Problem 50 - Precalculus

Question

Solve the logarithmic equations exactly. $$\log _{2}(x+1)+\log _{2}(4-x)=\log _{2}(6 x)$$

Step-by-Step Solution

Answer

The solution is \( x = 1 \).

Step 1: Apply Logarithm Addition Law

Logarithms can be added using the property \(\log_b(m) + \log_b(n) = \log_b(m \cdot n)\). Apply this to the first part of the equation: \( \log_2(x+1) + \log_2(4-x) = \log_2((x+1)(4-x)) \).

Step 2: Equate Logarithmic Expressions

Now the equation simplifies to \( \log_2((x+1)(4-x)) = \log_2(6x) \). Since the bases are the same, we can equate the arguments: \( (x+1)(4-x) = 6x \).

Step 3: Simplify the Quadratic Equation

Expand the left side: \( 4x - x^2 + 4 - x = 6x \). Simplify to form a standard quadratic equation: \( -x^2 + 3x + 4 = 6x \). Rearrange to get \( -x^2 - 3x + 4 = 0 \). Then multiply through by -1 to get \( x^2 + 3x - 4 = 0 \).

Step 4: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 3, c = -4 \). Calculate the discriminant: \( 3^2 - 4 \cdot 1 \cdot (-4) = 9 + 16 = 25 \). So, \( x = \frac{-3 \pm 5}{2} \). This gives solutions \( x = 1 \) and \( x = -4 \).

Step 5: Verify Solutions in the Original Equation

Substitute \( x = 1 \) back into the original equations: \( \log_2(2) + \log_2(3) = \log_2(6) \), which holds true. Substitute \( x = -4 \): \( \log_2(-3) + \log_2(8) \), which is undefined because \( \log_2(-3) \) does not exist in the real number system.

Key Concepts

Logarithm Properties

Logarithms, much like exponents, follow specific properties that simplify mathematical expressions and solve equations. A crucial property of logarithms is the **addition law**, which states that the logarithm of a product is the sum of the logarithms of the factors.
This is expressed as \( \log_b(m) + \log_b(n) = \log_b(m \cdot n) \), where \( b \) is the base, and \( m \) and \( n \) are the arguments.
In the given exercise, we employ this property to simplify the left-hand side of the equation. Instead of dealing with each logarithmic term individually, we combine them into a single logarithmic expression by multiplying the arguments: \( \log_2((x+1)(4-x)) \).

• This decreases the complexity of the equation, making it easier to solve.
• It is essential to ensure that the arguments of the logarithms produce positive results, as logarithms are undefined for non-positive values.

Mastering these properties not only simplifies mathematical problems but also enhances our understanding of the logarithmic scale.

Quadratic Equation

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \).
In the solution of our exercise, after equating the logarithmic arguments, we arrive at a quadratic equation: \( x^2 + 3x - 4 = 0 \). Solving quadratic equations involves rearranging terms such that they match the standard form and then applying appropriate methods for finding the roots.

• The most common methods are factorization, completing the square, and the quadratic formula.
• In this instance, we utilize the quadratic formula, which is especially handy when the equation does not easily factorize.

It is essential to recognize quadratic equations during problem-solving, as these techniques provide the requisite tools to isolate the variable \( x \). By obtaining the values of \( x \), we can conclude possible solutions for the given equation.

Discriminant

The discriminant is a component of the quadratic formula that provides crucial information about the nature of the roots of a quadratic equation.
Represented by \( b^2 - 4ac \), where \( a, b, \) and \( c \) are coefficients of the quadratic equation \( ax^2 + bx + c = 0 \), it determines whether the roots are real or complex.

• If the discriminant is positive, the quadratic equation will have two distinct real roots.
• If it is zero, there will be exactly one real root, also known as a repeated or double root.
• If the discriminant is negative, the equation will have two complex roots.

For the problem at hand, the discriminant is calculated as \( 3^2 - 4 \times 1 \times (-4) = 25 \), which is positive. This shows that the equation has two real roots. Understanding the discriminant's role is pivotal in predicting the type and number of solutions of a quadratic equation and aids in selecting the appropriate method for solving it.