this solution to ensure it is valid within the domain of the original logarithmic equation.

Step 4: Check for Extraneous Solutions

The final, and perhaps most crucial, step is to check our solution in the original logarithmic equation. This is because logarithmic equations have domain restrictions; the argument of a logarithm must be positive. In our original equation, $\log _5(x+20)-\log _5(x-4)=2$, we have two logarithmic terms: $\log _5(x+20)$ and $\log _5(x-4)$. For these terms to be defined, we must have:

x + 20 > 0$ and $x - 4 > 0

Solving these inequalities, we get:

x > -20$ and $x > 4

Thus, the domain of the original equation is $x > 4$. Our potential solution, x = 5, satisfies this condition since 5 > 4. Now, we need to substitute x = 5 back into the original equation to ensure it holds true:

log5(5+20)−log5(5−4)=2\\\\log _5(5+20)-\\\\log _5(5-4) = 2

log5(25)−log5(1)=2\\\\log _5(25)-\\\\log _5(1) = 2

We know that $\\log _5(25) = 2$ because $5^2 = 25$, and $\\log _5(1) = 0$ because $5^0 = 1$. So, the equation becomes:

2−0=22 - 0 = 2

2=22 = 2

Since the equation holds true, x = 5 is a valid solution.

Conclusion

In summary, solving the logarithmic equation $\log _5(x+20)-\log _5(x-4)=2$ involves several steps: applying the quotient rule to combine logarithmic terms, converting the equation to exponential form, solving the resulting algebraic equation, and, most importantly, checking for extraneous solutions. By following these steps methodically, we can confidently arrive at the correct solution, which in this case is x = 5. This process highlights the importance of understanding the properties of logarithms and the domain restrictions they impose. Mastering these concepts is crucial for success in algebra and beyond.

Additional Practice Problems

To further solidify your understanding of solving logarithmic equations, here are a few additional practice problems:

  1. Solve: $\log_2(x + 3) + \log_2(x - 1) = 5$
  2. Solve: $\log_3(2x - 1) - \log_3(x - 2) = 1$
  3. Solve: $\log(x) + \log(x - 3) = 1$

Remember to always check your solutions against the domain of the original logarithmic expressions. Practice is key to mastering any mathematical concept, and logarithmic equations are no exception. By working through a variety of problems, you will develop a deeper understanding of the properties of logarithms and become more proficient at solving these types of equations. Good luck!