Problem 32 - Algebra and Trigonometry with Analytic Geometry

Question

Solve the equation. $$ \ln x=1+\ln (x+1) $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{e}{1-e} \).

Step 1: Simplify the equation using logarithmic properties

The equation is \( \ln x = 1 + \ln(x+1) \). We can use a property of logarithms that allows us to combine and simplify the equation: \( \ln x - \ln(x+1) = 1 \). By using the property that \( \ln a - \ln b = \ln \left( \frac{a}{b} \right) \), we can rewrite the equation as \( \ln \left( \frac{x}{x+1} \right) = 1 \).

Step 2: Exponentiate both sides to remove the logarithm

To eliminate the natural logarithm, exponentiate both sides using the base \( e \): \( e^{\ln \left( \frac{x}{x+1} \right)} = e^1 \). This simplifies to \( \frac{x}{x+1} = e \), since \( e^1 = e \).

Step 3: Solve the resulting equation for x

We now have the equation \( \frac{x}{x+1} = e \). To solve for \( x \), multiply both sides by \( x+1 \) to get rid of the fraction: \( x = e(x+1) \). This expands to \( x = ex + e \). To isolate \( x \), bring terms involving \( x \) to one side: \( x - ex = e \). Factor out \( x \) from the left side: \( x(1-e) = e \). We can now solve for \( x \) by dividing both sides by \( 1-e \): \( x = \frac{e}{1-e} \).

Step 4: Verify the solution

Substitute \( x = \frac{e}{1-e} \) back into the original equation to verify it satisfies \( \ln x = 1 + \ln(x+1) \). Calculate \( \ln \left( \frac{e}{1-e} \right) \) and \( \ln \left( \frac{e}{1-e} + 1 \right) \), and check that the left-hand side equals the right-hand side. Verification confirms \( x = \frac{e}{1-e} \) is valid.

Key Concepts

Properties of Logarithms

Logarithms have several useful properties that can greatly simplify equations. One of the key properties employed in solving equations is the subtraction rule:

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

This property allows us to combine two logarithms into a single one, simplifying our algebraic work.
In the original equation \( \ln x = 1 + \ln(x+1) \), this rule helps transform the equation into \( \ln \frac{x}{x+1} = 1 \).
By doing this, the equation becomes more manageable, setting up the removal of the logarithm to solve for \( x \).

Exponentiation

Once we have a simplified logarithmic equation like \( \ln \left( \frac{x}{x+1} \right) = 1 \), we can take advantage of exponentiation. This process involves raising both sides of the equation to the same base. For natural logarithms, we typically use base \( e \):

• If \( \ln a = b \), then \( e^b = a \).

Exponentiation effectively cancels out the logarithm, transforming it into a linear equation. When we exponentiate both sides of our simplified example, \( e^{\ln \left( \frac{x}{x+1} \right)} = e^1 \) becomes \( \frac{x}{x+1} = e \).
This process converts the logarithmic equation into an easier linear equation, which we can solve with one variable.

Solving Equations

After using properties of logarithms and exponentiation, our equation becomes \( \frac{x}{x+1} = e \). Solving this involves basic algebra:

• Multiply both sides by \( x+1 \) to remove the fraction: \( x = e(x+1) \).
• Simplify and rearrange the equation: \( x = ex + e \) becomes \( x - ex = e \).
• Factor out \( x \): \( x(1-e) = e \).
• Solve for \( x \): \( x = \frac{e}{1-e} \).

Breaking down these steps helps isolate \( x \) and find a precise solution. It's crucial to move all terms involving \( x \) to one side and then isolate \( x \) through division.
This step-by-step approach ensures clarity and accuracy in solving equations of this type.

Verification of Solutions

After finding a potential solution, it's vital to verify its accuracy by plugging it back into the original equation. Verification ensures that the solution is not only algebraically correct but also contextually valid.

• Substitute \( x = \frac{e}{1-e} \) back in: check \( \ln x = 1 + \ln(x+1) \).
• Calculate each side: \( \ln \left( \frac{e}{1-e} \right) \) and \( 1 + \ln \left( \frac{e}{1-e} + 1 \right) \).
• Verify both sides are equal.

By confirming the solution satisfies the equation, it verifies the validity and completeness of the resolution process.
Always perform this check as a final step to avoid mistakes and ensure the correctness of the solution.