Problem 9 - Thomas Calculus

Question

In Exercises \(5-10,\) solve for \(y\) in terms of \(t\) or \(x,\) as appropriate. $$ \ln (y-1)-\ln 2=x+\ln x $$

Step-by-Step Solution

Answer

\( y = 2xe^x + 1 \).

Step 1: Use Logarithmic Properties

Apply the property of logarithms where the difference of two natural logs can be written as the natural log of their division. Rewrite the equation: \( \ln(y-1) - \ln 2 = \ln\left(\frac{y-1}{2}\right) = x + \ln x \).

Step 2: Eliminate the Logarithm by Exponentiation

To remove the logarithm, exponentiate both sides of the equation. This yields: \( \frac{y-1}{2} = e^{x + \ln x} \). Since \( e^{\ln x} = x \), rewrite it as: \( \frac{y-1}{2} = xe^x \).

Step 3: Isolate the Expression for y

Multiply both sides of the equation by 2 to isolate \( y - 1 \): \( y - 1 = 2xe^x \).

Step 4: Solve for y

Finally, add 1 to both sides to solve for \( y \): \( y = 2xe^x + 1 \).

Key Concepts

Logarithmic Properties

Logarithms transform multiplication into addition, and division into subtraction, which can simplify the manipulation of equations. For natural logarithms (denoted as \( \ln \)), a key property used in the original exercise is:

• \( \ln a - \ln b = \ln \left(\frac{a}{b}\right) \)

This property vastly simplifies the given expression \( \ln(y-1) - \ln 2 \) to \( \ln\left(\frac{y-1}{2}\right) \).
This reduction is essential in solving logarithmic equations because it reduces two terms into a single logarithmic expression.
By combining these terms, it becomes easier to exponentiate and simplify further.

Exponentiation

Exponentiation is the process of raising a number to a power. In the context of solving equations with logarithms, exponentiation is often used to "undo" a logarithm. If \( \ln(a) = b \), then exponentiating both sides gives \( a = e^b \), where \( e \) is the base of the natural log.

In our exercise, once we have \( \ln\left(\frac{y-1}{2}\right) = x + \ln x \), exponentiating both sides results in \( \frac{y-1}{2} = e^{x + \ln x} \).
On the right side, due to properties of exponents, \( e^{x + \ln x} \) simplifies to \( xe^x \) because \( e^{\ln x} = x \). This transformation reduces the complexity, bringing us closer to solving for \( y \).

Equations

Solving an equation involves isolating the variable you are solving for. In this exercise, once the logarithm is converted, the equation is \( \frac{y-1}{2} = xe^x \).

The next step requires the isolation of \( y \):

• Multiply both sides by 2: \( y - 1 = 2xe^x \).
• Add 1 to both sides to solve for \( y \): \( y = 2xe^x + 1 \).

This sequence of operations involves basic algebraic manipulation, ensuring the desired variable is isolated, allowing for an explicit solution in terms of other variables.

Natural Logarithm

The natural logarithm, \( \ln \), is a crucial concept in calculus and mathematical analysis. It is the logarithm to the base \( e \), where \( e \) is approximately 2.71828. Natural logs are particularly useful because they simplify many mathematical expressions involving growth processes.

In the provided exercise, the natural logarithm allowed us to simplify the subtraction of logs into a division within a single \( \ln \).
By converting \( \ln \) expressions using exponentiation, equations relying on logs can be linearized or solved more straightforwardly. Such tools are valuable for tackling a broad range of calculus problems, simplifying complex expressions into manageable forms.