Problem 30 - A Graphical Approach to College Algebra

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. $$3(1.4)^{x}-4=60$$

Step-by-Step Solution

Answer

Exact solution is \(x = \frac{\log\left(\frac{64}{3}\right)}{\log(1.4)}\). Approximate solution is \(x \approx 9.110\).

Step 1: Isolate the Exponential Expression

The given equation is \(3(1.4)^{x} - 4 = 60\). First, we need to get the exponential term by itself on one side of the equation. To do this, add 4 to both sides:\[ 3(1.4)^{x} = 64 \]

Step 2: Divide to Isolate the Base

Next, divide both sides of the equation by 3 to further isolate \( (1.4)^x \): \[ (1.4)^{x} = \frac{64}{3} \approx 21.333 \]

Step 3: Take the Logarithm of Both Sides

Apply the natural logarithm or common logarithm to both sides of the equation to bring the power down: \[ \log((1.4)^{x}) = \log\left(\frac{64}{3}\right) \] Utilizing the logarithm power rule \(\log(a^b) = b \cdot \log(a)\), we rewrite:\[ x \cdot \log(1.4) = \log\left(\frac{64}{3}\right) \]

Step 4: Solve for x

To solve for \(x\), divide both sides by \(\log(1.4)\):\[ x = \frac{\log\left(\frac{64}{3}\right)}{\log(1.4)} \] Using a calculator, compute:\[ \log\left(\frac{64}{3}\right) \approx 1.329 \quad \text{and} \quad \log(1.4) \approx 0.146 \]Thus, \[ x \approx \frac{1.329}{0.146} \approx 9.110 \]

Step 5: Round the Solution

Round the solution to the nearest thousandth. Therefore, the solution is approximately:\[ x \approx 9.110 \]

Key Concepts

Logarithms

When solving exponential equations, logarithms are a powerful tool. They help manage equations where the variable is in the exponent, such as \( (1.4)^x = \frac{64}{3} \). By applying a logarithm to both sides, you can bring the variable down from the exponent.Here's how it works:

• The power rule of logarithms: \( \log(a^b) = b \cdot \log(a) \). This is crucial because it allows you to transform \( (1.4)^x \) into something more handleable: \( x \cdot \log(1.4) \).
• Natural or common logarithms can be used; both have unique applications depending on your calculator or preference.

By using logarithms, you simplify complicated exponential relationships and make them solvable just like linear equations.

Exact and Approximate Solutions

With many mathematical problems, finding both exact and approximate solutions is essential.For the equation \(3(1.4)^{x} - 4 = 60\), we find the exact form of \(x\) initially by setting up the expression \( x = \frac{\log\left(\frac{64}{3}\right)}{\log(1.4)} \).

• Exact Solution: This involves leaving the expression as a fraction of logarithms. It's precise but not always immediately understandable.
• Approximate Solution: Here, we compute the numerical values of the logarithms to three decimal places. This is where a calculator becomes handy to crunch numbers such as \( \log\left(\frac{64}{3}\right) \approx 1.329 \) and \( \log(1.4) \approx 0.146 \), then rounding the result to \( x \approx 9.110 \).

Approximations give us practical answers, helpful in real-world settings where complex results need interpreting quickly.

Calculator Use

Calculators are indispensable when working with exponential equations, especially for finding approximate solutions.They simplify the process of computing logarithms and division, like in solving for \( x \approx 9.110 \). Here's how to maximize your calculator's functionalities:

• Logarithm Functions: Utilize the built-in log functions. Many calculators have buttons for both natural \( \log \) and common \( \log \) to assist in conversions.
• Precision: Set your calculator to display results to the required decimal places, such as thousandths, for accurate approximations.
• Step-by-Step Verification: After each calculation step, check the results to ensure they're logical before moving on. Re-calculating can avoid errors, especially in exams or homework.

By efficiently using a calculator, students achieve precise results swiftly and confidently, making them more effective learners.