Solving Logarithmic Equations Step By Step Guide

This article delves into the methods for solving various logarithmic equations, providing step-by-step solutions and explanations for each. We'll tackle equations with different logarithmic bases and complexities, ensuring a comprehensive understanding of the underlying principles. The key to solving logarithmic equations lies in understanding the properties of logarithms and applying them strategically to isolate the variable. Remember, it's crucial to check for extraneous solutions, as logarithmic functions are only defined for positive arguments. Let's explore these techniques through a series of examples.

(a) ${ \log_5 (7x + 1) = \log_5 (3x + 29) }$

To solve this logarithmic equation, we'll leverage the fundamental property that if ${ \log_b a = \log_b c }$, then ${ a = c }$, provided that ${ b }$ is a positive number not equal to 1 and both ${ a }$ and ${ c }$ are positive. This property allows us to eliminate the logarithms and focus on the algebraic equation formed by their arguments.

• Step 1: Apply the Property:

  Since we have the same base (5) on both sides of the equation, we can equate the arguments:
  ${ 7x + 1 = 3x + 29 }$

• Step 2: Solve for x:

  Now, we have a simple linear equation to solve. Subtract 3x from both sides: ${ 4x + 1 = 29 }$ Subtract 1 from both sides: ${ 4x = 28 }$ Divide both sides by 4: ${ x = 7 }$

• Step 3: Check for Extraneous Solutions:

  It's crucial to verify that our solution doesn't result in taking the logarithm of a non-positive number. We substitute ${ x = 7 }$ back into the original equation: ${ \log_5 (7(7) + 1) = \log_5 (3(7) + 29) }$ ${ \log_5 (50) = \log_5 (50) }$ Since both arguments are positive, our solution ${ x = 7 }$ is valid.

• Conclusion: The solution to the equation ${ \log_5 (7x + 1) = \log_5 (3x + 29) }$ is ${ x = 7 }$. This process highlights the importance of not only applying logarithmic properties but also verifying the solutions within the original context of the equation. Checking for extraneous solutions is a critical step in solving logarithmic equations, ensuring the validity of the results. The application of the one-to-one property of logarithms simplifies the equation, transforming it into a solvable algebraic form. Always remember to substitute the solution back into the original equation to confirm its validity.

(b) ${ \ln [2x(3x-11)] = \ln (x - 22) }$

This equation involves the natural logarithm, denoted by ${ \ln }$, which is the logarithm to the base e. Similar to the previous example, we'll use the property that if ${ \ln a = \ln c }$, then ${ a = c }$, provided both ${ a }$ and ${ c }$ are positive. Carefully handling the algebraic manipulations and checking for extraneous solutions are paramount in this case. The natural logarithm function plays a significant role in various mathematical and scientific fields, making the ability to solve equations involving it an essential skill. This example underscores the importance of understanding the domain restrictions of logarithmic functions.

• Step 1: Apply the Property:

  Equating the arguments of the natural logarithms, we get:
  ${ 2x(3x - 11) = x - 22 }$

• Step 2: Simplify and Solve the Quadratic Equation:

  Expanding the left side, we have: ${ 6x^2 - 22x = x - 22 }$ Move all terms to one side to form a quadratic equation: ${ 6x^2 - 23x + 22 = 0 }$ Now, we can solve this quadratic equation. Factoring the quadratic, we look for two numbers that multiply to ${ 6 imes 22 = 132 }$ and add up to -23. These numbers are -11 and -12. We can rewrite the middle term and factor by grouping: ${ 6x^2 - 12x - 11x + 22 = 0 }$ ${ 6x(x - 2) - 11(x - 2) = 0 }$ ${ (6x - 11)(x - 2) = 0 }$ Thus, we have two potential solutions: ${ x = \frac{11}{6} }$ or ${ x = 2 }$

• Step 3: Check for Extraneous Solutions:

  We need to check if these solutions satisfy the original equation and the domain restrictions of the logarithms. First, consider ${ x = \frac{11}{6} }$:

  • Check ${ 2x(3x - 11) }$:
    ${ 2(\frac{11}{6})(3(\frac{11}{6}) - 11) = \frac{11}{3}(\frac{11}{2} - 11) = \frac{11}{3}(\frac{-11}{2}) < 0 }$ Since the argument is negative, ${ x = \frac{11}{6} }$ is an extraneous solution.

  Now, consider ${ x = 2 }$:

  • Check ${ 2x(3x - 11) }$:
    ${ 2(2)(3(2) - 11) = 4(6 - 11) = 4(-5) = -20 }$ Again, the argument is negative, so ${ x = 2 }$ is also an extraneous solution.
  • Check ${ x - 22 }$:
    ${ 2 - 22 = -20 }$ This argument is also negative.

• Conclusion: Both potential solutions lead to negative arguments within the logarithms, making them extraneous. Therefore, there is no solution to the equation ${ \ln [2x(3x-11)] = \ln (x - 22) }$. This outcome emphasizes the critical importance of checking for extraneous solutions, especially when dealing with logarithmic equations. The process of solving such equations often involves algebraic manipulations that can introduce solutions that are not valid within the domain of the logarithmic functions.

(c) ${ \log_2 x + \log_2 (4x + 7) = 1 }$

This equation involves the sum of two logarithms with the same base. We can use the property ${ \log_b a + \log_b c = \log_b (ac) }$ to combine the logarithms into a single term. This step simplifies the equation and allows us to solve for x. The base 2 logarithm is commonly encountered in computer science and information theory, making its understanding crucial. Remember to always consider the domain of the logarithmic function when solving equations.

• Step 1: Combine the Logarithms:

  Using the logarithmic property, we combine the two logarithms on the left side: ${ \log_2 [x(4x + 7)] = 1 }$

• Step 2: Convert to Exponential Form:

  To eliminate the logarithm, we rewrite the equation in exponential form. Recall that ${ \log_b a = c }$ is equivalent to ${ b^c = a }$. Applying this to our equation: ${ 2^1 = x(4x + 7) }$ ${ 2 = 4x^2 + 7x }$

• Step 3: Solve the Quadratic Equation:

  Rearrange the equation to form a quadratic equation: ${ 4x^2 + 7x - 2 = 0 }$ We can solve this quadratic equation by factoring. We need two numbers that multiply to ${ 4 imes -2 = -8 }$ and add up to 7. These numbers are 8 and -1. We rewrite the middle term and factor by grouping: ${ 4x^2 + 8x - x - 2 = 0 }$ ${ 4x(x + 2) - 1(x + 2) = 0 }$ ${ (4x - 1)(x + 2) = 0 }$ This gives us two potential solutions: ${ x = \frac{1}{4} }$ or ${ x = -2 }$

• Step 4: Check for Extraneous Solutions:

  We must check both solutions in the original equation. First, consider ${ x = \frac{1}{4} }$: ${ \log_2 (\frac{1}{4}) + \log_2 (4(\frac{1}{4}) + 7) = \log_2 (\frac{1}{4}) + \log_2 (8) = -2 + 3 = 1 }$ So, ${ x = \frac{1}{4} }$ is a valid solution.

  Now, consider ${ x = -2 }$: ${ \log_2 (-2) + \log_2 (4(-2) + 7) }$ Since we cannot take the logarithm of a negative number, ${ x = -2 }$ is an extraneous solution.

• Conclusion: The only valid solution to the equation ${ \log_2 x + \log_2 (4x + 7) = 1 }$ is ${ x = \frac{1}{4} }$. This example demonstrates the importance of combining logarithmic terms to simplify equations and the necessity of checking solutions for validity within the logarithmic domain. Quadratic equations often arise in the process of solving logarithmic equations, requiring factoring or the quadratic formula to find potential solutions.

(d) ${ \lg (x^2 - 2x - 13) - \lg (x - 5) = \lg (2x - 1) }$

This equation involves the common logarithm, denoted by ${ \lg }$, which is the logarithm to the base 10. We'll use the property ${ \log_b a - \log_b c = \log_b (\frac{a}{c}) }$ to combine the logarithms on the left side. Then, we'll apply the property that if ${ \log_b a = \log_b c }$, then ${ a = c }$. The common logarithm has applications in various fields, including acoustics and chemistry. As always, verifying solutions for extraneous values is crucial.

• Step 1: Combine the Logarithms:

  Using the logarithmic property for subtraction, we combine the logarithms on the left side: ${ \lg \frac{x^2 - 2x - 13}{x - 5} = \lg (2x - 1) }$

• Step 2: Equate the Arguments:

  Since we have the same base (10) on both sides, we can equate the arguments: ${ \frac{x^2 - 2x - 13}{x - 5} = 2x - 1 }$

• Step 3: Solve the Equation:

  Multiply both sides by ${ x - 5 }$ to eliminate the fraction: ${ x^2 - 2x - 13 = (2x - 1)(x - 5) }$ Expand the right side: ${ x^2 - 2x - 13 = 2x^2 - 10x - x + 5 }$ ${ x^2 - 2x - 13 = 2x^2 - 11x + 5 }$ Move all terms to one side to form a quadratic equation: ${ 0 = x^2 - 9x + 18 }$ Factor the quadratic: ${ 0 = (x - 3)(x - 6) }$ This gives us two potential solutions: ${ x = 3 }$ or ${ x = 6 }$

• Step 4: Check for Extraneous Solutions:

  We must check both solutions in the original equation. First, consider ${ x = 3 }$:

  • Check ${ x^2 - 2x - 13 }$:
    ${ 3^2 - 2(3) - 13 = 9 - 6 - 13 = -10 }$ Since this is negative, ${ x = 3 }$ is an extraneous solution.

  Now, consider ${ x = 6 }$:

  • Check ${ x^2 - 2x - 13 }$:
    ${ 6^2 - 2(6) - 13 = 36 - 12 - 13 = 11 > 0 }$
  • Check ${ x - 5 }$:
    ${ 6 - 5 = 1 > 0 }$
  • Check ${ 2x - 1 }$:
    ${ 2(6) - 1 = 11 > 0 }$ Since all arguments are positive, ${ x = 6 }$ is a valid solution.

• Conclusion: The only valid solution to the equation ${ \lg (x^2 - 2x - 13) - \lg (x - 5) = \lg (2x - 1) }$ is ${ x = 6 }$. This example reinforces the importance of meticulous checking for extraneous solutions, as they can arise even after correctly applying logarithmic properties and algebraic manipulations. Quadratic equations are frequently encountered when solving logarithmic equations, requiring factoring or the quadratic formula to identify potential solutions.

(e) ${ 2 \lg x = \lg (6x + 8) - \lg 2 }$

This equation again involves the common logarithm. We'll use several properties of logarithms to solve it. First, we'll use the power rule, ${ n \log_b a = \log_b (a^n) }$, to simplify the left side. Then, we'll use the quotient rule, ${ \log_b a - \log_b c = \log_b (\frac{a}{c}) }$, to combine the logarithms on the right side. Finally, we'll equate the arguments and solve the resulting equation. This example provides a comprehensive review of logarithmic properties and their application in solving equations.

• Step 1: Apply the Power Rule:

  Use the power rule to rewrite the left side: ${ \lg (x^2) = \lg (6x + 8) - \lg 2 }$

• Step 2: Apply the Quotient Rule:

  Use the quotient rule to combine the logarithms on the right side: ${ \lg (x^2) = \lg \frac{6x + 8}{2} }$ ${ \lg (x^2) = \lg (3x + 4) }$

• Step 3: Equate the Arguments:

  Since we have the same base (10) on both sides, we can equate the arguments: ${ x^2 = 3x + 4 }$

• Step 4: Solve the Quadratic Equation:

  Rearrange the equation to form a quadratic equation: ${ x^2 - 3x - 4 = 0 }$ Factor the quadratic: ${ (x - 4)(x + 1) = 0 }$ This gives us two potential solutions: ${ x = 4 }$ or ${ x = -1 }$

• Step 5: Check for Extraneous Solutions:

  We must check both solutions in the original equation. First, consider ${ x = 4 }$: ${ 2 \lg 4 = \lg (6(4) + 8) - \lg 2 }$ ${ \lg (4^2) = \lg (32) - \lg 2 }$ ${ \lg 16 = \lg \frac{32}{2} }$ ${ \lg 16 = \lg 16 }$ So, ${ x = 4 }$ is a valid solution.

  Now, consider ${ x = -1 }$: Since we cannot take the logarithm of a negative number, ${ x = -1 }$ is an extraneous solution.

• Conclusion: The only valid solution to the equation ${ 2 \lg x = \lg (6x + 8) - \lg 2 }$ is ${ x = 4 }$. This final example underscores the importance of mastering logarithmic properties and applying them strategically to simplify equations. The power rule and quotient rule are essential tools for manipulating logarithmic expressions. As always, checking for extraneous solutions is a crucial step in the problem-solving process.