Problem 32 - Precalculus with Limits

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places. \(-14+3 e^{x}=11\)

Step-by-Step Solution

Answer

The value of \(x\) approximated to three decimal places is 2.121.

Step 1: Isolate the exponential term

The first step is to isolate the exponential term, \(e^{x}\), on one side of the equation. This is done by adding 14 to both sides of the equation. This results to \n\n\(3 e^{x}= 11+14\)\n\nwhich simplifies to\n\n\(3 e^{x} = 25\).

Step 2: Solve the exponential equation

Next, divide both sides of the equation by 3 to solve for \(e^{x}\).\n\nThis results to \n\n\(e^{x} =25/3=8.33\).

Step 3: Apply natural logarithm

In this equation, \(x\) is in the form of an exponent. To solve for \(x\), use the natural logarithm on both sides of the equation. The natural logarithm of \(e^{x}\) is \(x\), simplifying the equation. Therefore, \n\n\(\ln(e^{x})=\ln(8.33)\) \n\nsimplifies to \n\n\(x=\ln(8.33)\).

Step 4: Calculate the result

Finally, compute the value of \(x\) by evaluating the natural logarithm of 8.33, approximating the result to three decimal places. This gives \n\n\(x = 2.121\).

Key Concepts

Solving Exponential Equations Algebraically

When encountering an exponential equation, the goal is to determine the value of the variable in the exponent. Exponential equations are prevalent in scenarios modeling growth or decay, such as population studies or radioactive decay. The key to solving an exponential equation algebraically is to manipulate the equation into a form where the exponent can be isolated and then solved.

This process might involve basic algebraic operations, such as addition, subtraction, multiplication, or division. The objective is to get the term containing the exponent, typically in the form of \(e^{variable}\), on one side of the equation by itself. Once this term is isolated, further steps can be used to solve for the variable.

Isolate Exponential Term

The crucial step in dealing with an exponential equation is to isolate the exponential term. This means that we want that term, which contains the exponent, by itself on one side of the equation. For example, in the equation \( -14+3e^{x}=11 \), our goal is to isolate \( e^{x} \).

This involves performing operations that, in effect, undo the arithmetic around the exponential term. In the given example, we would add 14 to both sides to eliminate the -14 on the side with the exponential. This modification provides us a simplified form: \(3e^{x}=25\). Then, we divide both sides by 3 to get \(e^{x} = \frac{25}{3}\), successfully isolating \(e^{x}\). With the exponential term isolated, we are one step closer to finding the value of \(x\).

Natural Logarithm Application

Once we have isolated the exponential term, we employ the natural logarithm (denoted as \(ln\)) to solve for the variable in the exponent. The natural logarithm is the inverse operation to exponentiation with the base \(e\), Euler's number. It allows us to 'bring down' the exponent to a position where we can tactically isolate and solve for it algebraically.

For instance, applying the natural logarithm to both sides of the equation \(e^{x} = \frac{25}{3}\) gives us \(\ln(e^{x}) = \ln\left(\frac{25}{3}\right)\). Since the properties of logarithms tell us that \(\ln(e^{x}) = x\), we can simplify the equation to \(x = \ln\left(\frac{25}{3}\right)\). By applying the natural logarithm, we've effectively rendered the equation solvable for \(x\).

Approximate Results

In many real-world cases, the exact value of \(x\) when solving an exponential equation may be an irrational number that cannot be expressed precisely with a finite number of digits. Therefore, we often use a calculator to find an approximate value to a certain level of precision, typically denoted by the number of decimal places.

For the equation we solved using the natural logarithm, we find \(x = \ln\left(\frac{25}{3}\right)\). Calculators can estimate this to any number of decimal places. For practicality, especially in homework or exam settings, a three-decimal-place approximation is commonly used, which in this case results in \(x\approx 2.121\). While approximate values are not exact, they are very close and typically sufficiently accurate for most practical applications, enabling us to interpret the results within a meaningful context.