Problem 51 - Intermediate Algebra

Question

Solve each equation. $$ \ln (3 x+1)=\ln (x+7) $$

Step-by-Step Solution

Answer

The solution is \(x = 3\).

Step 1: Set the Equations Equal

Given that the natural logarithms of two expressions are equal, we can set the expressions inside the logarithms equal to each other. Therefore, we have:\(3x + 1 = x + 7\)

Step 2: Simplify the Equation

Subtract \(x\) from both sides to get all the terms involving \(x\) on one side:\(3x + 1 - x = x + 7 - x\)Which simplifies to:\(2x + 1 = 7\)

Step 3: Solve for x

Subtract 1 from both sides to isolate the term with \(x\):\(2x + 1 - 1 = 7 - 1\)This gives:\(2x = 6\)Now, divide both sides by 2 to solve for \(x\):\(x = \frac{6}{2}\)So, \(x = 3\).

Step 4: Verify the Solution

Substitute \(x = 3\) back into the original logarithmic expressions to ensure they are equal:\(\ln(3(3) + 1) = \ln(3 + 7)\)Calculate both sides:\(\ln(9 + 1) = \ln(10)\) and \(\ln(10) = \ln(10)\)Since both sides are equal, \(x = 3\) is indeed the correct solution.

Key Concepts

Logarithmic Equations

Logarithmic equations are equations in which a logarithm appears frequently, involving not just the logarithmic function itself but also algebraic manipulation to solve for a variable. These types of equations typically come in forms where the logarithms are equal to one another or are set equal to values. Understanding logarithmic equations helps us solve various real-world problems, such as those related to exponential growth and decay.

• Ensure that the domain of the logarithm is positive since the logarithm is only defined for positive values.
• Use the property that if \(\ln(a) = \ln(b)\), then \(a = b\).

In our example, we have \(\ln(3x+1) = \ln(x+7)\). By setting the arguments of the logarithms \(3x + 1\) and \(x + 7\) equal, we reduce the logarithmic problem to an algebraic one.

Equation Solving

Equation solving is a fundamental skill in algebra. It involves finding the value of the variable that makes the equation true. In the case of logarithmic equations, once you simplify them by equating the terms inside the logarithms, the task becomes an exercise in algebraic manipulation.

• Your goal is to isolate the variable on one side of the equation. This often involves combining like terms or using inverse operations.
• Every manipulation step should maintain the balance of the equation by performing the same operation on both sides.

For the equation \(3x + 1 = x + 7\), subtracting \(x\) from both sides simplifies it to \(2x + 1 = 7\).
Further steps involve straightforward arithmetic operations, leading to the solution \(x = 3\).

Natural Logarithms

Natural logarithms, denoted by \(\ln\), are logarithms with the base \(e\), where \(e\) is approximately equal to 2.71828. They are particularly useful because of their natural occurrence in many mathematical contexts and scientific applications, such as calculating time for continuous compounding interest, radioactive decay, and population dynamics.

• The natural logarithm of a number gives the power to which \(e\) must be raised to obtain that number.
• Important properties of natural logarithms include \(\ln(1) = 0\) and \(\ln(e) = 1\).

In equations, they simplify when using the identity rule \(\ln(a) = b\) implies \(a = e^b\). However, in our exercise, knowing that \(\ln(3x+1) = \ln(x+7)\) means \(3x + 1 = x + 7\), it's clear that the \(\ln\) on both sides cancels out to set us on the path of algebra.

Algebraic Manipulation

Algebraic manipulation refers to rearranging and simplifying equations to isolate the unknown variable. This skill is crucial in solving equations effectively.

• Start by applying basic arithmetic operations such as addition, subtraction, multiplication, and division to both sides of the equation.
• When handling terms, aim to simplify by combining like terms and eliminating any constants that can be moved directly to the other side.

Considering the equation \(2x + 1 = 7\), subtract 1 from both sides to get \(2x = 6\). Then, divide each side by 2 to solve for \(x\). This process results in the solution \(x = 3\).
Such manipulations not only solve the equation but also enhance understanding of how different operations affect the entire equation.