Problem 33 - Elementary and Intermediate Algebra

Question

Solve each equation. Give the exact solution and an approximation to four decimal places. See Example 4. $$ 2^{x+1}=3^{x} $$

Step-by-Step Solution

Answer

The exact solution is \( x = \frac{\ln(2)}{\ln(3) - \ln(2)} \) and the approximation is \( x \approx 1.7095 \).

Step 1: Express the equation with natural logarithms

Starting with the equation \( 2^{x+1} = 3^x \), take the natural logarithm (ln) of both sides to allow manipulation of the exponents: \[ \ln(2^{x+1}) = \ln(3^x) \]

Step 2: Apply logarithmic power rule

Use the power rule of logarithms, which states \( \ln(a^b) = b \cdot \ln(a) \), to simplify both sides:\[ (x+1) \cdot \ln(2) = x \cdot \ln(3) \]

Step 3: Expand and simplify the equation

Expand the left-hand side:\[ x \cdot \ln(2) + \ln(2) = x \cdot \ln(3) \] Subsequently, isolate the terms involving \( x \) by subtracting \( x \cdot \ln(2) \) from both sides:\[ \ln(2) = x \cdot \ln(3) - x \cdot \ln(2) \] Simplify further: \[ \ln(2) = x (\ln(3) - \ln(2)) \]

Step 4: Solve for x

Isolate \( x \) by dividing both sides by \( \ln(3) - \ln(2) \):\[ x = \frac{\ln(2)}{\ln(3) - \ln(2)} \] This expression gives the exact solution for \( x \).

Step 5: Approximate the solution

Approximate the result by calculating numerical values:1. Compute \( \ln(2) \approx 0.6931 \)2. Compute \( \ln(3) \approx 1.0986 \)3. Substitute these into the expression:\[ x \approx \frac{0.6931}{1.0986 - 0.6931} \]Calculate the denominator:\[ 1.0986 - 0.6931 = 0.4055 \]Then, divide to get:\[ x \approx \frac{0.6931}{0.4055} = 1.7095 \]

Key Concepts

Natural Logarithms

Natural logarithms, often denoted by \( \ln(x) \), are a special kind of logarithm where the base is the mathematical constant \( e \approx 2.71828 \). They are widely used in equations involving exponential functions because of their ability to simplify the process of solving for unknown variables. By transforming exponentials into multiplications, natural logarithms make the calculations more manageable. In our exercise, to solve the equation \( 2^{x+1} = 3^x \), we applied the natural logarithm to both sides. This conversion was crucial because it enabled us to manipulate the exponents easily. Such manipulation is necessary due to the difficulty of working with exponential equations directly. The intrinsic properties of logarithms allow us to bring down the exponents, making it possible to isolate the variable \( x \). Understanding natural logarithms can significantly simplify problems involving exponential terms. These kinds of transformations are essential in various fields, including mathematics, physics, and engineering.

Logarithmic Power Rule

The logarithmic power rule is an essential tool when dealing with logarithms. According to this rule, for any positive number \( a \) and any real number \( b \), the expression \( \ln(a^b) \) can be transformed into \( b \cdot \ln(a) \). This rule is particularly helpful when dealing with exponential functions because it allows us to "bring down" the exponent.In our exercise example, after taking the natural logarithm of both sides of the equation \( 2^{x+1} = 3^x \), we applied the power rule. This converted our equation from \( \ln(2^{x+1}) = \ln(3^x) \) to \( (x+1) \cdot \ln(2) = x \cdot \ln(3) \).The logarithmic power rule is crucial in rearranging and simplifying logarithmic expressions in equations. It can transform a seemingly complex problem into a simpler linear form, which can then be solved for the unknown variables. Mastering this rule will open up a wide array of strategies for handling exponential and logarithmic equations.

Exact Solutions

Exact solutions in mathematics refer to answers that use exact numbers rather than approximations. These often involve irrational numbers or numbers expressed in terms of roots or logarithms, as in our exercise example.To find the exact solution to the equation \( 2^{x+1} = 3^x \), we manipulated the equation using logarithms, eventually isolating \( x \) and arriving at \( x = \frac{\ln(2)}{\ln(3) - \ln(2)} \). This expression represents the exact solution, as it is expressed in terms of logarithms without resorting to numerical approximation.Finding exact solutions relies on algebraic manipulation and understanding the properties of logarithms and exponential functions. These solutions are significant, especially in theoretical analysis, as they give precise and general results not bound to specific numerical values. They serve as a solid foundation for further calculations that might require high precision.

Approximation Methods

When an exact solution is cumbersome or unnecessary, approximation methods become invaluable. They provide a way to estimate the solution with a specified degree of accuracy. In our context, after deriving the exact solution \( x = \frac{\ln(2)}{\ln(3) - \ln(2)} \), we approximated \( x \) to four decimal places.To approximate, we substituted known values of the logarithms into the expression:

• \( \ln(2) \approx 0.6931 \)
• \( \ln(3) \approx 1.0986 \)

Thus, the exact expression becomes \( x \approx \frac{0.6931}{1.0986 - 0.6931} \), which yields \( x \approx 1.7095 \) upon calculation.Approximation methods are practical in scenarios where precision is flexible, especially in engineering or computational fields, and when the exact solutions are too complex to be useful for everyday calculations. They bridge the gap between theoretical mathematics and real-world application, providing a usable result quickly and efficiently.