Problem 66 - College Algebra

Question

Solve each logarithmic equation in Exercises \(49-92 .\) 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 \(x = e^{-6}\) and the decimal approximation is \(x = 0.0025\).

Step 1: Express the equation in a simplified form

First, rewrite the given equation \(\ln \sqrt{x}+4=1\) in a simplified logarithmic form. A square root can be rewritten as a fraction exponent, so \(\sqrt{x}\) becomes \(x^{1/2}\). Therefore, the equation becomes: \(\ln (x^{1/2})+4=1\).

Step 2: Isolate the logarithm

The next step is to isolate the logarithmic term. Subtract 4 from both sides of the equation: \(\ln (x^{1/2}) = 1 - 4 = -3\).

Step 3: Convert the logarithmic equation into an exponential equation

Use the conversion rule which states that \(\ln a = b\) is equivalent to \(e^b = a\), to convert the equation \(\ln (x^{1/2}) = -3\) into an exponential equation: \(e^{-3} = x^{1/2}\).

Step 4: Solve for \(x\)

To find \(x\), square both sides of the equation: \(e^{-6} = x\).

Step 5: Check for domain restrictions

Since the base of the natural logarithm function is \(x > 0\), any solution that satisfies this restriction is valid. Thus, \(x = e^{-6}\) is a valid solution.

Step 6: Get a decimal approximation

To find the decimal approximation, plug \(x = e^{-6}\) into a calculator, and round to two decimal places. Thus, \(x = 0.0025\).

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a special type of logarithm that uses the number \( e \) (approximately equal to 2.71828) as its base. This makes it particularly useful in mathematics, especially in calculus and complex equations. In the context of our problem, the natural logarithm helps to express relationships involving exponential growth or decay, which are common in the real world.

The symbol \( \ln(x) \) essentially asks: "To what power must \( e \) be raised to obtain \( x \)?" Understanding this meaning is vital for solving equations that involve natural logarithms, such as turning them into exponential equations when needed.

Domain Restrictions

When dealing with logarithmic equations, it is crucial to consider the domain restrictions. The domain of the natural logarithm function \( \ln(x) \) is limited to positive real numbers, meaning \( x > 0 \). This is because you cannot take the logarithm of a non-positive number, as it does not produce a real number result.

In solving \( \ln \sqrt{x} + 4 = 1 \), ensuring that our solution respects this domain restriction is important. This leads us to realize that when \( \ln \sqrt{x} = -3 \), \( x \) must be positive. Checking for domain restrictions ensures that any proposed solution remains valid under the mathematical constraints of the original problem.

Exponential Equation

Converting a logarithmic equation into an exponential equation is a powerful technique to find solutions. Using the definition \( \ln a = b \rightarrow e^b = a \), you can switch the approach to solving the equation.

In our example, once the logarithm was isolated as \( \ln (x^{1/2}) = -3 \), it was transformed into an exponential equation: \( e^{-3} = x^{1/2} \). Solving this required squaring both sides to remove the square root and solve for \( x \). This process showcases how understanding both exponential and logarithmic forms can lead to solutions that might not be apparent initially.

Decimal Approximation

After finding the exact solution, \( x = e^{-6} \), it's often helpful to provide a decimal approximation, especially in real-world applications where exact values might be cumbersome.

Using a calculator, we find that \( e^{-6} \approx 0.00247875 \). Rounding this to two decimal places, the solution simplifies to \( x \approx 0.0025 \). This approximation is handy because it allows for a more intuitive understanding of the size or magnitude of the value in question, making it easier to communicate and apply practically.