Example: 3e^x = 12 - 5e^-x

Introduction

Key Concepts

Understanding Exponential Equations

Exponential equations involve variables in the exponent, making them distinct from linear or polynomial equations. The general form of an exponential equation is:

$$ a \cdot e^{kx} + b \cdot e^{-kx} = c $$

Solving these equations often requires techniques that can linearize or simplify the exponential terms. One effective method is the quadratic substitution, where a substitution transforms the equation into a quadratic form.

Quadratic Substitution Method

The quadratic substitution is a strategic approach to solving exponential equations by introducing a substitution that converts the equation into a quadratic form, which is easier to solve. Let's explore this method step-by-step using the example equation:

$$ 3e^{x} = 12 - 5e^{-x} $$

First, recognize that e^-x is the reciprocal of e^x. To simplify, let us set:

$$ u = e^{x} $$

Therefore, e^-x becomes:

$$ e^{-x} = \frac{1}{e^{x}} = \frac{1}{u} $$

Substituting these into the original equation:

$$ 3u = 12 - \frac{5}{u} $$

To eliminate the fraction, multiply both sides by u:

$$ 3u \cdot u = (12 - \frac{5}{u}) \cdot u \\ 3u^{2} = 12u - 5 $$

This simplifies to a standard quadratic equation:

$$ 3u^{2} - 12u + 5 = 0 $$

Solving the Quadratic Equation

With the quadratic equation established, we proceed to solve for u using the quadratic formula:

$$ u = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} $$

For our equation, a = 3, b = -12, and c = 5. Plugging these values into the formula:

$$ u = \frac{-(-12) \pm \sqrt{(-12)^{2} - 4 \cdot 3 \cdot 5}}{2 \cdot 3} \\ u = \frac{12 \pm \sqrt{144 - 60}}{6} \\ u = \frac{12 \pm \sqrt{84}}{6} \\ u = \frac{12 \pm 2\sqrt{21}}{6} \\ u = \frac{6 \pm \sqrt{21}}{3} $$

Therefore, the solutions for u are:

$$ u = 2 \pm \frac{\sqrt{21}}{3} $$

Reverting to the Original Variable

Recall that u = e^{x}. To find x, take the natural logarithm of both sides:

$$ x = \ln(u) $$

Substituting the values of u:

$$ x = \ln\left(2 + \frac{\sqrt{21}}{3}\right) \quad \text{and} \quad x = \ln\left(2 - \frac{\sqrt{21}}{3}\right) $$

However, since the logarithm of a negative number is undefined in the real number system, we must ensure that the arguments are positive:

• 2 + \frac{\sqrt{21}}{3} > 0
• 2 - \frac{\sqrt{21}}{3} > 0

Calculating the second expression:

$$ 2 - \frac{\sqrt{21}}{3} \approx 2 - 1.5275 = 0.4725 > 0 $$>

Thus, both solutions are valid, and we have:

$$ x = \ln\left(2 + \frac{\sqrt{21}}{3}\right) \quad \text{or} \quad x = \ln\left(2 - \frac{\sqrt{21}}{3}\right) $$

Verification of Solutions

It's crucial to verify the solutions by substituting them back into the original equation to ensure they satisfy the equation. This step eliminates extraneous solutions that may arise during the solving process.

Substituting x back into the original equation:

$$ 3e^{x} = 12 - 5e^{-x} $$>

For each value of x, calculate both sides to confirm equality.

Graphical Interpretation

Graphing both sides of the equation can provide a visual verification of the solutions. Plotting y = 3e^x and y = 12 - 5e^-x on the same graph allows us to identify the points of intersection, which correspond to the solutions of the equation.

The function y = 3e^x is an increasing exponential function, while y = 12 - 5e^-x is a decreasing function as e^-x decreases with increasing x. Their intersection points represent the values of x that satisfy the original equation.

Applications of Exponential Equations

Exponential equations are prevalent in various real-world scenarios, including population growth, radioactive decay, and financial calculations involving compound interest. Understanding how to solve such equations equips students with the tools to model and analyze dynamic systems effectively.

For instance, in population dynamics, the growth rate of a population can be modeled using exponential functions, making the ability to solve exponential equations essential for predicting future population sizes.

Advanced Concepts

Derivation of the Quadratic Substitution Technique

The quadratic substitution technique is not merely a heuristic but has a foundational basis in algebraic manipulation. To derive this method, consider an exponential equation where the variable appears both in the exponent and its reciprocal, as in:

$$ a \cdot e^{x} + b \cdot e^{-x} = c $$>

Letting u = e^{x} transforms the equation into:

$$ a \cdot u + \frac{b}{u} = c $$>

Multiplying both sides by u eliminates the denominator:

$$ a \cdot u^{2} + b = c \cdot u $$>

Rearranging terms results in a quadratic equation:

$$ a \cdot u^{2} - c \cdot u + b = 0 $$>

This demonstrates that exponential equations of this form can be reduced to quadratic equations, facilitating their solution using familiar algebraic methods.

Solving Higher-Degree Exponential Equations

While the quadratic substitution is effective for certain exponential equations, higher-degree or more complex equations may require additional techniques. For example:

$$ e^{2x} + e^{x} - 6 = 0 $$>

Here, setting u = e^{x} leads to:

$$ u^{2} + u - 6 = 0 $$>

Solving this quadratic equation yields:

$$ u = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2} \\ u = 2 \quad \text{or} \quad u = -3 $$>

Since u = e^{x} > 0, we discard u = -3 and solve for x:

$$ x = \ln(2) $$>

This example illustrates the versatility of the substitution method in handling various forms of exponential equations.

Complex Problem-Solving: Mixed Exponential and Polynomial Equations

Consider an equation that combines exponential and polynomial terms, such as:

$$ e^{x} + x^{2} = 5 $$>

Solving such equations analytically can be challenging due to their transcendental nature. In many cases, solutions may require numerical methods or approximations.

One approach is to graph both sides of the equation:

• y = e^x + x²
• y = 5

The points of intersection indicate the solutions. Alternatively, iterative methods like the Newton-Raphson technique can be employed to approximate the roots.

Interdisciplinary Connections: Exponential Equations in Physics

Exponential equations are integral to various branches of physics. For instance, the decay of radioactive materials is modeled by exponential functions. The number of undecayed nuclei at time t is given by:

$$ N(t) = N_{0} \cdot e^{-\lambda t} $$>

Here, N₀ is the initial quantity, and λ is the decay constant. Solving equations involving radioactive decay often involves exponential equations similar to the example discussed.

Understanding how to manipulate and solve these equations is crucial for students pursuing more advanced studies in physics and engineering disciplines.

Applications in Finance: Compound Interest Calculations

Exponential equations are also foundational in financial mathematics, particularly in calculating compound interest. The future value A of an investment compounded continuously is given by:

$$ A = P \cdot e^{rt} $$>

Where:

• P = Principal amount
• r = Annual interest rate
• t = Time in years

Students can apply quadratic substitution techniques when dealing with equations that arise from equating future and present values or solving for time when both growth and decay factors are involved.

Exploring Lambert W Function

For more advanced studies, the Lambert W function provides a solution to equations of the form:

$$ x \cdot e^{x} = k $$>

This function is particularly useful for solving equations where the variable appears both in the base and the exponent. While the Lambert W function is beyond the scope of the Cambridge IGCSE curriculum, it represents an extension of exponential equation solving techniques in higher-level mathematics.

Comparison Table

Aspect	Quadratic Substitution	Lambert W Function
Applicability	Effective for equations reducible to quadratic form via substitution.	Used for equations where variables are both in base and exponent.
Complexity	Accessible at the Cambridge IGCSE level.	Advanced mathematical concept beyond IGCSE.
Solution Method	Algebraic manipulation and quadratic formula.	Special function application requiring higher mathematics.
Real-World Applications	Used in solving engineering and physics problems requiring exponential solutions.	Applicable in specialized fields like computational biology and quantum physics.
Ease of Use	Generally straightforward with proper substitution.	Requires understanding of advanced mathematical functions.

Summary and Key Takeaways