Problem 61 - Precalculus

Question

Solve the logarithmic equations. Round your answers to three decimal places. $$\ln (2 x+3)=-2$$

Step-by-Step Solution

Answer

The solution is \( x \approx -1.433 \).

Step 1: Understand the logarithmic equation

We are given the equation \( \ln(2x + 3) = -2 \). The task is to solve for \( x \), where the natural logarithm function is defined.

Step 2: Exponentiate both sides

To eliminate the natural logarithm, we exponentiate both sides using the exponential function \( e \) because \( e^{\ln(y)} = y \). This gives us: \[ e^{\ln(2x + 3)} = e^{-2} \] Simplifying the left side, we have: \[ 2x + 3 = e^{-2} \]

Step 3: Calculate \( e^{-2} \)

Calculate \( e^{-2} \) using a calculator: \[ e^{-2} \approx 0.135 \] (Round this value to three decimal places.)

Step 4: Isolate \( x \)

To solve for \( x \), we need to isolate it on one side of the equation. Subtract 3 from both sides: \[ 2x = e^{-2} - 3 \] Substitute \( e^{-2} \approx 0.135 \): \[ 2x \approx 0.135 - 3 \] \[ 2x \approx -2.865 \]

Step 5: Solve for \( x \)

Now divide both sides by 2 to solve for \( x \): \[ x \approx \frac{-2.865}{2} \] \[ x \approx -1.4325 \] Round the result to three decimal places: \[ x \approx -1.433 \]

Key Concepts

Natural Logarithm

In mathematical terms, the natural logarithm is the logarithm to the base of the mathematical constant \( e \), which is approximately equal to 2.71828. Represented as \( \ln \), it provides a way to express the exponent needed to raise \( e \) to a certain number. This function is particularly important in calculus and has many applications in solving real-world problems, such as exponential growth or decay.

• The notation \( \ln(y) = x \) means that \( e^x = y \).
• It's commonly used in mathematic equations when dealing with continuous growth or compounding continuously.
• One of the key properties of the natural logarithm is that \( e^{\ln(y)} = y \).

In the context of solving equations, understanding natural logarithms helps in isolating variables when equations are given in logarithmic form. It transforms multiplication into addition, which simplifies the process of handling exponential relationships.

Exponential Function

The exponential function is a mathematical function denoted by \( e^x \), which expresses how a quantity increases or decreases at a rate proportional to its own value. It's widely used in various fields such as finance, physics, and biology to model exponential growth or decay.

• In an equation, \( e^x \) means "\( e \) raised to the power of \( x \)".
• Exponentiation is the opposite process of taking a logarithm.
• The function \( e \) serves as the base of natural logarithms and plays a critical role in continuous compounding.

When solving equations involving logarithms, we often need to "undo" the logarithm to solve for the variable. This involves using the exponential function to both sides. By exponentiating both sides of the equation, we convert a logarithmic expression back into its original form, paving the way to isolate and solve for the unknown variable.

Isolation of Variable

Isolation of variable is a key algebraic technique used in solving equations. Its essence lies in manipulating an equation in order to get the variable of interest by itself on one side of the equation. Understanding this process is essential because it directly leads to finding solutions to equations.

• Begin by simplifying the equation to clearly outline the variable you need to isolate.
• Use inverse operations to "undo" any operations applied to the variable like addition, subtraction, multiplication, or division.
• When dealing with logarithmic equations, first address the natural logarithm or exponential function to reveal the underlying relationship.

In the original problem, after addressing the natural logarithm by using the exponential function, the next step is then to express the equation in simpler terms, allowing an easy isolation of \( x \). Often, this involves simple arithmetic steps, such as subtracting or dividing both sides of the equation, as demonstrated in the rounding and simplification processes used to reach the final solution.