Problem 45 - Algebra 2

Question

Solve each equation. Round to the nearest ten-thousandth. $$ \ln (x+2)=5 $$

Step-by-Step Solution

Answer

x ≈ 146.4132

Step 1: Understand the Problem

We need to solve for \(x\) in the equation \(\ln (x+2) = 5\). The natural logarithm function can be inversely solved using the exponential function.

Step 2: Apply the Exponential Function

Exponentiate both sides of the equation to eliminate the natural logarithm. This gives us \(e^{\ln(x+2)} = e^5\). Since \(e^{\ln(x+2)} = x+2\), the equation simplifies to \(x + 2 = e^5\).

Step 3: Isolate x

Subtract 2 from both sides of the equation to solve for \(x\). This results in \(x = e^5 - 2\).

Step 4: Calculate the Numerical Value

Now, calculate \(e^5\) using a calculator. \(e^5 \approx 148.4132\). Thus, \(x = 148.4132 - 2 = 146.4132\).

Step 5: Round to the Nearest Ten-Thousandth

Round the result to the nearest ten-thousandth. This means rounding 146.4132 to four decimal places, resulting in \(146.4132\).

Key Concepts

Exponential Function

In mathematics, the exponential function is a powerful tool prominently featured in various equations, including logarithmic equations. The exponential function, denoted as \(e^x\), where \(e\) is Euler's number approximately equal to 2.71828, grows rapidly and is fundamental in calculus. In our problem, when we have \(\ln(x+2) = 5\), we use the property that the exponential function is the inverse of the natural logarithm. This means applying the exponential function to both sides of the equation can "undo" the logarithm, leading us to \(e^{\ln(x+2)} = e^5\). This simplifies to \(x+2 = e^5\). Understanding this relationship between logarithms and exponentials is crucial for solving such equations effectively.

Solving Logarithmic Equations

Logarithmic equations like \(\ln (x+2) = 5\) are solved by using the inverse property of logarithms. A logarithmic equation involves a log function with the unknown variable inside the logarithm. To solve it, one transforms the log equation into an exponential equation. By exponentiating both sides of the equation, we turn the logarithmic form into a more straightforward form where we can solve for the unknown variable. Thus, with the equation \(\ln (x+2) = 5\), exponentiating gives \(x+2 = e^5\). From here, solving becomes simple arithmetic with isolating \(x\) on one side by subtracting 2, resulting in \(x = e^5 - 2\). This process efficiently isolates the variable, permitting a direct path to a solution.

Numerical Approximation

Numerical approximation involves estimating the value of expressions that are not easily computed exactly. In cases involving natural logarithms or exponential functions, such estimation is commonplace. For instance, to find \(x = e^5 - 2\), one needs \(e^5\), which is not a simple integer. Calculators or mathematical software help calculate \(e^5\). Here, \(e^5 \approx 148.4132\). Using numerical approximation, we subtract 2 to find \(x \approx 146.4132\). Lastly, rounding helps make the number more digestible or useful in practical contexts. We round \(146.4132\) to four decimal places, producing a neat result. This step is crucial when precision to many decimal places is needed, ensuring accuracy in mathematical and real-world applications.

Inverse Functions

Understanding the concept of inverse functions is vital in solving complex equations. Inverse functions essentially "reverse" the effect of an original function. A perfect example is how the exponential function \(e^x\) and the natural logarithm \(\ln x\) are inverse to each other. If you apply both to a number sequentially, they nullify each other's effects. This property is utilized in our problem where \(\ln(x+2) = 5\). Here, by exponentiating both sides, we employ the inverse effect to simplify \(x+2 = e^5\). Recognizing and applying inverse relationships enables us to transition between different mathematical forms efficiently, making it easier to isolate and solve for variables in diverse types of mathematical problems.