Problem 38 - College Algebra and Calculus: An Applied Approach

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(5\left(10^{x-6}\right)=7\)

Step-by-Step Solution

Answer

Using a calculator to approximate to three decimal places, \(x = \log_{10}(\frac{7}{5}) + 6 \approx 6.146\)

Step 1: Isolate the Exponential Term

First, isolate the term with the exponent. This can be done by dividing both sides of the equation by 5 to get \(10^{x-6} = \frac{7}{5}\)

Step 2: Apply Logarithm

Next, apply a logarithm to both sides. Recall that the logarithm base 10 (also known as the common logarithm) is the inverse operation to raising 10 to an exponent. Applying logarithm base 10 to both sides, the equation simplifies to \(x-6 = \log_{10}(\frac{7}{5})\)

Step 3: Find the Value of x

Finally, solve for x by adding 6 to both sides. This gives \(x = \log_{10}(\frac{7}{5}) + 6\)

Key Concepts

Logarithms

Logarithms are mathematical operations that are the inverse of exponentiation. They allow us to solve equations where the unknown variable is an exponent. To understand logarithms, consider the equation where the base of an exponent is raised to a power to achieve a certain number. A logarithm answers the question: To what exponent must we raise the base to get that number?
For example, in the equation \(10^y = z\), the logarithm is written as \(y = \log_{10}(z)\). In the context of exponential equations, using logarithms is crucial because it lets us transform the equation into an algebraic form, which is easier to solve.

• Base 10 Logarithm: Also known as the common logarithm, this is often implied when no base is written. It's used in this exercise, precisely because our base is 10.
• Properties: Utilizing properties such as \(\log_{10}(ab) = \log_{10}(a) + \log_{10}(b)\) can simplify equations further, although not directly needed here.

Algebraic Solutions

Algebraic solutions consist of manipulating the equation to isolate the unknown variable step by step. When solving exponential equations algebraically, one of the initial tasks is to simplify the equation so that you can apply the right algebraic methods. The goal is to position the equation in a form where logarithms can be applied.

For instance, in our exercise, after isolating the exponential term, we transformed it into a simpler equation using a logarithm. The resulting equation \(x - 6 = \log_{10}(\frac{7}{5})\) becomes more manageable as a linear equation.

• Transformation: Each manipulation requires reversing previous operations using algebraic rules. The decision to use a logarithm was guided by the need to simplify our exponential equation into a linear one.
• Estimation: Finally, once the variable is isolated, we approximate the solution to detect more practical answers, which is especially important for exercises demanding decimal results.

Isolation of Variables

Isolation of variables is a significant concept in solving equations. It involves rearranging the equation so that one side contains the variable to be solved for, while the other side contains all other terms. In exponential equations, this often involves initially isolating the term that includes the exponential function. For our problem, the first step involved dividing the entire equation by 5. This action left us with the exponential form \(10^{x-6}\) on one side.

• Exponential Term Isolation: Critical in transforming the equation to a form that allows the usage of logarithms. This step is foundational because it sets up subsequent moves involving inverse operations, like logarithms.
• Practicality: Isolating the variable helps clarify the target of our solution strategy. Each action in isolation reflects a reversal of steps used to build the original equation, moving us back to the variable incrementally.

The concept strengthens problem-solving by enabling systematic solving of direct and more complex equations.