Problem 32 - Contemporary Precalculus

Question

Solve the equation for \(x\). $$\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}=t$$

Step-by-Step Solution

Answer

Question: Solve the equation \(e^x - e^{-x} = t(e^x + e^{-x})\) for x. Answer: To solve the given equation, we first combine the exponential terms and expand the expression to get \(e^{2x} - e^{-2x} = t(e^x + e^{-x})\). Then, we introduce a substitution, \(u = e^x\), and rewrite the equation as \(u^4 - tu^3 - tu + 1 = 0\). Solve the quartic equation for u and then find x using the initial substitution: \(x = \ln(u)\).

Step 1: Combine the exponential terms on both sides

First, we will eliminate the fractions by combining the exponential terms on both sides. To do this, multiply both sides of the equation by the denominator of the left side, which is \((e^x + e^{-x})\). This will give: $$ (e^x - e^{-x})(e^x + e^{-x}) = t (e^x + e^{-x}) $$

Step 2: Expand the expression

Now expand the left side using the distributive property or FOIL technique: $$ e^{2x} + e^{x}e^{-x} - e^{-x}e^x - e^{-2x} = t(e^x + e^{-x}) $$ Since \(e^a e^b = e^{a+b}\) and \(e^{a}e^{-a}=1\) for any a and b, the left side simplifies to: $$ e^{2x} + 1 - 1 - e^{-2x} = t(e^x + e^{-x}) $$ This simplifies further to: $$ e^{2x} - e^{-2x} = t(e^x + e^{-x}) $$

Step 3: Introduce substitution

To simplify our equation further, let's introduce a substitution. Let \(u = e^x\). Then, the reciprocal of \(u\) will be equal to \(e^{-x}\): $$ e^{-x} = \frac{1}{u} $$ Substitute these expressions for \(e^x\) and \(e^{-x}\) into our equation: $$ u^2 - \frac{1}{u^2} = t(u + \frac{1}{u}) $$

Step 4: Solve for u

Let's solve the equation for u. To eliminate the fractions, multiply both sides of the equation by \(u^2\): $$ u^4 - 1 = t(u^3 + u) $$ Now it is a quartic equation. To solve it, we could either use a quartic solver or find a way to transform it into a quadratic equation.

Step 5: Equate the equations to zero

Rearrange the equation such that all terms are on the left-hand side and the right-hand side is equal to zero: $$ u^4 - tu^3 - tu + 1 = 0 $$

Step 6: Solve for x

Since we have a quartic equation in u, we'll either need to use a quartic solver to find the roots or factor the equation into simpler expressions. Once we have the value(s) of u, we can find x using the initial substitution: $$ x = \ln(u) $$ If there is more than one u value, we get more than one x value. Use the natural logarithm to find each x value. This will give the final solution(s) for x in the given equation.

Key Concepts

Understanding Exponential Functions

Exponential functions are mathematical expressions that involve a constant base raised to a variable exponent. In these functions, the rate of change increases or decreases across the function's domain. In the given problem, the exponential functions are denoted as e^x and e^{-x}, where e is the base of the natural logarithm, approximately equal to 2.71828.

These functions are crucial in various fields, from population growth in biology to compound interest in finance. One distinctive feature of exponential functions is their inverses - logarithms, particularly the natural logarithm, which allows for the manipulation and solving of equations involving exponentials, as we will see in a later step in our problem-solving process. The exponential functions exhibit continuous growth or decay and never hit zero, which makes them pivotal in understanding changes over time.

Quartic Equations: A Deeper Look

Quartic equations are polynomial equations of the fourth degree. They have the general form ax^4 + bx^3 + cx^2 + dx + e = 0, where a, b, c, d, and e are constants, and a ≠ 0. Solving quartic equations can be quite complex and, unlike quadratic equations, may involve intricate formulas or require numerical methods for their solutions.

In our exercise, a quartic equation is obtained as part of the transformation process involving exponential functions. Through a clever substitution, we convert an unwieldy exponential equation into a more manageable quartic equation. This is a crucial step, as it allows us to employ various strategies for solving quartics, such as factoring, using synthetic division or applying the quartic formula, to ultimately find all possible values for the variable of interest.

The Role of Natural Logarithm in Solving Equations

The natural logarithm, denoted as ln, is the logarithm to the base e, where e is the mathematical constant approximately equal to 2.71828. It is the inverse operation to exponentiation, just as subtraction is the inverse of addition. In the context of solving equations, the natural logarithm is particularly useful for finding the original exponent in expressions of the form e^x.

Once we have isolated e^x on one side, taking the natural logarithm of both sides of the equation allows us to 'bring down' the exponent, turning a seemingly complicated problem into a simpler one. In the given problem, after we've determined the values of u from the quartic equation, we revert the substitution process by taking the natural logarithm of u to find the corresponding values of x. This reverses the initial operation of raising e to the power of x, providing us with the solution to the original problem.