Problem 55 - Precalculus with Limits

Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. \(\ln x-\ln (x+1)=2\)

Step-by-Step Solution

Answer

The solution to the equation is approximately \(x=1.582\).

Step 1: Apply the Properties of Logarithms

You can simplify the given equation \(\ln x-\ln (x+1)=2\) by applying one of the fundamental properties of logarithms which states that \(\ln(a) – \ln(b) = \ln(a/b)\). It then becomes \(\ln \left(\frac{x}{x+1}\right) = 2\)

Step 2: Remove the Logarithm

You remove the logarithm by exponentiating both sides of the equation using the base of the natural logarithm, which is e. This gives the equation \(\ e^{\ln \left(\frac{x}{x+1}\right)}=e^2\), which then simplifies to \(\left(\frac{x}{x+1}\right)=e^2\).

Step 3: Solve the Resulting Equation

With \(\left(\frac{x}{x+1}\right)=e^2\), you can cross multiply to get rid of the fraction resulting in the quadratic equation \(x = e^2(x+1)\). Expanding and simplifying this results in \(x-e^2x = e^2\), and further simplifying leads to \(x(e^2-1) = e^2\). Solving for x gives \(x=\frac{e^2}{e^2-1}\).

Step 4: Approximate the Result

The final step is to calculate the value of \(x=\frac{e^2}{e^2-1}\) and round it to three decimal places.

Key Concepts

Properties of Logarithms

The properties of logarithms are essential in understanding and solving logarithmic equations. Just like properties of exponents, logarithms have their own set of rules which make computations more manageable.

One key property that we use in the example exercise is the quotient rule of logarithms. It states that the logarithm of a quotient is equal to the difference of the logarithms, mathematically defined as \( \log(a/b) = \log(a) - \log(b) \). Another property that often comes in handy is the product rule, which says that the logarithm of a product is the sum of the logarithms: \( \log(ab) = \log(a) + \log(b) \).

Understanding these properties allows you to manipulate logarithmic expressions and often make them simpler to solve, just as in the step by step solution where \( \ln x - \ln (x+1) \) was simplified to \( \ln(x/(x+1)) \) using the quotient rule.

Solving Logarithmic Equations

Solving logarithmic equations involves isolating the logarithmic expression and then finding the value of the variable contained within it. To solve a logarithmic equation, you must be adept at working with the properties of logarithms.

Once the equation is simplified using the properties of logarithms, as seen with \( \ln \left(\frac{x}{x+1}\right) = 2 \) in our exercise, you often need to exponentiate both sides to remove the logarithm. This step usually involves using the base of the logarithm to both sides of the equation, effectively 'canceling out' the log. This technique was illustrated in Step 2 of the solution, where we exponentiated both sides of the equation with the base of the natural log, \( e \).

This step resulted in \( \frac{x}{x+1} = e^2 \) and allowed us to solve for \( x \) without the logarithm present, leading to an equation where \( x \) can be isolated and found in terms of \( e \) and its powers.

Natural Logarithm Exponentiation

Natural logarithm exponentiation refers to the process of raising \( e \), the base of the natural logarithm, to a power that is given by a logarithm. This is a critical step in solving natural logarithmic equations.

For instance, from the example provided, exponentiating both sides of the simplified logarithmic equation \( \ln \left(\frac{x}{x+1}\right) = 2 \) with \( e \), leads to the equation \( e^{\ln \left(\frac{x}{x+1}\right)} = e^2 \). According to the inverse properties of logarithms and exponentiation, \( e^{\ln a} = a \), which simplifies our original equation to \( \frac{x}{x+1} = e^2 \).

This is a fundamental concept in solving logarithmic equations because it allows us to transition from logarithmic form to exponential form, thereby enabling us to isolate and solve for the variable in question. In our exercise, after exponentiation, we were able to cross multiply and solve for \( x \) to arrive at \( x = \frac{e^2}{e^2 - 1} \) which is the critical final step before approximating the solution to three decimal places.