Problem 18 - Algebra for College Students

Question

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

Step-by-Step Solution

Answer

The solution is approximately \(x \approx 7.83\).

Step 1: Take the Natural Logarithm of Both Sides

Given the equation \(3^{x-1} = 2^{x+3}\), we want to solve for \(x\). To do so, we first take the natural logarithm of both sides to bring the exponents down: \[\ln(3^{x-1}) = \ln(2^{x+3})\] Using the logarithm power rule, we rewrite the equation as:\[(x-1)\ln(3) = (x+3)\ln(2)\]

Step 2: Expand and Rearrange the Equation

We expand the equation:\[x\ln(3) - \ln(3) = x\ln(2) + 3\ln(2)\] Rearrange the equation to consolidate terms involving \(x\) on one side and constants on the other:\[x\ln(3) - x\ln(2) = \ln(3) + 3\ln(2)\]

Step 3: Factor Out \(x\) and Solve for \(x\)

Factor \(x\) out of the left-hand side:\[x(\ln(3) - \ln(2)) = \ln(3) + 3\ln(2)\]Now, solve for \(x\):\[x = \frac{\ln(3) + 3\ln(2)}{\ln(3) - \ln(2)}\]

Step 4: Calculate Using a Calculator

Using a calculator, we compute the necessary natural logarithms and perform the arithmetic:\[\ln(3) \approx 1.0986, \quad \ln(2) \approx 0.6931\]Substitute these values into the equation:\[x \approx \frac{1.0986 + 3 \times 0.6931}{1.0986 - 0.6931}\]\[x \approx \frac{1.0986 + 2.0793}{0.4055} \approx 7.83\]

Step 5: Round to the Nearest Hundredth

The calculated value of \(x\) is approximately 7.83, which is already rounded to the nearest hundredth.

Key Concepts

Natural Logarithm

In mathematics, the natural logarithm is a powerful concept when working with exponential equations. It is denoted as \( \ln \) and is the logarithm to the base \( e \), where \( e \approx 2.71828 \). Using the natural logarithm helps to simplify exponential equations, especially when the terms involve variables in the exponents.

In the given equation \( 3^{x-1} = 2^{x+3} \), taking the natural logarithm of both sides is critical to bringing down the exponents via the logarithm power rule, allowing for straightforward manipulation and solving.

• The natural logarithm is particularly useful for equations involving exponential growth and decay.
• It transforms multiplicative relationships of exponents into additive relationships, making algebraic manipulation easier.

Understanding how to apply the natural logarithm is vital for solving complex exponential equations.

Logarithm Power Rule

The logarithm power rule is an essential property that enables us to handle exponents effectively. This rule states that for any positive number \( a \) and logarithm base \( b \), \( \log_b(a^n) = n \cdot \log_b(a) \).

In the context of the exercise, applying the logarithm power rule converts the equation \( \ln(3^{x-1}) = \ln(2^{x+3}) \) into \((x-1)\ln(3) = (x+3)\ln(2)\). This transformation is crucial as it allows the exponents to be written as multiples, making algebraic manipulation possible.

• The power rule is helpful in linearizing exponential relationships, a necessary step when isolating the variable \( x \).
• It allows for the separation of variables and constants, facilitating easier solutions.

This rule is a fundamental tool in dealing with any expression where the exponent is the primary unknown.

Approximate Solutions

After algebraic manipulation of the equations, finding precise numerical solutions often isn’t practical or necessary. Instead, approximations to a certain degree of accuracy are used, usually specified in the problem.

In this exercise, we rounded the calculated value of \( x \) to the nearest hundredth. Approximations, often based on computational tools or calculators, allow us to express complex numbers in a more understandable form.

• Approximations are crucial when dealing with irrational numbers, which are common when using logarithms.
• Maintaining a consistent level of precision is essential for clarity and accuracy.

The process aids students in understanding that achieving exact equality is often unrealistic, and approximate answers are just as valuable in practical applications.

Factorization

Factorization is a technique for breaking down equations into simpler components. When dealing with equations like \( x \ln(3) - x \ln(2) = \ln(3) + 3 \ln(2) \), factorization becomes a handy method.

In this equation, factored expressions simplify to \( x(\ln(3) - \ln(2)) = \ln(3) + 3\ln(2) \). By isolating \( x \), it becomes easier to solve for the variable.

• Factoring helps consolidate variables, reducing the equation's complexity.
• It is often used in combination with other algebraic techniques to isolate and solve for variables.

In exponential equations, factorization allows for a clearer step-by-step breakdown, showcasing logical progressions from one transformation to another.

Algebraic Manipulations

Algebraic manipulations play a crucial role in solving equations, especially those with exponential and logarithmic components. These manipulations involve rearranging equations, expanding them, consolidating like terms, and more.

For instance, in rearranging \( x\ln(3) - \ln(3) = x\ln(2) + 3\ln(2) \) into \( x(\ln(3) - \ln(2)) = \ln(3) + 3\ln(2) \), algebraic manipulations make it feasible to solve for \( x \) using simpler means.

• Combining and switching terms in equations often provide avenues to reach the solution step.
• Efficient algebraic manipulations result in clearer, more understandable results.

Proficiency in these methods is key in navigating complex problems with multiple variables and operations.