Problem 49 - College Algebra

Question

Exercises \(45-52\) involve equations with natural logarithms. Solve each equation by isolating the natural logarithm and exponentiating both sides. Express the answer in terms of \(e\) Then 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

The solution to the equation is \(x = e^{-0.5}\), which approximates to 0.61 when calculated to two decimal places.

Step 1: Isolate the natural logarithm

First, ensure that the natural logarithm is isolated on one side of the equation. To do this, subtract 6 from both sides of the equation: \(2 \ln x = 5 - 6 = -1\)

Step 2: Simplify the logarithmic expression

Next, divide by 2 on both sides to simplify the logarithmic expression further to eliminate the coefficient of the logarithm: \(\ln x = -1 / 2 = -0.5\)

Step 3: Exponentiate both sides

Exponentiate both sides of the equation to eliminate the logarithm. Here, 'e' is the base of the natural logarithm, hence we raise both sides of the equation to the power of 'e': \(x = e^{-0.5}\)

Step 4: Calculate decimal approximation

Finally, use a calculator to compute the decimal approximation of the solution. Keep in mind that the approximation should be correct to two decimal places.

Key Concepts

Isolating the Natural Logarithm

When you encounter a natural logarithm equation—like in the example where we have an equation involving \( \ln x \)—the goal is to make \( \ln x \) the subject. This process, known as isolating the logarithm, involves performing algebraic operations in order to have the natural logarithm by itself on one side of the equation.

In our exercise, the natural logarithm is bundled up with other numbers, so the first step is to get rid of those. Subtracting numbers or dividing coefficients attached to the \( \ln \) term achieves this. After you've subtracted 6 from both sides, you end up with \( 2 \ln x = -1 \). The \( \ln \) is not quite isolated yet because of the coefficient 2. Dividing both sides by 2 gives \( \ln x = -0.5 \), which is the logarithm in its most simplified form.

Exponentiating Both Sides

Once we have the natural logarithm on its own, the next puzzle to solve is how to 'free' the \( x \) from the confines of the logarithm. This is where the concept of exponentiating comes into play. It's a fancy term that simply means raising both sides of the equation to a power—in this case, the base of the natural logarithm, which is \( e \).

To undo the natural logarithm, we employ the identity \( e^{\ln x} = x \), which follows one of the basic properties of logarithms. Thus, raising \( e \) to the power of both sides of \( \ln x = -0.5 \) gives us \( e^{\ln x} = e^{-0.5} \), which simplifies down to \( x = e^{-0.5} \). This transformation is a critical step in solving natural logarithm equations and can be applied to a variety of logarithmic forms.

Natural Logarithm Properties

Understanding the properties of the natural logarithm is essential for solving equations like these. A few key properties include:

• The product rule: \( \ln(ab) = \ln a + \ln b \), which is useful when dealing with multiplication within the logarithm.
• The quotient rule: \( \ln \left(\frac{a}{b}\right) = \ln a - \ln b \), helpful when you have divisions.
• The power rule: \( \ln(a^b) = b \ln a \), which allows you to move exponents in front of the logarithm.

Remembering these relationships makes manipulating logarithmic expressions easier. In the given exercise, these properties aren't directly used, but they build the foundation that leads us to the rule that \( e^{\ln x} = x \), allowing us to exponentiate effectively and isolate \( x \).

Decimal Approximation of 'e'

The number \( e \) is known as Euler's number, which is an irrational number approximately equal to 2.71828. When you need a decimal approximation of an expression involving \( e \), you'll often use a calculator to find the value. This is particularly true when dealing with natural logarithms in equations.

After exponentiating both sides, we found that \( x = e^{-0.5} \), and now we need a more concrete number. This is typically done by entering \( e^{-0.5} \) into a calculator and rounding to the desired precision. In our case, two decimal places are required. Remember, because \( e \) is irrational, it can't be represented exactly by a decimal, so we approximate. For this exercise, if you punched the numbers into a calculator, you'd find that \( x \) is approximately 0.61.