Problem 12 - Precalculus: A Concise Course

Question

Determine whether each \(x\) -value is a solution (or an approximate solution) of the equation. \(\ln (x-1)=3.8\) (a) \(x=1+e^{3.8}\) (b) \(x \approx 45.701\) (c) \(x=1+\ln 3.8\)

Step-by-Step Solution

Answer

The values \(x = 1+e^{3.8}\) and \(x \approx 45.701\) are solutions of the equation, while \(x = 1 + \ln 3.8\) is not a solution to the equation.

Step 1: Substitute \(x = 1+e^{3.8}\) into the equation

Substitute \(x = 1+e^{3.8}\) into the equation \(\ln (x-1)=3.8\). This gives \(\ln((1+e^{3.8})-1)=3.8\). Simplifying this gives \(\ln(e^{3.8})=3.8\) which is true as the natural logarithm of \(e^{3.8}\) equals \(3.8\). So, \(x =1+e^{3.8}\) is a solution to the equation.

Step 2: Substitute \(x \approx 45.701\) into the equation

Substitute \(x \approx 45.701\) into the equation \(\ln (x-1)=3.8\). This gives \(\ln((45.701)-1) \approx 3.8\). Evaluating the left-hand side gives a number very close to \(3.8\). As \(45.701\) was rounded, the left-hand side is an approximation of \(3.8\). So, \(x \approx 45.701\) is an approximate solution of the equation.

Step 3: Substitute \(x = 1+\ln 3.8\) into the equation

Substitute \(x = 1+\ln 3.8\) into the equation \(\ln(x-1) = 3.8\). This gives \(\ln((1+\ln 3.8)-1) = 3.8\). Simplifying this gives \(\ln(\ln 3.8)=3.8\). Evaluating the left-hand side doesn't give \(\ln(\ln 3.8)=3.8\), but it results in a different value. So, \(x=1+\ln3.8\) is not a solution to the equation.

Key Concepts

Logarithmic Equations

When solving logarithmic equations, the goal is to find the value of the variable that makes the equation true. This often involves leveraging the properties of logarithms, such as the fact that if \(\log_b(x) = y\), then \(b^y = x\). The natural logarithm, often represented as \(\ln\), is a specific type of logarithm where the base is \(e\), Euler's number, approximately equal to 2.71828.

In the exercise, \(\ln(x-1) = 3.8\), the key step is to isolate \(x\) by applying the inverse operation. Using the property that \(\ln(e^y)=y\), we can solve for \(x\) by raising \(e\) to the power of both sides, which is effectively the inverse operation of taking the natural logarithm.

Exponential Functions

An exponential function is a mathematical expression in the form \(b^x\), where \(b\) is a constant called the base and \(x\) is the exponent. The function rapidly increases as \(x\) grows when the base is greater than 1, which is a typical characteristic of exponential growth. In the context of the natural logarithm equation \(\ln (x-1) = 3.8\), when we rewrite the equation to find \(x\), we utilize the fundamental relationship between logarithms and exponentials: \(x = 1+e^{3.8}\) represents an exponential function where \(e\) is the base and \(3.8\) is the exponent. This shows the direct application of exponential functions in solving logarithmic equations.

Approximate Solutions

Approximate solutions are often used when a precise answer is either impossible or unnecessary. These solutions are close enough to the exact value for practical purposes. In many cases, complex equations or those involving irrational numbers like \(e\) require approximation. The textbook exercise provided an example of an approximate solution, \(x \approx 45.701\), derived from the natural logarithm equation. It's worth noting that approximation is a powerful tool in mathematics, allowing us to work with numbers and equations that might otherwise be unwieldy. When we evaluate \(\ln(x-1)\) using the approximate value, it yields a result close to the given number \(3.8\), demonstrating its validity as an approximation.

Mathematical Substitution

Mathematical substitution is a technique commonly used to solve equations. It involves replacing a variable with a numerical value or another expression to simplify the problem. This method can also reveal whether a particular value is a solution to the equation. In the original exercise, substitution is used in three different cases. For example, substituting \(x\) with \(1+e^{3.8}\) reduces the equation to \(\ln(e^{3.8})=3.8\), which is a true statement, indicating that the value is indeed a solution. On the other hand, substituting \(x\) with \(1+\ln 3.8\) does not satisfy the original logarithmic equation, leading to the conclusion that this value is not a solution. Substitution helps to clarify which values satisfy the given equation and which do not.