Problem 46 - Precalculus with Limits

Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. \(\ln x-7=0\)

Step-by-Step Solution

Answer

The solution to the equation \( \ln x-7 = 0 \) is \( x \approx 1096.633 \)

Step 1: Isolate the log expression

The first step is to isolate the logarithmic expression on one side of the equation by adding 7 to both sides. As a result our equation will be \( \ln x = 7 \)

Step 2: Eliminate the log

Now we can eliminate natural log by transforming this logarithmic expression into an exponential expression since base 'e' exponential function is equivalently inverse to the natural logarithm function. Doing this change will result in the equation \( x = e^7 \)

Step 3: Approximate the solution

The final step is to compute the exponent of 'e' to get the solution to 'x'. Due to specifics of the exercise the result should be approximated to three decimal places. Thus, our solution is \( x \approx 1096.633 \)

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a specific type of logarithm. It uses the constant \( e \) as its base, where \( e \approx 2.71828 \). The natural logarithm is useful in numerous fields, including calculus and higher-level algebra.

Natural logarithms follow many of the same properties as common logarithms:

• \( \ln(1) = 0 \) because \( e^0 = 1 \)
• \( \ln(e) = 1 \) because \( e^1 = e \)
• \( \ln(ab) = \ln(a) + \ln(b) \)
• \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
• \( \ln(a^b) = b \cdot \ln(a) \)

In solving the equation \( \ln x - 7 = 0 \), the natural logarithm \( \ln x \) was isolated first. This step allows us to manipulate the log expression more easily by transforming it into an exponential form.

Exponential Functions

Exponential functions are functions that involve the constant \( e \), where \( e \) is approximately 2.71828. These functions often take the form \( f(x) = a \cdot e^{bx} \), where \( a \) and \( b \) are constants. The importance of \( e \) lies in its properties related to growth and decay, among other natural phenomena.

Exponential functions are the inverse of natural logarithms, a relationship that we utilize in this equation. By applying the inverse, given \( \ln x = 7 \), we switched to an exponential form: \( x = e^7 \). The key idea here is that exponentiating both sides with base \( e \) cancels out the natural logarithm, allowing us to solve for \( x \). This exponential relationship makes it easy to calculate \( x \) once the logarithm is isolated.

Algebraic Manipulation

In mathematics, algebraic manipulation refers to the process of rearranging and simplifying expressions and equations to find solutions. For logarithmic equations, this process usually involves isolating the logarithmic part first.

In the exercise, we started with the equation \( \ln x - 7 = 0 \). Through algebraic manipulation, we performed the following steps:

• Added 7 to both sides to isolate the logarithm: \( \ln x = 7 \).
• By using the property of inverse functions, transformed the logarithmic equation into an exponential one: \( x = e^7 \).
• Finally, calculated the value of \( e^7 \) and rounded to three decimal places to get \( x \approx 1096.633 \).

These manipulative steps and the transition from log to exponential form illustrate the problem-solving process, making it easier to understand such equations.