Problem 24 - College Algebra

Question

Solve each exponential equation in Exercises \(1-26\) Express the solution set in terms of natural logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$e^{4 x}-3 e^{2 x}-18=0$$

Step-by-Step Solution

Answer

The solution for the exponential equation is \(x = \frac{ln(6)}{2}\) which is approximately x = 0.90 in decimal form.

Step 1: Re-expressing using a Substitution

Express the given equation \(e^{4 x}-3 e^{2 x}-18 = 0\) as a quadratic equation in 'y' where y = e^{2x}. This leads to \(y^2 - 3y - 18 = 0\).

Step 2: Solving the Quadratic Equation

Factor the quadratic equation: \(y^2 - 3y - 18 = 0\) can be factored into \((y - 6)(y + 3) = 0\). The solutions for y are then y = 6 and y = -3.

Step 3: Reverse Substitution

Substitute e^{2x} back for 'y'. This gives two equations: e^{2x} = 6 and e^{2x} = -3.

Step 4: Solving for 'x'

Solving \(e^{2x} = 6\), we get \(x = \frac{ln(6)}{2}\). The equation \(e^{2x} = -3\) has no real number solutions because the exponential function is always positive.

Step 5: Obtaining Decimal Approximation

Using a calculator, we find the decimal approximation of \(x = \frac{ln(6)}{2}\) to be approximately x = 0.90 correct to two decimal places.

Key Concepts

Natural Logarithms

When solving exponential equations, natural logarithms often play a pivotal role. Natural logarithms, denoted as 'ln', represent the inverse operation to raising the number 'e' to a power. In essence, while the exponential function with base 'e' takes a number and gives you its exponential growth, the natural logarithm function takes that growth and tells you the original number.

For instance, if we have the equation, \(e^x = b\), the natural logarithm allows us to solve for 'x' by applying \(ln\) to both sides, which yields \(x = ln(b)\). This process is crucial for understanding and finding solutions to exponential equations, as seen in the exercise above.

However, it's important to recognize that \(ln(x)\) is only defined for positive values of 'x', reflecting the fact that an exponential function with base 'e' will never produce a negative number or zero. This characteristic is what led to the conclusion in the textbook solution that \(e^{2x} = -3\) has no real solution, since the logarithm of a negative number is undefined in the realm of real numbers.

Quadratic Equations

Quadratic equations are a type of polynomial equation of the second degree, typically taking the form \(ax^2 + bx + c = 0\), where 'a', 'b', and 'c' are constants. To find solutions to quadratic equations, one can use various methods such as factoring, completing the square, or using the quadratic formula.

In the context of solving exponential equations, transforming the given exponential equation into a quadratic form by introducing a suitable substitution is a clever tactic. This is seen in the step-by-step solution from the exercise, where the substitution \(y = e^{2x}\) allows us to work with a more familiar quadratic form, \(y^2 - 3y - 18 = 0\).

Once in the quadratic form, factoring is a straightforward approach if the equation can be easily decomposed into two binomials, leading to clear solutions for the substituted variable. This demonstrated method can simplify the process of finding solutions to certain types of exponential equations, highlighting the interplay between different algebraic concepts.

Exponential Functions

Exponential functions, characterized by an equation of the form \(f(x) = a^x\), where 'a' is a positive constant, describe situations where a quantity grows or decays at a rate proportional to its current value.

Their unique property, the base 'a' raised to the power of 'x', allows exponential functions to describe a wide variety of phenomena in science, finance, and other fields. In our exercise, the base is the irrational number 'e', which is approximately 2.718. 'e' is a special base because it is the base rate of growth shared by all continually growing processes.

When you see an exponential equation like \(e^{4x}-3e^{2x}-18=0\), one of the strategies to solve it, as outlined in the provided solution, involves considering the expression \(e^{2x}\) as a common variable. This manipulation turns the complicated-looking exponential equation into a quadratic one, which is far simpler to solve with the methods you've already learned.

Understanding how exponential functions operate and interact with logarithms is essential in algebra, and practicing these problems helps build that foundational knowledge.