Problem 68 - 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. $$\left(16+\frac{0.878}{26}\right)^{3 t}=30$$

Step-by-Step Solution

Answer

The solution to the equation \((16+\frac{0.878}{26})^{3t}=30\) is approximately \(t = 0.372\), after rounding to three decimal places.

Step 1: Simplify the Base of the Exponential

The first step involves simplifying the base of the exponential, that is \(16+\frac{0.878}{26}\), resulting in approximately 16.0338. Now, rewrite the equation as \((16.0338)^{3t} = 30\).

Step 2: Apply Natural Logarithm on Both Sides

In order to solve for \(t\), apply natural logarithm on both sides of the equation:\(\ln((16.0338)^{3t}) = \ln(30)\3t \ln(16.0338) = \ln(30)\).

Step 3: Solve for Variable t

Now, isolate the variable \(t\) by dividing both sides by \(3\ln(16.0338)\).The solution to \(t\) will be \(t = \frac{\ln(30)}{3\ln(16.0338)}\).

Step 4: Calculate the numerical value of t

The rounded solution of \(t\) after calculation will be \(t \approx 0.372\) to three decimal places.

Step 5: Verify the Solution Using a Graphing Utility

You can verify this solution by plotting \((16.0338)^{3t}\) and \(30\) on a graph. If the solution is correct, the graph of \((16.0338)^{3t}\) will intersect with \(30\) at \(t = 0.372\).

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is a mathematical function that helps to solve equations involving exponential expressions. It is particularly useful in solving problems where you need to isolate a variable that serves as an exponent. The natural logarithm is the logarithm to the base \( e \), where \( e \) is an irrational constant approximately equal to 2.718. Using natural logarithms, complex exponential equations can be transformed into simpler linear equations.
In the given problem, applying the natural logarithm to both sides follows the property \( \ln(a^b) = b \cdot \ln(a) \). This property helps in bringing down the exponent, allowing us to deal more conveniently with the variable of interest. Specifically, for the equation \( (16.0338)^{3t} = 30 \), after taking the natural log, it becomes \( 3t \cdot \ln(16.0338) = \ln(30) \). This conversion makes solving for "\( t \)" straightforward by turning the exponential equation into a linear form.
Remember that applying logarithms simplifies the solving process tremendously by making the operations on the algebraic expressions more manageable.

Graphing Utility

A graphing utility is a tool used to visualize mathematical equations and verify solutions to complex problems. It can be a physical calculator or software like Desmos or GeoGebra that lets you plot graphs of various mathematical functions. When solving exponential equations, using a graphing utility helps confirm whether the mathematical solution derived is correct.
In the context of the original problem, once the equation \( y = (16.0338)^{3t} \) is graphed alongside the horizontal line \( y = 30 \), you can visually check the point of intersection. The point where these graphs meet corresponds to the value of \( t \), reinforcing the algebraic solution that you calculated formerly. For instance, if both graphs intersect when \( t = 0.372 \), it validates that your solution is accurate.
The graphical intersection is a powerful verification tool since it offers an intuitive visual confirmation of the intersection ensuring that your algebraic work is on the right track.

Rounding Decimals

Rounding decimals is a mathematical process used to simplify a number while maintaining a degree of accuracy based on the required significant figures. This is particularly useful when a precise answer isn't necessary, or when it is practically more efficient to communicate.
In the context of solving exponential equations, rounding the result to a specific number of decimal places is critical for precise yet readable results. For this problem, after finding \( t = \frac{\ln(30)}{3\ln(16.0338)} \), you arrive at \( t \approx 0.372 \) when rounded to three decimal places.
In practice, rounding involves looking at the digit immediately following the desired decimal place. If it's 5 or higher, you round up. Otherwise, you round down by leaving the last retained digit unchanged. Rounding ensures results are easily interpretable and practical for real-world applications without losing crucial accuracy in most scenarios.