Problem 45 - Algebra and Trigonometry

Question

Solve each exponential equation . Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$ e^{4 x}+5 e^{2 x}-24=0 $$

Step-by-Step Solution

Answer

The solution in terms of natural logarithms is \( x = \frac{1}{2} ln(3) \) and the decimal approximation is \( x = 0.55 \).

Step 1: Identify the Form of the Equation

Look at the given equation \( e^{4x} + 5e^{2x} - 24 = 0 \) and notice that it resembles a second-degree polynomial. We can manipulate it into a quadratic equation. But first, let \( y = e^{2x} \) then the given equation becomes \( y^2 + 5y - 24 = 0 \).

Step 2: Solve The Quadratic Equation

Next, solve the quadratic equation using the quadratic formula \( y = \frac{-b ± \sqrt{b^2 - 4ac}}{2a} \), where a = 1, b = 5, c = -24. This results in \( y = \frac{-5 ± \sqrt{(5)^2 - 4*1*(-24)}}{2*1} \). Simplifying, it will give \( y = \frac{-5 ± \sqrt{121}}{2} \), which further simplifies to \( y = 3, y = -8 \).

Step 3: Convert Back to Original Variable

Next, convert back to the original variable \( x \) by substituting \( y = e^{2x} \) into the solutions obtained. This gives \( e^{2x} = 3 \) and \( e^{2x} = -8 \). Note that \( e^{2x} \) is always greater than 0 for any real number \( x \), so we ignore \( e^{2x} = -8 \). Solving \( e^{2x} = 3 \) using natural logarithms gives \( x = \frac{1}{2} ln(3) \).

Step 4: Find the Decimal Approximation

Finally, use a calculator to find the decimal approximation for \( x = \frac{1}{2} ln(3) \). The solution rounded to two decimal places gives \( x = 0.55 \).

Key Concepts

Natural Logarithms

Natural logarithms are a fundamental concept in mathematics used to simplify the process of solving exponential equations. When working with equations that involve the natural exponential function, denoted as \( e \), natural logarithms can help us find solutions in terms of a more manageable form.

The natural logarithm, written as \( \ln(x) \), is the inverse function of the exponential function \( e^x \). This property allows us to convert an exponential equation into a linear one, which is much easier to solve. For example, in the problem \( e^{2x} = 3 \), we can apply the natural logarithm to both sides to yield:\[ 2x = \ln(3) \] Dividing both sides by 2 gives us \[ x = \frac{1}{2} \ln(3) \].

• Natural logarithms help in "un-doing" the exponent, making the equation easier to tackle.
• This conversion is crucial when dealing with equations that don't have straightforward numerical solutions.

Understanding how to work with natural logarithms is vital for students tackling exponential equations.

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree, usually in the form \( ax^2 + bx + c = 0 \). These equations are essential in various fields of study due to the nature of their solutions.

In our original exercise, we turned the exponential equation \( e^{4x} + 5e^{2x} - 24 = 0 \) into a quadratic form by letting \( y = e^{2x} \), transforming it into \( y^2 + 5y - 24 = 0 \).

To solve this quadratic equation, we used the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 5 \), and \( c = -24 \). This provides the solutions:

• \( y = 3 \)
• \( y = -8 \)

The negative solution \( y = -8 \) is disregarded since it does not suit \( e^{2x} \), where the output must be positive for real numbers \( x \). Thus, the valid solution arises from \( y = 3 \), simplifying how we solve exponential-like equations using quadratic methods.

Decimal Approximation

Decimal approximation is an essential step when the exact value yields mathematically complex or irrational numbers. With exponential equations, after finding a theoretical solution in terms of logarithms, it's common to convert these results to decimals for clarity or practical use.

After determining the solution \( x = \frac{1}{2} \ln(3) \), a calculator can evaluate \( \ln(3) \), and compute \( x \).
Using a calculator, we approximate \( \ln(3) \approx 1.09861 \). Thus, computing:\[ x = \frac{1}{2} \times 1.09861 \approx 0.5493 \] Rounded to two decimal places, we get \( x = 0.55 \).

• Decimal approximation provides a user-friendly number that is easy to interpret.
• It is essential in applications where precise numerical outputs are necessary, like engineering or physics.

In summary, being adept at converting calculated logarithmic values into decimals ensures practical application and deeper understanding.