Problem 60 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the exponential equation algebraically. Round your result to three decimal places. Use a graphing utility to verify your answer. $$100 e^{0.005 x}=125,000$$

Step-by-Step Solution

Answer

To get the final answer, it is necessary to substitute the values in the above equation and calculate x. Also, the answer is validated by substitifying x-value in the original equation or using the graphing utility.

Step 1: Express the equation in terms of exponent

The given equation is \(100 e^{0.005x} = 125,000\). First, let's simplify it by dividing both sides by 100. We get, \(e^{0.005x} = 1250\)

Step 2: Apply natural logarithm

Apply the natural logarithm (ln) on both sides of the equation to remove the exponential part. This yields \(\ln(e^{0.005x}) = \ln(1250)\). According to the logarithm property, we can bring the exponent out front, which gives \(0.005x*ln(e) = \ln(1250)\). Since ln(e) equals 1, we are left with \(0.005x = ln(1250)\)

Step 3: Solve for x

To isolate x, divide both sides of the equation by 0.005. This gives \(x = \frac{ln(1250)}{0.005}\)

Step 4: Calculate the value of x

Substitute the value of ln(1250) and divide by 0.005 to get the value of x. Keep the answer to three decimal places.

Step 5: Verify result using graphing utility

To validate the solution, substitute the value of x in the original equation, and check if both sides are equal. Also, one can plot the equation \(100 \(e^{0.005x}\) and \(125,000\) on a graphing utility and see if they intersect at the calculated x-value. A congruent result validates the answer.

Key Concepts

Understanding the Natural Logarithm

The natural logarithm, often written as \( \ln \), is a mathematical function that plays a crucial role in solving exponential equations. It is the logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.718. The natural logarithm is particularly useful because it helps us "undo" exponentials when solving equations. For example, if you have an equation of the form \( e^x = a \), taking the natural logarithm of both sides results in \( x = \ln(a) \). This is because \( \ln(e^x) = x \), due to the property \( \ln(e) = 1 \).
In the context of the problem, applying the natural logarithm allows us to simplify \( e^{0.005x} \) by transforming it into a multiplication, thus making it easier to solve for \( x \). Remember, the basic property \( \ln(e^a) = a \ln(e) \) simplifies to \( a \), since \( \ln(e) = 1 \).
Overall, understanding the natural logarithm and its properties provides a powerful tool for solving equations involving exponents.

Using a Graphing Utility for Verification

A graphing utility is an electronic tool or software that helps you visualize mathematical equations and functions by graphing them. It is particularly valuable in verifying solutions to algebraic equations. Once you find a solution algebraically, you can use a graphing utility to double-check your answer visually.

• Plot the left-hand side of the equation, \( 100 e^{0.005x} \), and the right-hand side, \( 125,000 \).
• The point where these two graphs intersect on the \( x \)-axis confirms the value of \( x \) obtained previously.

This method provides a second layer of verification to ensure that the calculated solution is indeed correct. Moreover, visualizing the equations can provide deeper insight into the behavior of the functions involved. If the intersection aligns with your computed result, it serves as strong evidence that your algebraic work is accurate.

Equation Solving with Exponential Functions

Solving exponential equations involves finding the value of the variable that makes the equation true. Exponential equations commonly include an expression with a variable in the exponent, such as \( e^{0.005x} = 1250 \). Solving these requires a clear understanding of logarithmic functions, specifically the natural logarithm.

• The first step is often to isolate the exponential expression. In the example, this was done by dividing both sides by 100, yielding \( e^{0.005x} = 1250 \).
• Next, the natural logarithm is applied to both sides to remove the exponent. This uses the property \( \ln(e^a) = a \ln(e) \), simplifying further to \( a \) as \( \ln(e) = 1 \).
• After applying \( \ln \), you simplify and solve the resulting equation for the variable. In this case, it leads to \( 0.005x = \ln(1250) \), and solving for \( x \) gives us \( x = \frac{\ln(1250)}{0.005} \).

It's essential to round the final solution as per the problem's requirements, ensuring precision and correctness in practical applications. Mastery of these steps can greatly enhance your ability to tackle various types of exponential equations.