Problem 19 - Contemporary Precalculus

Question

Solve the equation. First express your answer in terms of natural logarithms (for instance, \(x=(2+\ln 5) /(\ln 3)) .\) Then use a calculator to find an approximation for the answer. $$6 e^{-1.4 x}=21$$

Step-by-Step Solution

Answer

Question: Solve the equation \(6 e^{-1.4 x}=21\) for x. Answer: The solution to the given equation is approximately \(x \approx -1.006\).

Step 1: Isolate the exponential term

To isolate the exponential term \(e^{-1.4x}\), divide both sides of the given equation by 6. $$\frac{6 e^{-1.4 x}}{6} = \frac{21}{6} \Rightarrow e^{-1.4 x} = \frac{7}{2}$$

Step 2: Apply natural logarithm on both sides

Apply the natural logarithm (ln) on both sides of the equation to eliminate the exponential term. $$\ln(e^{-1.4x}) = \ln\left(\frac{7}{2}\right)$$

Step 3: Use properties of logarithm

Using the properties of logarithm, we can simplify the left side of the equation. Since \(\ln(e^x) = x\), we obtain $$-1.4x = \ln\left(\frac{7}{2}\right)$$

Step 4: Solve for x

Divide both sides of the equation by -1.4 to isolate x. $$x = \frac{\ln\left(\frac{7}{2}\right)}{-1.4}$$ Now, we will use a calculator to approximate the value of x.

Step 5: Approximate x using a calculator

Using a calculator, we can find the approximate decimal value of x. $$x \approx \frac{\ln(3.5)}{-1.4} \approx -1.006$$ So, the solution to the given equation is approximately \(x \approx -1.006\).

Key Concepts

Natural Logarithms

When working with exponential equations, one of your most powerful tools is the natural logarithm, denoted as \( \ln \). This particular logarithm is so named because it uses the constant \( e \), approximately equal to 2.71828, as its base.

In essence, the natural logarithm of a number \( x \) answers the question, 'To what power must \( e \) be raised to produce \( x \)?' That's why \( \ln(e^x) = x \), a rule which is crucial in solving exponential equations where \( e \) is the base of the exponential term. The natural logarithm also has a direct relationship with continuous growth processes, making it highly relevant in fields like economics, biology, and physics.

Moreover, while working with natural logarithms, calculators become very handy to find numerical solutions. It's important for students to understand the theory behind the natural logarithm, but also to get comfortable with using the tool to find approximations, which was exactly the case in the given exercise. In our exercise, the final answer first involved expressing \( x \) in terms of \( \ln \) before reaching an approximation.

Properties of Logarithms

Grasping the properties of logarithms can turn a unwieldy equation into something more manageable. These properties are based on the fundamental nature of logarithms as exponents. Here are a few essential properties:

• Product Rule: \( \ln(ab) = \ln(a) + \ln(b) \)
• Quotient Rule: \( \ln(\frac{a}{b}) = \ln(a) - \ln(b) \)
• Power Rule: \( \ln(a^x) = x \cdot \ln(a) \)

In the exercise, the Power Rule was particularly crucial because it allowed us to convert the term \( \ln(e^{-1.4x}) \) into \( -1.4x \). This step is vital – if you don't apply the property correctly, you can't isolate the variable and solve the equation. After applying the property, all that’s left is purely algebraic manipulation to solve for \( x \). As seen in the exercise, this understanding transformed a seemingly complex exponential equation into a solvable algebraic equation.

Exponential Functions

An exponential function is a powerful mathematical concept where a constant base is raised to a variable exponent. In its general form, the function is represented as \( y = b^x \), where \( b \) is a positive number other than 1.

The function defined by \( f(x) = e^x \) is the most commonly used exponential function, due to its unique properties and natural occurrence in various scientific laws and real-world applications, including compound interest and population growth.

When we encounter an equation like \( 6e^{-1.4 x} = 21 \) from the original exercise, the challenge is to unravel the exponent, \( -1.4x \). A fundamental property of an exponential function is that the only way to 'remove' the base \( e \) from the expression is by applying a natural logarithm, because the two functions are inverse operations. This tug-of-war between exponential functions and logarithms is at the crux of solving such equations. Armed with these concepts and the rules of logarithms, we can skillfully find the solution to equations that model exponential growth or decay, just as we did in our original exercise.