Problem 97 - Intermediate Algebra

Question

For Problems \(81-97\), solve each of the equations. $$ \log _{2}(x+2)=1-\log _{2}(x+3) $$

Step-by-Step Solution

Answer

The solution is \( x = -1 \).

Step 1: Isolate One Logarithm

We start with the original equation \( \log_{2}(x+2) = 1 - \log_{2}(x+3) \). To solve for \( x \), first isolate \( \log_{2}(x+2) \) on one side of the equation. You can do this by moving \( \log_{2}(x+3) \) to the left: \[ \log_{2}(x+2) + \log_{2}(x+3) = 1 \].

Step 2: Combine Logarithms

Use the property of logarithms that states \( \log_{b}(m) + \log_{b}(n) = \log_{b}(m \cdot n) \) to combine the left side: \[ \log_{2}((x+2)(x+3)) = 1 \].

Step 3: Convert Logarithmic Equation to Exponential Form

Since logarithms are the inverse of exponents, convert the equation into exponential form. This gives us: \[ (x+2)(x+3) = 2^{1} \].

Step 4: Expand the Quadratic

Expand the left side of the equation: \[ x^2 + 5x + 6 = 2 \].

Step 5: Solve the Quadratic Equation

Rearrange the equation to set it to zero: \[ x^2 + 5x + 4 = 0 \].Next, factor the quadratic equation: \[ (x+4)(x+1) = 0 \].

Step 6: Find Solutions for x

Set each factor to zero and solve for \( x \):1. \( x + 4 = 0 \Rightarrow x = -4 \)2. \( x + 1 = 0 \Rightarrow x = -1 \).

Step 7: Check for Valid Solutions

Both solutions need to be checked in the original logarithmic expressions to ensure the expressions are defined. Substitute:1. \( x = -4 \Rightarrow \log_{2}(-4+2), \log_{2}(-4+3) \) are undefined (negative arguments).2. \( x = -1 \Rightarrow \log_{2}(-1+2)=\log_{2}(1), \log_{2}(-1+3)=\log_{2}(2) \), which check out.

Key Concepts

Exponential Form

The concept of exponential form is essential when dealing with logarithmic equations. It involves transforming a logarithmic expression into its equivalent in the form of an exponent. This switch is based on the fact that logarithms and exponents are inverse operations.

In simple terms, if you have a logarithmic equation like \( \log_b(a) = c \), you can express it in exponential form as \( b^c = a \). This transformation makes it easier to solve equations, especially when the logarithm is isolated on one side.

• Logarithmic form: \( \log_2((x+2)(x+3)) = 1 \)
• Exponential form: \( (x+2)(x+3) = 2^1 \)

By converting to exponential form, you now deal directly with the relationship between the variable and the base. This makes the process of finding the variable's value straightforward as you work towards simplifying the expression.

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \) and solving them is crucial for finding values of \( x \) that satisfy certain conditions. When dealing with transformed logarithmic equations, we sometimes encounter these quadratics.

Once a logarithmic equation is transformed into an exponential form, it often leads to a quadratic equation. From there, the process involves simplifying and solving the quadratic equation. The common methods of solving a quadratic equation include factoring, completing the square, and using the quadratic formula.

For the problem at hand, expanding \( (x+2)(x+3) \) resulted in the equation:

• \( x^2 + 5x + 4 = 0 \)

This equation is set to zero, a typical characteristic of quadratics. Solving this requires factoring into simpler linear expressions:

• \( (x+4)(x+1) = 0 \)

After factoring, you solve for \( x \) by setting each factor equal to zero, yielding possible solutions of \( x = -4 \) and \( x = -1 \).

Properties of Logarithms

The properties of logarithms are tools that make it easier to simplify and manipulate logarithmic expressions. These properties are often used together with exponential forms to solve logarithmic equations.

Some key properties include:

• Product Property: \( \log_b(m) + \log_b(n) = \log_b(m \cdot n) \)
• Quotient Property: \( \log_b(m) - \log_b(n) = \log_b\left(\frac{m}{n}\right) \)
• Power Property: \( \log_b(m^n) = n \cdot \log_b(m) \)

In the original equation \( \log _{2}(x+2) = 1-\log _{2}(x+3) \), the product property is used to combine terms, resulting in:

• \( \log_{2}((x+2)(x+3)) = 1 \)

This property's use allowed transforming two logarithmic expressions into one, paving the way for a straightforward conversion to exponential form. When solving logarithmic equations, understanding and applying these properties ensure the logical flow and accuracy of the solution process.