Problem 52 - Elementary and Intermediate Algebra

Question

Solve each equation. See Example 7 . $$ \ln \left(x^{2}+4 x\right)=\ln \left(x^{2}+16\right) $$

Step-by-Step Solution

Answer

The solution is \( x = 4 \).

Step 1: Understand the problem

The problem involves solving the equation \( \ln(x^2 + 4x) = \ln(x^2 + 16) \). The natural logarithm function \( \ln \) implies that if \( \ln(a) = \ln(b) \), then \( a = b \) given the property of logarithms.

Step 2: Remove the logarithm

Use the property of logarithms that states if \( \ln(a) = \ln(b) \), then \( a = b \). Set the arguments equal to each other: \[ x^2 + 4x = x^2 + 16 \]

Step 3: Simplify the equation

Subtract \( x^2 \) from both sides of the equation:\[ x^2 + 4x - x^2 = x^2 + 16 - x^2 \] This simplifies to:\[ 4x = 16 \]

Step 4: Solve for \( x \)

Divide both sides of the equation by 4 to solve for \( x \): \[ x = \frac{16}{4} = 4 \]

Step 5: Verify the solution

Substitute \( x = 4 \) back into the original equation to verify it satisfies both sides:First compute \( \ln(x^2 + 4x) \):\( x^2 + 4x = 4^2 + 4(4) = 16 + 16 = 32 \).Then compute \( \ln(x^2 + 16) \):\( x^2 + 16 = 4^2 + 16 = 16 + 16 = 32 \).Both expressions are \( \ln(32) \), so \( x = 4 \) is a valid solution.

Key Concepts

Natural Logarithms

Natural logarithms, represented as \( \ln \), are special logarithms based on the constant \( e \), which is approximately 2.71828. This constant \( e \) is known as Euler's number and is the base of natural logarithms. When you see \( \ln(x) \), it means the power to which you must raise \( e \) to get \( x \).

This function is crucial in mathematics because of its properties that simplify many types of problems, especially those involving exponential functions.

Here are some key points to remember:

• The natural logarithm of a number enlarges quickly as the number increases.
• Unlike other logarithms, \( \ln(1) = 0 \) since \( e^0 = 1 \).
• The derivative of \( \ln(x) \) is \( \frac{1}{x} \), making it valuable in calculus.
• Natural logarithms are used extensively in science and engineering to model natural phenomena.

Understanding natural logarithms can make solving equations like \( \ln(x^2 + 4x) = \ln(x^2 + 16) \) much simpler.

Equation Solving

Equation solving is about finding the value of the variable that satisfies a given equation. The importance of solving equations lies in understanding how different mathematical expressions can describe the same state, condition, or pattern.

With logarithmic equations, you often rely on setting the arguments equal, once you eliminate the logarithms. Here's a simple guide on how to approach solving such logarithmic equations:

• First, understand the equation and identify the type of logarithm involved, such as natural logarithms.
• Next, if the logarithmic expressions have the same base, use the property \( \ln(a) = \ln(b) \Rightarrow a = b \) to eliminate the logarithm and focus on the argument.
• Simplify the equation resulting from setting these arguments equal and solve for the variable.
• Finally, verify your solution by substituting it back into the original equation to ensure both sides are equal.

This step-by-step procedure can effectively solve equations like the one in the exercise, ensuring a comprehensive understanding and accurate solution.

Properties of Logarithms

Logarithms possess unique properties that make them incredibly useful for simplifying calculations and solving equations. When dealing with natural logarithms, these properties become particularly advantageous.

Here are some fundamental properties of logarithms that can assist you:

• Product Property: \( \ln(a \cdot b) = \ln(a) + \ln(b) \). Use this to split logarithms of multiplied terms into a sum of two logarithms.
• Quotient Property: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \). This property helps break down logarithms of fractions into a difference of two logarithms.
• Power Property: \( \ln(a^b) = b \cdot \ln(a) \). It simplifies the logarithm of an exponent by bringing the exponent in front as a multiplier.
• Equality Property: If \( \ln(a) = \ln(b) \), then \( a = b \). This is immensely useful in solving logarithmic equations, as seen in the original exercise where \( \ln(x^2 + 4x) = \ln(x^2 + 16) \) implies \( x^2 + 4x = x^2 + 16 \).

Mastering these properties facilitates the manipulation and resolution of complex logarithmic expressions with ease.