Problem 63 - 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. $$\ln \sqrt{x+3}=1$$

Step-by-Step Solution

Answer

The exact solution is \( x = e^2 - 3 \). After rounding off to two decimal places, the approximate value is \( x = 4.39 \).

Step 1: Remove the natural logarithm

To remove the natural log from the equation, one must exponentiate both sides with base \(e\) (Euler's number) to get a standard quadratic equation. This gives: \(e^{\ln \sqrt{x+3}} = e^1 \). Simplifying that, we get \( \sqrt{x+3} = e \).

Step 2: Square both sides

The square root can be eliminated by squaring both sides of the equation obtained in step 1. This gives: \( (\sqrt{x+3})^2 = e^2 \). Simplifying, we get \( x + 3 = e^2 \).

Step 3: Solve for \(x\)

Subtract 3 from both sides of the equation to isolate \( x \). This gives us \( x = e^2 - 3 \).

Step 4: Decimal Approximation

Substitute \( e^2 \) with its value of approximately 7.39 to get the decimal solution. Thus, \( x \) is approximately 7.39 - 3 = 4.39.

Step 5: Check the Domain

Finally, we need to check if \( x = e^2 - 3 \) lies in the domain of the original logarithmic expression, since the arguments of logarithms are always positive. The necessary condition is \( x+3 > 0 \). Substituting \( x = e^2 - 3 \), it's valid because \( e^2 > 3 \), thus it does lie in the domain of the original logarithmic equation.