Problem 23 - 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. $$e^{4 x}+5 e^{2 x}-24=0$$

Step-by-Step Solution

Answer

The solution for the given equation is \(x \approx 0.55\).

Step 1: Rewrite the equation

To solve this equation, it would be easier to handle if it could be rewritten in a format that can be factored. Let's set \(y = e^{2x}\). This results to a new equation: \(y^2 + 5y - 24 = 0\).

Step 2: Factorise the equation

Next, factor the quadratic equation. It should be divisible into two binomial expressions. The factored form of the equation would be: \((y - 3)(y + 8) = 0\).

Step 3: Solve for 'y'

Set each factor equal to zero and solve for 'y'. This leads to two solutions for 'y': \(y = 3\) and \(y = -8\).

Step 4: Substitute 'y' with 'e^{2x}'

Replace 'y' with 'e^{2x}' in each solution. So the solutions become: \(e^{2x} = 3\) and \(e^{2x} = -8\). However, since e to any power is never negative, \(e^{2x} = -8\) has no solutions.

Step 5: Solve for 'x'

Now, solve \(e^{2x} = 3\) for 'x', you can take natural logarithm on both sides to get: \(2x = \ln(3)\). Therefore, \(x = 0.5\ln(3)\).

Step 6: Simplify the solution

Simplify the solution further by plugging \(\ln(3)\) into a calculator and rounding to two decimal places: \(x \approx 0.55\).

Key Concepts

Natural Logarithms

Natural logarithms are useful when dealing with exponential equations. They provide a way to express the exponent in terms of a base related to the number e, approximately equal to 2.71828. The natural logarithm of a number is denoted by \(\ln\). For example, solving an equation like \(e^{2x} = 3\) requires us to find the value of \(x\) by using natural logs.

• Taking the natural logarithm of both sides of the equation, we have \(\ln(e^{2x}) = \ln(3)\).
• This simplifies to \(2x = \ln(3)\) because \(\ln(e^y) = y\) for any \(y\).
• Thus, \(x = \frac{\ln(3)}{2}\). This provides an exact form involving natural log, which is crucial for deriving the solution in terms of known mathematical constants.

It is important to understand that natural logs translate exponential growth into linear growth, simplifying many problems in calculus and algebra.

Factoring Quadratic Equations

Factoring is a method used to simplify complex quadratic equations into more manageable linear equations. For instance, from our problem, we converted the exponential into a quadratic form: \(y^2 + 5y - 24 = 0\), by substituting \(y = e^{2x}\).

• To factor the equation, we look for two numbers whose product is \(-24\) and whose sum is \(5\).
• These numbers are \(8\) and \(-3\).
• So, we factor the equation as \((y - 3)(y + 8) = 0\).

Once factored, each part is set to zero, giving potential solutions for \(y\). Factoring decodes the quadratic’s hidden roots, revealing solutions that can be directly worked into other formats like linear equations.

Solution Sets

A solution set is a collection of all possible solutions that satisfy a given equation. When working with equations, particularly those transforming variables, it's essential to identify valid solutions.

• After factoring \((y - 3)(y + 8) = 0\), solve for \(y\), giving two potential solutions: \(y = 3\) and \(y = -8\).
• Replace each \(y\) with \(e^{2x}\) leading to \(e^{2x} = 3\) and \(e^{2x} = -8\).
• Since \(e^{2x}\) cannot be negative, discard \(e^{2x} = -8\). Thus, the valid solution set is \(\{ e^{2x} = 3 \}\).

Understanding solution sets allows you to filter out invalid solutions, ensuring only feasible answers are given in a mathematical context.

Decimal Approximations

Decimal approximations are used when a solution requires a more practical, numeric form rather than a symbolic one. For our exercise, we simplify \(x = \frac{\ln(3)}{2}\) by calculating its decimal equivalent.

• Use a calculator to find \(\ln(3)\) which approximates to around 1.0986.
• Next, divide by 2 to solve for \(x\), giving \(x \approx 0.5493\).
• Round the result to two decimal places as instructed: \(x \approx 0.55\).

Decimal approximations provide usability in real-world problems, allowing for a better grasp and applicability of the solutions in practical scenarios, while still acknowledging that the exact value is retained in its symbolic form.