Problem 23 - Beginning and Intermediate Algebra

Question

Solve the exponential equation algebraically. Then check using a graphing calculator. $$(3.9)^{x}=48$$

Step-by-Step Solution

Answer

To solve the exponential equation \((3.9)^x = 48\) algebraically, apply the natural logarithm to both sides: \(\ln((3.9)^x) = \ln(48)\). With logarithm rules, the equation becomes \(x \cdot \ln(3.9) = \ln(48)\). Now, isolate \(x\) by dividing both sides by \(\ln(3.9)\), yielding \(x \approx 2.58\). Verify the solution using a graphing calculator by plotting the function \(y = (3.9)^x\) and checking if the point \((2.58, 48)\) lies on the graph. If the point is on the graph, the solution is confirmed: \(x \approx 2.58\).

Step 1: Take the natural logarithm of both sides

To algebraically solve for \(x\), one must isolate it. In order to do this for the given exponential equation, take the natural logarithm of both sides of the equation: \[\ln\left((3.9)^x\right) = \ln(48)\] The logarithm rules allow us to take exponential variable associations out of the logarithm, meaning the equation becomes: \[ x \cdot \ln(3.9) = \ln(48)\]

Step 2: Solve for x

In this step, the goal is to isolate \(x\) on one side of the equation. That can be done by dividing both sides of the equation by \(\ln(3.9)\): \[x = \frac{\ln(48)}{\ln(3.9)}\]

Step 3: Calculate the value of x

Finally, calculate the numerical value of \(x\) using the function of a calculator capable of computing logarithms: \[x \approx 2.58\]

Step 4: Verify the solution using a graphing calculator

Now, the obtained solution \(x \approx 2.58\) can be verified using a graphing calculator. Firstly, graph the function \(y= (3.9)^{x}\), keeping in mind the value of \(x\) from step 3. Then, verify if the point \((2.58, 48)\) lies on the graph. If it does, this confirms that the solution \(x \approx 2.58\) is correct. Please remember that the calculator may not show an exact match between the plotted point and the graph due to rounding and the accuracy of the calculator display. Consequently, once the point \((2.58, 48)\) has been verified to be on the graph, the solution is confirmed as \(x \approx 2.58\).

Key Concepts

Natural Logarithms

Understanding how to use natural logarithms is crucial for solving exponential equations. The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \) is approximately equal to 2.71828. It serves as the inverse function of the exponential function when the base is itself \( e \). For instance:

• \( \ln(e^x) = x \) because when \( e \) is raised to the power of \( x \), it makes the expression \( e^x \).
• Conversely, \( e^{\ln x} = x \) expresses the same in reverse.

In our original problem, using the natural logarithm allowed us to isolate the variable \( x \) in the equation \( (3.9)^x = 48 \). By taking \( \ln \) of both sides, the properties of logarithms let us turn the exponential equation into a solvable linear equation, \( x \cdot \ln(3.9) = \ln(48) \). This simplification is the cornerstone of solving exponential equations algebraically.

Graphing Calculator

Graphing calculators are useful tools for confirming algebraic solutions. They can display functions and let you visually verify if solutions make sense in a graphical context. For instance, after solving an exponential equation like \((3.9)^x = 48\), you'd graph the function \(y = (3.9)^x\). With a graphing calculator, locate where this graph intersects the line \(y = 48\), which should ideally align with the algebraically determined \(x\)-value. To do so, access the graph function mode, enter \(y = (3.9)^x\), and also the horizontal line \(y = 48\) for reference. Then, use your graphing calculator's trace or intersection functions to find the approximate intersection point.While technology can often verify your calculations, remember that different settings or rounding may slightly affect your display. Overall, using graphing calculators gives a visual confirmation of your algebraic solution.

Exponential Functions

Exponential functions, characterized by expressions where the variable is in the exponent, have a wide range of applications in mathematics and beyond. They manifest as \(f(x) = b^x\), where \( b \) is a positive constant base. Important properties include:

• If \( b > 1 \), the function is increasing, which means as \(x\) gets larger, \(f(x)\) grows exponentially.
• If \( 0 < b < 1 \), the function is decreasing, leading to a decay in \(f(x)\) as \(x\) grows.

In practical terms, increasable functions like population growth, compound interest, and radioactivity often follow exponential growth, similar to the equation \((3.9)^x = 48\) we solved earlier. Recognizing these patterns helps in anticipating function behavior and solving equations involving exponential terms.

Solving Equations Algebraically

Solving equations algebraically involves manipulation to isolate the variable without numerical estimation or graphing initially. For exponential equations, like \((3.9)^x = 48\), you typically:

• Apply logarithms to simplify, allowing the exponent to be derived linearly, for instance, taking the natural logarithm: \( \ln((3.9)^x) = \ln(48) \).
• Use the property \( \ln(a^b) = b \cdot \ln(a) \) to transform into a multiplication problem: \( x \cdot \ln(3.9) = \ln(48) \).
• Solve for the desired variable by isolating it, like \( x = \frac{\ln(48)}{\ln(3.9)} \).

This method provides exact answers conditional upon understanding algebraic rules and logarithms. Such step-by-step techniques are systematic approaches to tackling a variety of equations.