Problem 22 - Precalculus with Limits

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places. \(6^{x}+10=47\)

Step-by-Step Solution

Answer

The solution to the equation \(6^{x}+10=47\) approximated to three decimal places is 2.246.

Step 1: Isolate the Exponential Expression

First, subtract 10 from both sides of the equation to isolate the exponential expression on one side:\n\(6^{x}+10-10=47-10 \Rightarrow 6^{x}=37.\)

Step 2: Use Logarithms to Solve for the Unknown

Use a logarithm to solve for the unknown, transforming the equation to the form x = log_b(a)/log_b(c). Here, we get \(x = \log(37)/\log(6)\).

Step 3: Approximate the Solution

Finally, use a calculator to approximate the solution to three decimal places. The solution is approximately 2.246, although it might vary slightly depending on the calculator used.

Key Concepts

Exponential Expression Isolation

When solving an exponential equation, the first crucial step is isolating the exponential expression. This process involves manipulating the equation so that the term with the exponential function is on one side of the equation by itself. This isolation is typically achieved through basic algebraic operations, such as addition, subtraction, multiplication, or division.

For example, with the equation \(6^{x}+10=47\), we aim to get \(6^{x}\) by itself on one side. To achieve this, we subtract 10 from both sides to get \(6^{x} = 37\). Isolating the exponential expression simplifies the equation, making it easier to apply logarithmic operations to solve for the variable. It's essential for the students to understand this step as it sets the stage for employing logarithms.

Logarithmic Problem Solving

After isolating the exponential expression, the next step in solving the equation is logarithmic problem solving. Logarithms are the inverse operations of exponentiation, meaning they allow us to 'undo' exponential functions and solve for the unknown exponent.

To apply logarithms to the equation \(6^{x} = 37\), we can take the logarithm of both sides, typically using the common logarithm (log base 10) or natural logarithm (ln, which is log base \(e\)). In our example, we use the common logarithm to get \(x = \frac{\log(37)}{\log(6)}\). This step transforms the exponential equation into a form where the variable can be solved using basic algebra. It's key for students to grasp that logarithms are tools that make managing exponential terms possible.

Approximating Solutions

The last stage in solving an exponential equation is approximating solutions. After transforming our equation with logarithms, we often reach a point where exact solutions are not easily obtainable, and a calculator becomes necessary.

Using a calculator to evaluate \(x = \frac{\log(37)}{\log(6)}\), we can find an approximate solution to the equation. It's important to round the solution to the specified degree of precision, which in the given problem is three decimal places. Therefore, our solution is approximately 2.246. Students should understand that the accuracy of the approximation can vary slightly depending on calculator models and settings. Double-checking this approximation can help ensure its accuracy and reliability in the context of the problem.