Problem 47 - College Algebra and Calculus: An Applied Approach

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(6\left(2^{3 x-1}\right)-7=9\)

Step-by-Step Solution

Answer

The solution is \(x = 0.232\), when approximation is done to three decimal places.

Step 1: Simplify the equation

The aim is to isolate the term with \(x\). So, the initial step is to simplify the equation using basic algebra operations. That involves adding 7 to both sides of the equation which gives: \(6\left(2^{3 x-1}\right) = 9 + 7\).

Step 2: Further simplification

After the simplification done in Step 1, the equation becomes: \(6\left(2^{3 x-1}\right) = 16\). Divide both sides of the equation by 6 leading to: \(\left(2^{3 x-1}\right) = \frac{16}{6}\). Simplify the right side of the equation, resulting in \(\left(2^{3 x-1}\right) = \frac{8}{3}\).

Step 3: Apply logarithm on both sides

Apply natural logarithm (\(\ln\)) to both sides of the equation to get rid of the exponent: this results in \(3x - 1 = \ln\left(\frac{8}{3}\right)\).

Step 4: Isolate \(x\)

Solve for \(x\) by first adding 1 to both sides: \(3x = \ln\left(\frac{8}{3}\right) + 1\). Then divide both sides of the equation by 3 to finally isolate \(x\): \(x = \frac{\ln\left(\frac{8}{3}\right) + 1}{3}\).

Step 5: Approximate the result

The last step is to use a calculator to evaluate the final result. Round off to the nearest three decimal places.

Key Concepts

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to isolate the variable of interest. In this exercise, the goal is to solve for \(x\) in the given exponential equation.

First, add or subtract terms to both sides of the equation to simplify it. This is done by moving constants around. For example, adding 7 to both sides resulted in \(6\left(2^{3x-1}\right) = 16\).

Next, to further simplify, divide both sides by the coefficient of the exponential term, which here is 6, leading us to \(2^{3x-1} = \frac{8}{3}\).

Algebraic manipulation is a powerful tool, allowing us to transform complex expressions into simpler forms, making it easier to find solutions.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is crucial for dealing with exponential equations. It helps us eliminate exponentials by converting exponential expressions into linear ones.

Applying the natural logarithm to both sides of the equation \(2^{3x-1} = \frac{8}{3}\) transforms it into \(3x - 1 = \ln\left(\frac{8}{3}\right)\).

This transformation works because logarithms have the property \(\ln(a^b) = b \ln(a)\), allowing us to pull the exponent as a coefficient, which simplifies solving for the variable \(x\).

Solve for Variable

Solving for a variable involves isolating it on one side of the equation. After applying the natural logarithm, you have \(3x - 1 = \ln\left(\frac{8}{3}\right)\).

The next step is to get rid of constants surrounding \(x\) by performing inverse operations. Add 1 to both sides to cancel out the subtraction: \(3x = \ln\left(\frac{8}{3}\right) + 1\).

Finally, divide by 3, the coefficient of \(x\), resulting in: \(x = \frac{\ln\left(\frac{8}{3}\right) + 1}{3}\).

Performing these steps methodically helps in accurately isolating and solving for the variable.

Approximation Techniques

Approximation techniques are essential for finding a numerical solution when an exact analytical solution is cumbersome or impossible to achieve. In this particular exercise, after isolating \(x\), you get \(x = \frac{\ln\left(\frac{8}{3}\right) + 1}{3}\).

To compute this, a calculator is typically used to evaluate the expression \(\ln\left(\frac{8}{3}\right)\) and then apply the arithmetic operations.

Finally, round the result to three decimal places for a precise answer that fits the requirement.

• Approximation makes solutions practical and understandable.
• Rounding to specified decimal places ensures consistency in results.

These techniques simplify real-world problem-solving by converting complex numbers into usable figures.