Problem 35 - Precalculus with Limits

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places. \(2^{x}=3^{x+1}\)

Step-by-Step Solution

Answer

The solution to the equation is approximately \(x = -1.631\)

Step 1: Convert exponential equation to logarithmic form

The given equation is \(2^{x}=3^{x+1}\). It can be rewritten in logarithmic form as: \(\log_2(2^x)=\log_2(3^{x+1})\). This means that x = \((x+1)*\log_2(3)\)

Step 2: Rearrange the equation

Rearrange the equation to get a linear equation. \x-(x+1)*\log_2(3) = 0 \x*(1-\log_2(3)) = -\log_2(3) \x = \(-\frac{\log_2(3)}{1 - \log_2(3)}\)

Step 3: Calculate the numerical value

Approximate the numerical value of x using calculator with rounding to three decimal places. x \(\approx -1.631\)

Key Concepts

Logarithmic Form

Converting an exponential equation into logarithmic form is a crucial step in solving equations where variables are in exponents. In the context of our exercise, we had the equation 2^{x}=3^{x+1}. To put this into logarithmic form, we utilize the property that a logarithm is the inverse operation of exponentiation.

For example, the statement y = b^x is equivalent to log_b(y) = x. Applying this understanding to our exercise leads us to express x as a function of a logarithm, transforming the complex exponential equation into a more manageable linear form. By using the base of the left exponent (in this case, 2), we can create an equation where the variable x is no longer an exponent but rather a coefficient, simplifying the process to find its value.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to isolate and solve for variables. In our exercise, once we converted the exponential equation into logarithmic form, we had to isolate the variable x. The key step was distributing x over the term log_2(3) and applying subtraction across the equation to get all terms with x on one side.

This manipulation yielded a linear equation in standard form, enabling us to solve for x directly. By understanding how to distribute, combine like terms, and isolate the variable, students can transform seemingly complex equations into ones that are easier to interpret. These algebraic techniques are essential for solving a wide variety of mathematical problems.

Numerical Approximation

Solving for exact values is not always feasible, especially when dealing with irrational numbers that cannot be expressed as a simple fraction. This is where numerical approximation comes in handy. In our example, after using algebra to solve for x, we considered the numerical approximation x \(\approx -1.631\).

This approximation was obtained by employing a calculator to evaluate the expression to the designated decimal places, providing a practical solution to real-world problems where precision to multiple decimal places is sufficient. Numerical approximation is an invaluable skill, especially for students who will encounter real-world scenarios where exact solutions may be impractical or unnecessary for the level of precision required.