Problem 66 - Algebra and Trigonometry

Question

Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ \ln \sqrt{x+4}=1 $$

Step-by-Step Solution

Answer

The solution is \(x = e^{2} - 4\), or approximately 3.39 when rounded to two decimal places.

Step 1: Simplify the Root

Since the square root is equivalent to raising to the power of 1/2, we can transform the root into the equivalent power representation. Therefore, we have: \(\ln (x+4)^{1/2} = 1\)

Step 2: Applying exponent rules for logarithms

The logarithm of a power can be rewritten as that power times the logarithm, so we can bring down the 1/2, which results in: \(\frac{1}{2}\ln {(x+4)} = 1\)

Step 3: Isolate logarithm term

Move the term with the \(x\) variable to one side by multiplying both sides of the equation by 2. Thus, we have: \(\ln (x+4) = 2\)

Step 4: Transform logarithmic equation to exponential form

Transform the equation from logarithmic form to exponential form. Remember that the base of the natural logarithm, \(\ln\), is \(e\). Therefore, our equation becomes: \(e^{2} = x+4\)

Step 5: Solve for x

Subtract 4 from both sides: \(x = e^{2} - 4\)

Step 6: Check for domain restrictions

The domain for \(\ln (x+4)\) is \(x+4 > 0\). This gives \(x > -4\). Since \(e^{2} - 4 = 3.39\), which is more than -4, the solution is valid.

Step 7: Approximate the value

Using a calculator, you can find that \(e^{2} - 4\) equals approximately 3.39.

Key Concepts

Domain of Logarithms

The domain of a logarithmic function is a crucial aspect to consider. For the natural logarithm function, written as \( \ln(x) \), the expression inside the log must be greater than zero. This means that the expression \( x + 4 \) in \( \ln(x + 4) \) must satisfy the inequality \( x + 4 > 0 \).

By solving this inequality, we find that the domain of \( x + 4 \) is \( x > -4 \). This condition ensures that the value inside the logarithm is always positive, making the function valid.

If this rule is not followed, we end up with invalid results, so always check the domain when working with logarithms.

• Ensure expressions inside \( \ln \) are positive.
• Verify solutions meet the domain requirements to avoid rejection.

Natural Logarithm

The natural logarithm, represented by \( \ln \), is a logarithm with the base \( e \), where \( e \approx 2.718 \, \).

This makes \( \ln \) especially important in calculus and exponential growth models.

For instance, in the equation \( \ln \sqrt{x+4} = 1 \), we use properties of the natural logarithm to simplify and solve problems. The property \( \ln(a^b) = b \cdot \ln(a) \) comes in handy here.

Understanding that \( \ln \) works with the special base \( e \) allows transformations between different forms as seen in solving equations.

• \( \ln \) uses base \( e \), a significant mathematical constant.
• Acknowledge fundamental properties like \( \ln(a^b) = b \cdot \ln(a) \).

Solving Logarithmic Equations

Solving logarithmic equations involves several fundamental steps.

In the given exercise, the expression \( \ln (x+4)^{1/2} = 1 \) requires using the relationship between powers and logarithms. Reducing it to \( \frac{1}{2}\ln(x+4) = 1 \) simplifies the problem.

The goal is to isolate the logarithmic term by applying basic algebraic operations, such as multiplying both sides by necessary factors to simplify and eventually isolate "\( \ln(x+4) \)."

Understanding these steps helps address the key idea that every manipulation stays in line with established mathematical principles.

• Break down the equation step-by-step, isolating log terms.
• Use algebraic transformations to solve smoothly.

Exponential Form Transformation

Converting a logarithmic equation to exponential form is a pivotal step. The equation \( \ln(x+4) = 2 \) can be transformed using \( e \), the base for natural logs.

This changes \( \ln(x+4) = 2 \) into the exponential form \( e^{2} = x+4 \), where the solution to the equation becomes a straightforward algebraic problem.

Subtracting 4 yields the solution \( x = e^{2} - 4 \). This solution aligns with the natural exponent base's power form.

Learning to switch between logarithmic and exponential forms allows for efficient problem-solving.

• Recognize \( \ln \) as a counterpart to base \( e \) exponentiation.
• Apply \( a = \ln(b) \) as \( e^a = b \) for transformations.