Solving Logarithmic And Radical Equations Step-by-Step

by ADMIN 55 views

This article delves into the intricate world of solving logarithmic and radical equations, offering a step-by-step approach to tackle these mathematical challenges. We will explore various techniques and strategies, providing detailed explanations and illustrative examples to enhance understanding. Whether you're a student grappling with homework problems or a math enthusiast seeking to expand your knowledge, this guide will equip you with the tools and insights necessary to conquer logarithmic and radical equations.

1. Solving the Exponential Equation: x^(log₅ x) = 125

Let's embark on our journey by tackling the exponential equation x^(log₅ x) = 125. This equation presents a unique challenge due to the variable exponent, requiring us to employ logarithmic properties to unravel its solution. The key to solving equations like this lies in strategically applying logarithms to both sides, transforming the exponential form into a more manageable algebraic expression. By understanding the interplay between logarithms and exponents, we can effectively isolate the variable and arrive at the desired solution.

A Step-by-Step Approach

To solve x^(log₅ x) = 125, we can take the logarithm base 5 of both sides. This strategic move leverages the property of logarithms that allows us to bring the exponent down as a coefficient. Applying this property, we transform the equation into a more accessible form, paving the way for algebraic manipulation and simplification.

  1. Apply logarithm base 5 to both sides: log₅(x^(log₅ x)) = log₅(125)

    This step is crucial as it allows us to utilize the power rule of logarithms, which states that logₐ(b^c) = c * logₐ(b). By applying this rule, we can bring the exponent, log₅ x, down as a coefficient, simplifying the equation.

  2. Use the power rule of logarithms: (log₅ x) * (log₅ x) = log₅(125)

    Now we have (log₅ x) multiplied by itself, which can be written as (log₅ x)². This transformation is a significant step towards isolating the variable. The right side of the equation, log₅(125), can be simplified further since 125 is a power of 5.

  3. Simplify log₅(125): Since 125 = 5³, log₅(125) = 3

    This simplification is crucial for making the equation easier to solve. Replacing log₅(125) with 3 gives us a cleaner equation to work with.

  4. Rewrite the equation: (log₅ x)² = 3

    We now have a quadratic equation in terms of log₅ x. This form is much easier to solve than the original exponential equation.

  5. Take the square root of both sides: log₅ x = ±√3

    Remember to consider both the positive and negative square roots when taking the square root of both sides of an equation. This step is essential for capturing all possible solutions.

  6. Solve for x: We now have two separate equations to solve:

    • log₅ x = √3 => x = 5^(√3)
    • log₅ x = -√3 => x = 5^(-√3)

    To solve for x, we rewrite the logarithmic equations in exponential form. This is the inverse operation of taking the logarithm and allows us to isolate x.

The Solutions

Therefore, the solutions to the equation x^(log₅ x) = 125 are x = 5^(√3) and x = 5^(-√3). These solutions represent the values of x that satisfy the original equation. It's always a good practice to verify these solutions by plugging them back into the original equation to ensure they hold true. This step helps to catch any potential errors made during the solving process.

2. Solving the Logarithmic Equation: 2 log x = log 2 + log (3x - 4)

Next, we tackle the logarithmic equation 2 log x = log 2 + log (3x - 4). This equation involves logarithms with the same base, allowing us to leverage logarithmic properties to combine and simplify terms. A key aspect of solving logarithmic equations is ensuring that the arguments of the logarithms are positive, as logarithms are only defined for positive values. By carefully applying logarithmic rules and considering domain restrictions, we can effectively solve this equation.

Unveiling the Solution

To solve this equation, we will use properties of logarithms to simplify and combine terms. This will allow us to eliminate the logarithms and obtain an algebraic equation that we can solve for x.

  1. Use the power rule of logarithms: log(x²) = log 2 + log(3x - 4)

    The power rule, logₐ(b^c) = c * logₐ(b), allows us to rewrite 2 log x as log(x²). This step is crucial for combining the logarithmic terms on the left side of the equation.

  2. Use the product rule of logarithms: log(x²) = log[2(3x - 4)]

    The product rule, logₐ(b) + logₐ(c) = logₐ(b * c), allows us to combine the two logarithmic terms on the right side of the equation into a single logarithm. This step simplifies the equation significantly.

  3. Equate the arguments: x² = 2(3x - 4)

    Since the logarithms on both sides of the equation have the same base (base 10, if not explicitly stated), we can equate their arguments. This step eliminates the logarithms and transforms the equation into an algebraic equation.

  4. Expand and rearrange: x² = 6x - 8 x² - 6x + 8 = 0

    Expanding the right side and rearranging the terms gives us a quadratic equation in standard form. This form is easily solvable using factoring or the quadratic formula.

  5. Factor the quadratic equation: (x - 4)(x - 2) = 0

    Factoring the quadratic equation allows us to find the values of x that make the equation equal to zero.

  6. Solve for x: x = 4 or x = 2

    Setting each factor equal to zero gives us two potential solutions: x = 4 and x = 2.

Checking for Extraneous Solutions

It is crucial to check these solutions in the original equation to ensure they are valid. Logarithms are only defined for positive arguments, so we must ensure that both x and 3x - 4 are positive.

  • For x = 4: log 4 is defined, and 3(4) - 4 = 8, so log 8 is defined. Thus, x = 4 is a valid solution.
  • For x = 2: log 2 is defined, and 3(2) - 4 = 2, so log 2 is defined. Thus, x = 2 is a valid solution.

The Solutions

Therefore, the solutions to the equation 2 log x = log 2 + log (3x - 4) are x = 4 and x = 2. Both solutions satisfy the original equation and the domain restrictions of the logarithms.

3. Navigating Logarithmic Equations: log₆ x³ - 2 log₆ 6 = 1

Now, let's tackle the logarithmic equation log₆ x³ - 2 log₆ 6 = 1. This equation involves logarithmic terms with the same base, allowing us to utilize logarithmic properties to simplify and solve for x. A key aspect of solving logarithmic equations is understanding and applying the various logarithmic rules effectively. By carefully manipulating the equation and isolating the variable, we can arrive at the solution.

Deciphering the Equation

To solve log₆ x³ - 2 log₆ 6 = 1, we'll employ logarithmic properties to simplify the equation and ultimately isolate x.

  1. Use the power rule of logarithms: log₆(x³) - log₆(6²) = 1

    The power rule, logₐ(b^c) = c * logₐ(b), allows us to rewrite 2 log₆ 6 as log₆(6²). This step simplifies the second term in the equation.

  2. Simplify 6²: log₆(x³) - log₆(36) = 1

    Simplifying 6² to 36 makes the equation cleaner and easier to work with.

  3. Use the quotient rule of logarithms: log₆(x³/36) = 1

    The quotient rule, logₐ(b) - logₐ(c) = logₐ(b/c), allows us to combine the two logarithmic terms into a single logarithm. This step significantly simplifies the equation.

  4. Convert to exponential form: x³/36 = 6¹

    Converting the logarithmic equation to exponential form is the key to eliminating the logarithm and isolating x. The equation logₐ(b) = c is equivalent to a^c = b.

  5. Solve for x³: x³ = 6 * 36 x³ = 216

    Multiplying both sides by 36 isolates x³.

  6. Solve for x: x = ∛216 x = 6

    Taking the cube root of both sides gives us the solution for x.

The Solution

Therefore, the solution to the equation log₆ x³ - 2 log₆ 6 = 1 is x = 6. This solution satisfies the original equation and the domain restrictions of the logarithm.

4. Tackling Radical Equations: √(12 - x) + √(x + 1) = √(13 + 4x)

Finally, let's confront the radical equation √(12 - x) + √(x + 1) = √(13 + 4x). This equation involves square roots, requiring us to employ a strategy of squaring both sides to eliminate the radicals. However, it's crucial to remember that squaring both sides can introduce extraneous solutions, so we must diligently check our answers in the original equation. By carefully squaring, simplifying, and checking for extraneous solutions, we can successfully solve this radical equation.

Dissecting the Radical

To solve √(12 - x) + √(x + 1) = √(13 + 4x), we'll square both sides strategically to eliminate the square roots.

  1. Square both sides: [√(12 - x) + √(x + 1)]² = [√(13 + 4x)]²

    Squaring both sides is the first step in eliminating the square roots. However, it's important to remember that squaring a binomial requires careful expansion.

  2. Expand the left side: (12 - x) + 2√(12 - x)√(x + 1) + (x + 1) = 13 + 4x

    Expanding the left side using the formula (a + b)² = a² + 2ab + b² is crucial. The term 2√(12 - x)√(x + 1) is often missed, but it's essential for the correct solution.

  3. Simplify: 13 + 2√[(12 - x)(x + 1)] = 13 + 4x

    Simplifying the equation by combining like terms makes it easier to isolate the remaining square root term.

  4. Isolate the radical: 2√[(12 - x)(x + 1)] = 4x

    Isolating the radical term is necessary before squaring both sides again.

  5. Divide by 2: √[(12 - x)(x + 1)] = 2x

    Dividing both sides by 2 simplifies the equation further.

  6. Square both sides again: (12 - x)(x + 1) = 4x²

    Squaring both sides again eliminates the remaining square root.

  7. Expand and rearrange: 12 + 11x - x² = 4x² 5x² - 11x - 12 = 0

    Expanding the left side and rearranging the terms gives us a quadratic equation in standard form.

  8. Factor the quadratic equation: (5x + 4)(x - 3) = 0

    Factoring the quadratic equation allows us to find the potential solutions for x.

  9. Solve for x: x = -4/5 or x = 3

    Setting each factor equal to zero gives us two potential solutions: x = -4/5 and x = 3.

The Crucial Check for Extraneous Solutions

Now, we must check these solutions in the original equation to ensure they are valid. This is a critical step in solving radical equations, as squaring both sides can introduce extraneous solutions.

  • For x = -4/5: √(12 - (-4/5)) + √((-4/5) + 1) = √(64/5) + √(1/5) = (8/√5) + (1/√5) = 9/√5 √(13 + 4(-4/5)) = √(49/5) = 7/√5 Since 9/√5 ≠ 7/√5, x = -4/5 is an extraneous solution.
  • For x = 3: √(12 - 3) + √(3 + 1) = √9 + √4 = 3 + 2 = 5 √(13 + 4(3)) = √25 = 5 Since 5 = 5, x = 3 is a valid solution.

The Solution

Therefore, the only solution to the equation √(12 - x) + √(x + 1) = √(13 + 4x) is x = 3. The solution x = -4/5 is an extraneous solution and must be discarded.

Conclusion

In this comprehensive guide, we have explored the methods for solving logarithmic and radical equations. We have demonstrated how to apply logarithmic properties, manipulate equations, and check for extraneous solutions. By mastering these techniques, you can confidently tackle a wide range of logarithmic and radical equations. Remember to always verify your solutions to ensure accuracy and avoid extraneous results. With practice and a solid understanding of the underlying principles, you can excel in solving these mathematical challenges.

Keywords: logarithmic equations, radical equations, solving equations, exponential equations, logarithmic properties, extraneous solutions, mathematics.