Problem 24 - Algebra and Trigonometry

Question

Find the solution of the exponential equation, correct to four decimal places. $$ \frac{10}{1+e^{-x}}=2 $$

Step-by-Step Solution

Answer

The solution is approximately \(x \approx -1.3863\).

Step 1: Multiply Both Sides by the Denominator

We start by removing the fraction. Multiply both sides of the equation by \(1 + e^{-x}\):\[10 = 2(1 + e^{-x})\]

Step 2: Expand and Simplify the Equation

Distribute the 2 on the right-hand side:\[10 = 2 + 2e^{-x}\]Subtract 2 from both sides to isolate the term with the exponential:\[8 = 2e^{-x}\]

Step 3: Solve for the Exponential Term

Divide both sides by 2 to solve for \(e^{-x}\):\[e^{-x} = 4\]

Step 4: Apply the Natural Logarithm

Take the natural logarithm on both sides to solve for \(-x\):\[-x = \ \ln 4\]

Step 5: Solve for x

Multiply both sides by -1 to solve for \(x\):\[x = -\ln 4\]Calculate this value using a calculator to get:\[x \approx -1.3863\]

Key Concepts

Natural Logarithm

The natural logarithm is an essential mathematical concept that helps us solve exponential equations. It is denoted as "ln" and is the inverse operation of the exponential function with the base of Euler's number, denoted as "e", which is approximately 2.71828.
When you take the natural logarithm of a number, you are essentially asking, "To what power must e be raised to yield this number?"
In the original equation, after isolating the term with the exponential expression, we had the equation \(e^{-x} = 4\). To "undo" the exponential function in this scenario, we took the natural logarithm of both sides to find the value of \(-x\). This step is based on the property that \( ext{if} \, e^y = z, \, ext{then} \, y = ext{ln}(z) \).
The natural logarithm simplifies the process of clearing the exponential, allowing us to transform this nonlinear equation into a more manageable algebraic form.

Exponential Functions

Exponential functions are identified by their constant base, typically "e" for natural exponential functions, raised to a variable power. In mathematical terms, they look like \( f(x) = e^x \). These functions are known for their unique properties, such as continual growth or decay, depending on whether the exponent is positive or negative.
In our exercise, we dealt with an exponential term \(e^{-x}\). The negative exponent denotes a decay function, implying that the function's value decreases as the variable \(x\) increases.
Understanding how to manipulate exponential functions is crucial since they frequently appear in equations modeling real-world phenomena like population growth, radioactive decay, and compound interest. Importantly, they're often involved in exponential equations where finding the unknown variable requires algebraically manipulating and inverting the exponential using logarithms.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to simplify or solve them. This skill is vital in solving exponential equations since it allows for the isolation of terms for calculus and other advancements in solving.
Let's look at how we used this in our example problem:

• Initially, we had \( \frac{10}{1+e^{-x}}=2 \). To eliminate the fraction, we multiplied through by \(1 + e^{-x}\).
• The equation was expanded to \(10 = 2 + 2e^{-x}\), after which we subtracted 2 from both sides to begin isolating the exponential term.
• To solve for \(e^{-x}\), we divided the entire equation by 2. This isolation was crucial in preparing to take the logarithm.
• Finally, negating the equation after applying the natural logarithm helped us solve for \(x\).

Algebraic manipulation simplifies complex equations and reveals solutions that aren't immediately visible, playing a fundamental role in mathematical problem-solving.