Problem 28 - A Graphical Approach to Precalculus with Limits

Question

Solve each exponential equation. Express the solution set so that (a) solutions are in exact form and, if irrational, (b) solutions are approximated to the nearest thousandth. Support your solutions by using a calculator. $$5(1.2)^{3 x-2}+1=11$$

Step-by-Step Solution

Answer

The solution is \(x \approx 1.934\), rounded to the nearest thousandth.

Step 1: Isolate the Exponential Term

Begin by isolating the exponential expression. Subtract 1 from both sides of the equation: \[ 5(1.2)^{3x-2} = 10 \] Then, divide both sides by 5: \[ (1.2)^{3x-2} = 2 \] This leaves us with the exponential term on one side.

Step 2: Take the Logarithm of Both Sides

To solve for \(x\), take the natural logarithm (or common logarithm) of both sides. Using the natural logarithm: \[ \ln((1.2)^{3x-2}) = \ln(2) \] Apply the power rule for logarithms: \[ (3x-2) \cdot \ln(1.2) = \ln(2) \] This reformulates the equation in a linear form involving \(x\).

Step 3: Solve for x

Divide both sides by \(\ln(1.2)\): \[ 3x - 2 = \frac{\ln(2)}{\ln(1.2)} \] Add 2 to both sides: \[ 3x = \frac{\ln(2)}{\ln(1.2)} + 2 \] Finally, divide by 3 to solve for \(x\): \[ x = \frac{\frac{\ln(2)}{\ln(1.2)} + 2}{3} \]

Step 4: Calculate the Exact Solution

Use a calculator to compute the value: 1. Calculate \( \ln(2) \approx 0.693147 \) and \( \ln(1.2) \approx 0.182322 \).2. Compute \( \frac{0.693147}{0.182322} \approx 3.800743 \).3. Add 2: \( 3.800743 + 2 = 5.800743 \).4. Divide by 3: \( \frac{5.800743}{3} \approx 1.933581 \).Thus, \(x \approx 1.933581\) is the exact form.

Step 5: Approximate the Solution to the Nearest Thousandth

Round the exact solution to three decimal places: \[ x \approx 1.934 \]This approximation matches to the nearest thousandth.

Key Concepts

Logarithms

Logarithms play a crucial role in solving exponential equations, as they help us transform these non-linear expressions into more manageable forms. In the exercise, we encountered the exponential equation involving the term \((1.2)^{3x-2}\). To solve this equation, it's effective to use logarithms to bring the variable exponent down, turning the problem into a linear equation.

The power rule for logarithms states that \(\log_b(a^n) = n \cdot \log_b(a)\), meaning we can move the exponent to the front as a coefficient. This is helpful in isolating the variable "\(x\)" in the equation. By employing the natural logarithm (denoted as \(\ln\)), we reformulated the equation \(\ln((1.2)^{3x-2}) = \ln(2)\) to allow direct computation. Logarithms thus transform the complex exponential manipulation into straightforward arithmetic

Exact Solutions

To find an exact solution for exponential equations, we start by manipulating the equation with algebraic principles. The first step is isolating the exponential term, which ensures that the exponential part stands alone on one side of the equation.

After applying logarithms, the equation \((3x-2) \cdot \ln(1.2) = \ln(2)\) becomes linear in \(x\). The resulting expression \(3x - 2 = \frac{\ln(2)}{\ln(1.2)}\) is straightforward to handle with elementary algebra.

To find the exact form of the solution, use a calculator for precision, ensuring that each step respects the properties of logarithms and basic algebra. The final computation \(x = \frac{\frac{\ln(2)}{\ln(1.2)} + 2}{3}\) yields a more exact numerical solution when calculated precisely, which can be critical in certain scientific and mathematical applications.

Approximation

Once we determine the exact solution using a calculator, the next common step is approximation. Approximation simplifies the result to a user-friendly form, often required in real-world scenarios where precision beyond a few decimal places is unnecessary.

The solution we found, \(x \approx 1.933581\), is quite precise. However, many applications require the solution rounded to a specific decimal place, such as the thousandth. Rounding involves identifying the decimal to the left of the required position and following standard rounding rules.

In this problem, we approximated the result to the nearest thousandth, resulting in \(x \approx 1.934\). This method streamlines the result while maintaining a balance between precision and simplicity in practical use cases.