Problem 73 - College Algebra

Question

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth. $$ 7 e^{3 x-5}+7.9=47 $$

Step-by-Step Solution

Answer

The solution is approximately \(x = 2.2396\).

Step 1: Isolate the Exponential Term

Start by isolating the exponential term on one side of the equation. You have:\[7 e^{3x-5} + 7.9 = 47\]Subtract 7.9 from both sides of the equation to get:\[7 e^{3x-5} = 39.1\]

Step 2: Solve for the Exponential Term

Now, divide both sides of the equation by 7 to isolate the exponential expression:\[e^{3x-5} = \frac{39.1}{7}\]Simplifying the right-hand side gives:\[e^{3x-5} = 5.5857\]

Step 3: Take the Natural Logarithm of Both Sides

To solve for \(x\), apply the natural logarithm to both sides of the equation:\[\ln(e^{3x-5}) = \ln(5.5857)\]Using the property of logarithms, \(\ln(e^y) = y\), this simplifies to:\[3x - 5 = \ln(5.5857)\]

Step 4: Solve for x

First, calculate \(\ln(5.5857)\) using a calculator. This gives you approximately 1.7187.Now, solve for \(x\) by adding 5 to both sides and then dividing by 3:\[3x = 1.7187 + 5 \x = \frac{6.7187}{3}\]This gives:\[x \approx 2.2396\]

Step 5: Verify the Solution

Substitute \(x = 2.2396\) back into the original equation to ensure correctness:Calculate \(e^{3(2.2396)-5}\) and multiply it by 7, then add 7.9. You should get approximately 47, confirming the solution's accuracy.

Key Concepts

Understanding Natural Logarithm

When solving exponential equations, the natural logarithm often comes into play. Let's explore what it is. A natural logarithm, represented as \( \ln \), is a special type of logarithm that uses the base \( e \), Euler's number, approximately 2.718. Here's a notable property:

• \( \ln(e^x) = x \)

This means the natural logarithm is the inverse of the exponential function with base \( e \).
In exponential equations, this property allows us to "undo" the exponential, simplifying complex expressions.When applied to each side of a balance, the equation remains unchanged, but often easier to handle. In our original problem, after isolating the exponential part, we take the natural logarithm. This step transforms \( e^{3x-5} \) into a simple linear expression, facilitating the next phase of solving for \( x \).

Isolation of Variables

Isolating variables is a crucial step in solving equations, including exponential ones. It involves making sure the variable you're solving for is by itself.
This concept simplifies equations and helps you focus on what matters.Here's how to do it in a structured way:

• First, move all terms not involving the variable to the other side of the equation. For our exercise, this meant subtracting 7.9 from both sides.
• Next, isolate the term involving the variable by dividing by any coefficients. In our example, this involved dividing both sides by 7 to manage the exponential term \( e^{3x-5} \).

Through these steps, the variable \( x \) becomes the main focus, making it easier to apply further operations, like logarithms. In this manner, isolating simplifies complex problems, forming the groundwork for finding the solution.

Mastering Rounding Calculations

Rounding is an essential skill in mathematics, especially when solutions demand precision. This exercise specifically instructs us to round to the nearest ten-thousandth.Here's a quick guide:

• Identify the number up to four decimal places. For our calculation, 6.7187 is already at this precision level.
• Observe the number in the fifth decimal place. If it is 5 or higher, round up the fourth decimal place by one. If less than 5, keep the fourth place as is.

In our exercise, the rounded value led to the solution \( x \approx 2.2396 \), ensuring a balance between accuracy and simplicity.
Being mindful of precision ensures consistency, especially in detailed calculations like these, helping to verify correctness in the final answers.

Verification of Solution

Verification is key to ensuring that a solution is correct. After finding a solution, checking it through substitution back into the original equation confirms its validity. Verification involves a straightforward approach:

• Insert the solved \( x \) value back into the original variables to re-calculate the outcome.
• Ensure that both sides of the equation are approximately equal after the substitution.

In our example, replacing \( x = 2.2396 \) back into the exponential expression and performing the operations: calculate \( e^{3(2.2396)-5} \), multiply by 7, and add 7.9. The result should closely approach 47.This verification step not only provides confidence in the numerical answer but also reassures understanding of the method and its accuracy. Having this safeguard ensures that errors are caught and corrected before finalizing your work.