Problem 45 - A Graphical Approach to Precalculus with Limits

Question

Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. $$2 \ln (x-1)+30=34$$

Step-by-Step Solution

Answer

The solution is \(x = e^2 + 1\).

Step 1: Isolate the Logarithm

The given equation is \(2 \ln (x-1) + 30 = 34\). First, subtract 30 from both sides to isolate the logarithmic term. This gives:\[ 2 \ln (x-1) = 4. \]

Step 2: Divide to Solve for the Logarithm

Divide both sides by 2 to solve for \(\ln(x-1)\): \[ \ln(x-1) = 2. \]

Step 3: Exponentiate to Eliminate the Logarithm

To eliminate the natural logarithm, exponentiate both sides using base \(e\). This means use \(e\) as the base to power both sides: \[ e^{\ln(x-1)} = e^2. \] Because \(e^{\ln(x-1)}\) simplifies to \(x-1\), we have \[ x-1 = e^2. \]

Step 4: Solve for x

Add 1 to both sides of the equation to solve for \(x\): \[ x = e^2 + 1. \]

Step 5: Verify the Solution

Use a calculator to check the exact form solution. First compute \(e^2\) and then add 1. This should satisfy the original equation. The exact form of the solution is consistent in all verifications.

Key Concepts

Natural Logarithm

A natural logarithm is a logarithm with the base of the irrational number \(e\), which is approximately equal to 2.718. When we see \( \ln(x) \), it is equivalent to saying the logarithm of \(x\) to the base \(e\). Natural logarithms are denoted by \( \ln \) instead of the usual \( \log \), which is used for base 10. Understanding this is crucial when solving logarithmic equations because it provides a way to handle exponential growth or decay.

In the exercise, we encountered \(2 \ln(x-1)\). The purpose of isolating the logarithm, as seen in Step 1 and 2, is to simplify terms so that we can easily exponentiate and manage the equation. This simplification makes subsequent operations straightforward.

• Logarithms transform multiplicative operations into additive ones, which helps in solving the equation.
• The property \( a \cdot \ln(b) = \ln(b^a) \) simplifies expressions by absorbing constants into the logarithm.

Exponentiation

Exponentiation is the process of raising a number to a power. In the context of logarithms, it's the key tool used to "undo" or eliminate a logarithmic operation. In the step-by-step solution, exponentiation is used to eliminate the natural logarithm once \( \ln(x-1) \) was isolated. By raising \(e\) to the power on both sides, the equation is simplified.

This is possible due to the property \( e^{\ln(b)} = b \), which highlights that the exponential and logarithmic functions are inverses. This concept is fundamental because it allows us to solve for \(x\) by rewriting the equation in a solvable form.

• Exponentiation reverses the effect of the logarithm, converting back to its original form.
• It simplifies expressions to a form where algebraic methods can be applied.

When you see \( \ln(x-1) = 2 \), converting it to \( e^2 \) helps unravel the unknown \(x\), making the isolate-and-solve technique effective.

Exact Form Solution

An exact form solution retains mathematical expressions in their original, uncluttered format, avoiding approximations or decimals unless calculation requires. It's about expressing the answer succinctly and in a pure mathematical form, which is necessary when precision is crucial.

In our exercise, the exact form solution was \( x = e^2 + 1 \). This means instead of estimating \(e^2\) as a decimal (which can go on indefinitely), we retain it as \(e^2\) in our final answer. This approach ensures accuracy and easily allows conversion to decimals if needed in future calculations.

• Exact form is crucial in fields requiring precision like engineering and physics.
• It provides a versatile format that can convert to any needed form later, such as numeric computation on a calculator.

By verifying with a calculator, we confirm that the mathematical expression works in both exact form and numerical approximations, aiding students in learning the importance of accuracy in solutions.