Problem 19 - Algebra for College Students

Question

Solve each exponential equation and express approximate solutions to the nearest hundredth. $$ 3^{2 x+1}=2^{3 x+2} $$

Step-by-Step Solution

Answer

The approximate solution is \( x \approx 2.44 \).

Step 1: Isolate Variables

Start by rewriting the equation as it is: \( 3^{2x+1} = 2^{3x+2} \). Notice both sides are powers with different bases. The typical approach for solving exponential equations involves using logarithms to isolate the variable.

Step 2: Apply Logarithms

Take the natural logarithm (ln) of both sides to get \( \ln(3^{2x+1}) = \ln(2^{3x+2}) \). Use the logarithmic identity \( \ln(a^b) = b \cdot \ln(a) \) to simplify: \((2x+1) \cdot \ln(3) = (3x+2) \cdot \ln(2)\).

Step 3: Expand and Rearrange

Expand both sides to distribute the logarithms: \( 2x \cdot \ln(3) + \ln(3) = 3x \cdot \ln(2) + 2 \cdot \ln(2) \). Rearrange the equation to bring all terms involving \(x\) to one side: \( 2x \cdot \ln(3) - 3x \cdot \ln(2) = 2 \cdot \ln(2) - \ln(3) \).

Step 4: Factor and Solve for x

Factor out the \(x\) from the left-hand side: \( x(2 \cdot \ln(3) - 3 \cdot \ln(2)) = 2 \cdot \ln(2) - \ln(3) \). Solve for \(x\) by dividing both sides: \( x = \frac{2 \cdot \ln(2) - \ln(3)}{2 \cdot \ln(3) - 3 \cdot \ln(2)} \).

Step 5: Calculate the Value of x

Use a calculator to evaluate each logarithm and perform the arithmetic: \( x \approx \frac{2(0.6931) - 1.0986}{2(1.0986) - 3(0.6931)} \). Compute the approximation: \( x \approx \frac{1.3862 - 1.0986}{2.1972 - 2.0793} \approx \frac{0.2876}{0.1179} \). This results in \( x \approx 2.44 \).

Step 6: Round to the Nearest Hundredth

The computed value of \(x\) is already approximated, so rounding it to the nearest hundredth does not change the value: \( x \approx 2.44 \).

Key Concepts

Logarithms

Logarithms are a critical mathematical tool for solving exponential equations. They help transform the expression into a simpler form, making it easier to isolate the variable. A logarithm answers the question: "To what exponent must the base be raised to produce a given number?" For example, in base-10 logarithms, if we say \( \log_{10}(100) \), it equals 2 because \( 10^2 = 100 \).

Logarithms make it possible to convert multiplication into addition, thanks to their properties. One such useful property is the power rule, which allows us to bring exponents down in front as coefficients. This is precisely how they help in solving equations like the one in the original exercise. By transforming the original problem into an equation with logarithms, it becomes much more manageable and solvable.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a logarithm with the base \( e \), where \( e \approx 2.71828 \). It is a frequent feature in calculus due to its intrinsic characteristics related to continuous growth and natural occurring phenomena.

When solving equations involving exponential terms, using the natural logarithm is beneficial due to its base being \( e \). It simplifies computations when dealing with complex mathematical models. In problems like the given exercise, taking the natural logarithm of both sides converts an exponential equation into a polynomial equation, a much easier form to handle.

Solving Equations

Solving equations, particularly those involving exponents, often begins with simplifying the equation to isolate the variable of interest. In the given exercise, we see a complex exponential equation where direct calculation isn't feasible without transforming both sides using logarithms.

The approach involves eliminating the exponent by converting the equation through logarithms, allowing you to handle it like a linear equation. Once transformed, the equation can be expanded and rearranged to isolate the variable \( x \) neatly. Mathematics often require turning an abstract representation into an arithmetic operation, as shown when we assess both structures by applying logarithms. This logical breakdown into manageable steps enables us to arrive at the solution accurately.