Problem 25 - Algebra and Trigonometry

Question

Find the solution of the exponential equation, rounded to four decimal places. \(\frac{50}{1+e^{-x}}=4\)

Step-by-Step Solution

Answer

The solution is approximately \(x = -2.4423\).

Step 1: Isolate the Exponential Term

First, we want to get the exponential function on one side by itself. Start by multiplying both sides of the equation by \(1+e^{-x}\) to eliminate the fraction: \[50 = 4(1 + e^{-x})\]Next, distribute the 4 on the right:\[50 = 4 + 4e^{-x}\]

Step 2: Isolate the Exponential

Subtract 4 from both sides to isolate the exponential term:\[46 = 4e^{-x}\]Then, divide both sides by 4:\[e^{-x} = rac{46}{4}\]Simplify:\[e^{-x} = 11.5\]

Step 3: Eliminate the Exponent

To solve for \(x\), take the natural logarithm (\(\ln\)) of both sides. Since the natural logarithm and the exponential function are inverses, applying \(\ln\) will eliminate the exponent:\[\ln(e^{-x}) = \ln(11.5)\]This simplifies to:\[-x = \ln(11.5)\]

Step 4: Solve for x

Since we have \(-x\) on one side, multiply both sides by \(-1\) to solve for \(x\):\[x = -\ln(11.5)\]Using a calculator for the logarithm, approximate \(\ln(11.5)\) and round the result to four decimal places:\[x \approx -2.4423\]

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is a fundamental mathematical function that serves as the inverse of the exponential function \(e^x\). It is particularly suited for solving exponential equations because it can undo the effect of the exponential function.
When dealing with equations that involve exponentials, applying the natural logarithm helps to isolate the variable of interest. In the context of the given problem, taking the natural logarithm on both sides of the equation enables us to eliminate the exponential component: \(\ln(e^{-x})\).
Here, the property \(\ln(e^x) = x\) is used, simplifying the process by allowing us to directly solve for \(x\) once the exponential term is isolated. It is crucial to understand that the base of the natural logarithm is \(e\), which is approximately 2.718, and this makes \(\ln\) especially useful in natural growth processes and calculus.

Solving Equations

Solving equations involves finding the value of the unknown variable that makes the equation true. For exponential equations, this typically requires several algebraic manipulations to isolate the variable of interest.
In this exercise, we start by eliminating any fractions to simplify the equation. The process involves multiplying through by the denominator of the fraction to clear it, a common strategy in algebra. Once fractions are removed from the equation, the next goal is to move all terms involving the variable to one side of the equation.
Finally, once the variable is isolated, solving may involve applying functions like the natural logarithm to further simplify the equation. Each step should be carefully calculated and verified, as minor arithmetic errors can lead to incorrect solutions.

Exponent Rules

Exponent rules are a set of guidelines for simplifying expressions involving powers. They are vital when working with exponential equations, especially when terms include exponentials with negative exponents or fractional bases.
Some crucial rules include:

• \(e^{a+b} = e^a \cdot e^b\)
• \(e^{-x} = \frac{1}{e^x}\)
• \((e^a)^b = e^{ab}\)

In the provided problem, knowing that \(e^{-x}\) can be written as \(\frac{1}{e^x}\) helps us understand how to isolate this term during the solution process.
When isolating exponents, combining like terms or factoring out common elements can simplify complex expressions, enabling easier application of further mathematical operations like logarithms.

Inverse Functions

Inverse functions, as their name implies, reverse the effect of an original function. This property makes them invaluable for solving equations. For example, the natural logarithm \(\ln\) is the inverse of the exponential function \(e^x\). This means that applying \(\ln\) to \(e^x\) returns the original input: \(x\).
This relationship is crucial when solving equations because it allows us to "undo" an exponential expression. In our problem, taking \(\ln\) on both sides of the equation \(e^{-x} = 11.5\) gives \(-x = \ln(11.5)\), effectively removing the exponential term and isolating \(-x\).
Understanding inverse functions enables us to approach equations strategically, using one function to cancel out the effect of another, bringing us closer to finding the solution.