Problem 66 - College Algebra Essentials

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 exact solution to the equation is \( x=e^2-4 \). The decimal approximation for the solution is approximately \( x \approx 3.39 \)

Step 1: Rewrite the logarithmic equation

Rewrite \( \ln \sqrt{x+4}=1 \) in exponential form. The base for natural logarithms is \( e \), so \( e^1 = \sqrt{x+4} \). Therefore, the equation is \( e=\sqrt{x+4} \).

Step 2: Solve for x

To solve for x, first square both sides to get rid of the square root. \( (e)^2=(\sqrt{x+4})^2 \) simplifies to \( e^2=x+4 \). Then, subtract 4 from both sides. \( x=e^2-4 \).

Step 3: Check the solution

Now, check that the solution is in the domain of the original equation. The domain of a logarithmic function is \( x > 0 \). So, for \( x+4 \) to be positive, \( x \) must be greater than -4. Obviously, \( e^2-4 \) is greater than -4, meaning the solution is within the domain.

Step 4: Decimal approximation

Find a decimal approximation for the solution up to two decimal places using a calculator. Substituting \( e \approx 2.72 \) into \( e^2-4 \), a decimal approximation for \( x \) is 3.39.