Problem 63 - 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. $$ 6+2 \ln x=5 $$

Step-by-Step Solution

Answer

\[ x \approx 0.61 \]

Step 1: Isolate the logarithmic term

First, we start by isolating the logarithmic expression on one side of the equation. We do this by subtracting six from both sides. This yields: \(2 \ln x = 5 - 6\) or \(2 \ln x = -1\).

Step 2: Solve for logarithmic variable

Now we need to get just \( \ln x \) on one side by dividing both sides by two: \( \ln x = -1 / 2\) or \( \ln x = -0.5\).

Step 3: Convert into exponential form and solve

We now convert the equation from logarithmic to exponential form. In general, the logarithm base b of a number x: \( \ln x = n \) translates to \( e^n = x \). Our equation becomes \( e^{-0.5} = x \)

Step 4: Check the domain and calculate the solution

Logarithm terms are undefined for arguments less than or equal to zero. However, as \( e^{-0.5} \) gives a positive value, it is within the domain. Hence, the solution to the equation is \( x = e^{-0.5} \). Because we need a decimal approximation, calculate the value of \( e^{-0.5} \) to get the decimal solution. \( x \approx 0.61 \) correct to two decimal places.

Key Concepts

exponential functions

Exponential functions are mathematical expressions that involve a constant base raised to a variable exponent. In the context of this exercise, we deal with the function \( e^x \), where \( e \) is a special mathematical constant approximately equal to 2.71828. It is the base of natural logarithms and is crucial in calculus and complex analysis.

Exponential functions display rapid growth or decay, depending on whether the exponent is positive or negative. This is visible in our problem when we converted \( \ln x = -0.5 \) to exponential form, giving us \( e^{-0.5} = x \).

Here's why exponential functions are significant:

• They model growth and decay processes, such as population growth, radioactive decay, and compound interest.
• They have unique properties, such as a constant percentage rate of change.
• They provide solutions to differential equations widely used in physics and engineering.

logarithmic functions

Logarithmic functions are the inverse of exponential functions. If an exponential function is represented as \( y = b^x \), then its logarithmic form is \( x = \log_b(y) \). For natural logarithms, where the base \( b \) is \( e \), the function is written as \( \ln x \).

Logarithms have several key applications:

• They transform multiplicative processes into additive ones, simplifying calculations.
• They are used in solving exponential equations, as done in our exercise.
• They appear in various fields, including statistics, to measure order of magnitude and growth rates.

For example, in the given exercise, the equation \( 2 \ln x = -1 \) allows for simplifying and understanding the growth rate by converting it to an exponential form:\( e^{-0.5} = x \). This conversion is possible thanks to the properties of logarithms, making them a crucial mathematical tool.

domain of a function

The domain of a function refers to all possible input values (or 'x' values) that will not lead to undefined mathematical expressions. For logarithmic functions like \( \ln x \), the domain consists of all positive real numbers, \( x > 0 \). This is because the logarithm of a non-positive number is undefined.

Understanding the domain is essential:

• It ensures that the function is applicable and produces a real number result.
• It helps in identifying and rejecting any invalid solutions for the given equation, as stated in the original exercise instructions.
• It guides the problem-solving process by clarifying which values can be input into the function safely.

In our problem, we conclude that \( x = e^{-0.5} \approx 0.61 \) is an acceptable solution because it satisfies the domain condition \( x > 0 \). This confirmation step is vital for validating solutions and ensuring they adhere to the mathematical rules of logarithms.