Problem 17 - Beginning and Intermediate Algebra

Question

Solve the exponential equation algebraically. Then check using a graphing calculator. $$e^{t}=1000$$

Step-by-Step Solution

Answer

To solve the exponential equation \(e^{t} = 1000\) algebraically, apply the natural logarithm to both sides: \(\ln(e^{t}) = \ln(1000)\). Using logarithm properties, simplify the equation to \(t = \ln(1000)\), and evaluate the natural logarithm to find that \(t \approx 6.908\). Verify the solution using a graphing calculator by graphing both functions, \(y=e^{t}\) and \(y=1000\), and checking for their intersection point, which occurs near \(t=6.908\).

Step 1: (Step 1: Isolate the exponential term)

(We have to isolate the exponential term e^t on one side of the equation. In this case, it is already isolated, so let's move on to the next step.)

Step 2: (Step 2: Apply the natural logarithm)

(To solve for t, use the natural logarithm that is denoted as ln() which is the inverse function of e^(), to both sides of the equation.) \[ \ln(e^{t}) = \ln(1000) \]

Step 3: (Step 3: Use logarithm properties)

(Using the logarithm property that says \(\ln(a^{b}) = b\ln(a)\) and knowing that ln(e)=1, we simplify the equation.) \[ t \cdot \ln(e) = \ln(1000) \] \[ t = \ln(1000) \]

Step 4: (Step 4: Evaluate the natural logarithm)

(Let's calculate the value of t by evaluating the natural logarithm of 1000.) \[ t \approx 6.908 \]

Step 5: (Step 5: Verify the solution using a graphing calculator)

(To check the solution, graph y=e^t and y=1000 simultaneously on a graphing calculator. The intersection point of these curves represents the solution to the given equation.) - When graphing y=e^t, you will see that it is an increasing exponential function that starts near the x-axis and rises faster than linear functions. - When graphing y=1000, you will see that it is a horizontal straight line at 1000 level on the y-axis. - By looking at their intersection point, you can verify that the two functions meet near t=6.908, confirming our algebraic solution.

Key Concepts

Natural Logarithms

Natural logarithms are an essential tool in solving exponential equations like the one in our exercise, where we have the equation \( e^t = 1000 \). The natural logarithm is denoted as \( \ln(x) \) and is specifically the inverse of the base \( e \) exponential function. This means using \( \ln \) helps us "undo" the exponentiation by \( e \), making it easier to isolate the variable.
In our equation, applying the natural logarithm to both sides gives us \( \ln(e^t) = \ln(1000) \). This step is crucial, as it allows us to transition from an exponential to a more manageable algebraic equation. Understanding the natural logarithm's role is key in tackling such problems successfully.

Logarithm Properties

Logarithm properties are powerful tools for simplifying equations. In our exercise, after applying the natural logarithm, we use a specific property: \( \ln(a^b) = b \ln(a) \). This is known as the "power rule" for logarithms.
Thanks to this rule, we can simplify \( \ln(e^t) = t \cdot \ln(e) \). And since \( \ln(e) = 1 \), this simplifies further to just \( t = \ln(1000) \). These properties save us time and effort, allowing us to solve equations efficiently without getting caught up in complex calculations.

• Logarithm properties transform exponential functions into linear ones.
• They clarify and reduce the complexity of equations.
• Familiarity with properties like the power rule is crucial in algebra.

Graphing Calculator

A graphing calculator is a great visual tool to confirm the solutions of algebraic equations. After solving the equation \( e^t = 1000 \) and finding \( t \approx 6.908 \), we can use a graphing calculator to graph the functions \( y = e^t \) and \( y = 1000 \).
By observing the graph, we see \( y = e^t \) as an increasing curve starting near the x-axis and rising steeply, while \( y = 1000 \) is a horizontal line. The intersection point of these two graphs indicates the solution to our equation. This method not only verifies our algebraic solution but also provides a visual confirmation that enhances understanding.

Algebraic Solution Verification

Algebraic solution verification is both an important and reassuring part of solving equations. Once we find the value of \( t \) algebraically (\( t = \ln(1000) \approx 6.908 \)), it's crucial to check the accuracy of this result.
Verification can be carried out using a graphing calculator, as we did. However, another approach involves substituting back into the original equation. Here, you can compute \( e^{t} \) using the \( t \) value found, ensuring it indeed equals 1000. This two-step approach - algebra and a numerical check - develops reliability in mathematics and cultivates confidence in your solutions.