Problem 31 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 25-32: Solve the equation without using a calculator. $$ e^{2 x}+2 e^{x}-15=0 $$

Step-by-Step Solution

Answer

The solution is \(x = \ln(3)\).

Step 1: Identify the Equation Form

The equation given is \(e^{2x} + 2e^x - 15 = 0\). Notice that it is a quadratic equation in terms of \(e^x\). We can use substitution to simplify it.

Step 2: Substitute to Simplify the Equation

Let \(u = e^x\). Then \(e^{2x} = (e^x)^2 = u^2\). Substitute these into the equation to get \(u^2 + 2u - 15 = 0\).

Step 3: Factor the Quadratic Equation

Now, we need to factor the equation \(u^2 + 2u - 15 = 0\). This can be factored into \((u + 5)(u - 3) = 0\).

Step 4: Solve for 'u'

Set each factor equal to zero: \(u + 5 = 0\) gives \(u = -5\) and \(u - 3 = 0\) gives \(u = 3\). Since \(u = e^x\), discard \(u = -5\) because \(e^x\) cannot be negative.

Step 5: Solve for 'x'

With \(u = 3\), we have \(e^x = 3\). Solving for \(x\), we take the natural logarithm on both sides to get \(x = \ln(3)\).

Key Concepts

Exponential Functions

Exponential functions are a type of mathematical function where the variable appears as an exponent. They are commonly represented in the form \(f(x) = a^x\), where \(a\) is a constant called the base. In our exercise, we deal with the exponential base \(e\), the natural number approximately equal to 2.718. This specific function is \(e^{2x} + 2e^x - 15 = 0\). Here, the exponential functions are used to model growth processes such as population growth, compound interest, or radioactive decay.

• An essential feature of exponential functions is their rapid growth or decay, depending on the base being greater or less than 1.
• They appear frequently in real-life applications and mathematical models.
• The natural exponential function is unique due to its base, \(e\), relating closely to calculus.

Understanding exponential functions helps in solving various real-world problems and mathematical equations like the one outlined above.

Factoring Polynomials

Factoring polynomials involves breaking down a complex polynomial into simpler terms, or factors, that when multiplied together give back the original polynomial. The ability to factor polynomials is crucial in simplifying mathematical equations, especially quadratic equations.

• In our exercise, we convert \(e^{2x} + 2e^x - 15 = 0\) into a simpler quadratic form \(u^2 + 2u - 15 = 0\) by substitution.
• This transformed equation, \(u^2 + 2u - 15 = 0\), is factored into \((u + 5)(u - 3) = 0\).
• Factoring can help identify the roots of polynomial equations, making them easier to solve.

This transformation simplifies solving the equation by breaking it into manageable pieces, leading us to the solutions for \(u\).

Solving Equations

Solving equations is the process of finding the values of variables that satisfy a given mathematical equation. This concept is central in algebra and involves various techniques depending on the type of equation.

• The equation-solving process often begins with simplifying or rearranging the terms.
• In this exercise, we initially use substitution to handle the equation as a quadratic, making it easier to manage.
• Once factored, the equation \((u + 5)(u - 3) = 0\) is solved by equating each factor to zero, providing possible values for \(u\).

For \(u = e^x\), only the positive solution \(u = 3\) is valid since the exponential function \(e^x\) is never negative. This demonstrates our step-by-step approach towards finding the solution for \(x\).

Logarithms

Logarithms are the inverse operations of exponentiation, helping in solving equations where the variable is an exponent. Understanding logarithms is key to solving exponential equations.

• The natural logarithm, denoted \(\ln\), uses the base \(e\) and is crucial in dealing with natural exponential functions.
• To solve for \(x\) in \(e^x = 3\), we use the natural logarithm to rewrite it as \(x = \ln(3)\). This step leverages the property of logarithms that allows us to "bring down" the exponent as a multiplier.
• Logarithmic properties facilitate the transition from exponential form to a calculable linear form.

Applying logarithms in this context aids us in isolating the variable, leading directly to the solution \(x = \ln(3)\). Understanding and using logarithms is essential in solving and interpreting equations involving exponential components.