Problem 57 - A Graphical Approach to Precalculus with Limits

Question

Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. $$\log _{7}(4 x)-\log _{7}(x+3)=\log _{7} x$$

Step-by-Step Solution

Answer

The solution is \( x = 1 \).

Step 1: Use the properties of logarithms

Apply the properties of logarithms to simplify the expression: \( \log_{b}(a) - \log_{b}(c) = \log_{b}\left(\frac{a}{c}\right) \). For the given equation, this means \( \log_{7}(4x) - \log_{7}(x+3) = \log_{7}\left(\frac{4x}{x+3}\right) \). The equation now reads \( \log_{7}\left(\frac{4x}{x+3}\right) = \log_{7}(x) \).

Step 2: Set the insides of the logarithms equal

Since the logarithms with the same base are equal, their arguments must be equal as well. Therefore, \( \frac{4x}{x+3} = x \).

Step 3: Solve the equation

Clear the fraction by multiplying both sides by \( x+3 \): \( 4x = x(x+3) \). Expand the right side to get \( 4x = x^2 + 3x \). Move all terms to one side to form a quadratic equation: \( 0 = x^2 - x \). Factor the equation as \( x(x-1) = 0 \).

Step 4: Find the solutions

The factored equation \( x(x-1) = 0 \) gives two potential solutions: \( x = 0 \) and \( x = 1 \).

Step 5: Check validity of solutions

Substitute each solution back into the original logarithmic equation to check if they are valid. For \( x = 0 \), \( \log_{7}(4 \times 0) \) is undefined, so \( x = 0 \) is not valid. For \( x = 1 \), \( \log_{7}(4 \times 1) - \log_{7} (1 + 3) = \log_{7} (1) \), which simplifies to \( \log_{7} (4) - \log_{7} (4) = 0 \), so \( x = 1 \) is valid.

Key Concepts

Properties of Logarithms

Logarithms can initially seem intimidating, but understanding their properties can simplify your mathematical problems significantly. One such property is the logarithmic subtraction rule which states: \( \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right) \). This means that when you subtract logarithms of the same base, you can transform them into the logarithm of a division of their arguments.
This property is quite useful when dealing with equations that include logarithms. By converting subtraction into division, it often makes the equation more straightforward. For example, in the given exercise, the equation \( \log_7(4x) - \log_7(x+3) = \log_7(x) \) becomes \( \log_7\left(\frac{4x}{x+3}\right) = \log_7(x) \).

• This simplifies the expression by decreasing the number of logarithmic terms.
• It opens avenues to set up the arguments for equations (without logarithms) directly.
• Understanding and applying this property can transform complex logarithmic equations into forms that are easier to manipulate and solve.

Solving Equations

Once logarithmic terms have been simplified, the next step is to solve the resulting equations. In our example, after using the property of logarithms, the equation turns into \( \frac{4x}{x+3} = x \). Equations are typically solved by isolating the variable. Here, it's crucial to eliminate fractions by multiplying through by the denominator, which simplifies the equation.
For this exercise, multiply both sides by \( x+3 \) to eliminate the fraction. So we have \( 4x = x(x+3) \). This results in a more straightforward expression that's easier to solve:

• Expand the right (the product of \( x(x+3) \)).
• "Clear" equations by moving all terms to one side to set up a zero-equation that can often be simplified further.
• Recognize when you have a quadratic equation, which suggests specific solving methods such as factoring.

With all terms on one side, you can now easily transition into solving the quadratic nature of the problem.

Quadratic Equations

When solving equations, sometimes you will come across what are known as quadratic equations. These take the form of \( ax^2 + bx + c = 0 \) and can be solved using various methods, such as factoring, using the quadratic formula, or completing the square.
In our specific problem, the equation turns into \( x^2 - x = 0 \), which features no constant term. This is a telltale sign that factoring is a suitable method to find solutions. Factoring involves writing the equation as a product of its roots leading to: \( x(x-1) = 0 \).

• Set each factor equal to zero: \( x = 0 \) and \( x-1 = 0 \), deriving potential solutions.
• Always plug your solutions back into the original equation to verify validity, as with logarithmic equations, not all algebraic solutions may satisfy the original log constraints.
• Understand that while solving quadratic equations algebraically is straightforward, checking these solutions within the context of the original equation ensures they are practical and correct.

Quadratic equations offer structured solving pathways, but context checking, especially with logarithms, is essential in confirming realistic solutions.