Problem 64 - Introductory and Intermediate Algebra for College Students 4th

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 is \( x = e^2 - 4 \). The decimal approximation (to two decimal places) depends on the calculator used for it.

Step 1: Remove the logarithm

In the first step, we want to simplify the equation by eliminating the natural logarithm function: \( \ln \sqrt{x+4} = 1 \) We can rewrite \(\ln a = b\) in exponential form as \(a = e^b\), so: \( \sqrt{x+4} = e^1 = e \)

Step 2: Square both sides of the equation

We want to remove the square root, so we square both sides of the equation: \( (\sqrt{x+4})^2 = e^2 \) Which simplifies to: \( x + 4 = e^2 \)

Step 3: Solve for x

Subtract 4 from both sides to isolate \(x\): \( x = e^2 - 4 \)

Step 4: Check if the solution is in the domain

We need to check if this value of \( x \) lies in the domain of the original logarithmic expression. The domain is \( x > -4 \) (since \(x+4 > 0\)). As \( e^2 > 4 \), \(x\) will be a positive number and thus inside the domain of the original logarithmic expression.

Step 5: Find a decimal approximation

If needed, we can convert the exact solution to a decimal approximation (correct to two decimal places) using a calculator.