Problem 20 - Algebra and Trigonometry

Question

Find the solution of the exponential equation, correct to four decimal places. $$ 10^{1-x}=6^{x} $$

Step-by-Step Solution

Answer

The solution is approximately 0.8687.

Step 1: Take the Logarithm of Both Sides

To simplify the equation, take the logarithm of both sides. You can choose either natural logarithms (\( \ln \)) or common logarithms (\( \log \)), but here we'll use natural logarithms: \( \ln(10^{1-x}) = \ln(6^x) \).

Step 2: Apply Logarithm Rules

Using the power rule of logarithms (\( \ln(a^b) = b \ln(a) \)), the equation becomes: \((1-x) \ln(10) = x \ln(6) \).

Step 3: Rearrange the Equation

Reorganize the terms to isolate \( x \) on one side: \( \ln(10) - x \ln(10) = x \ln(6) \) becomes \( \ln(10) = x \ln(6) + x \ln(10) \).

Step 4: Factor Out x

Factor \( x \) from the terms on the right side: \( \ln(10) = x (\ln(6) + \ln(10)) \).

Step 5: Solve for x

Divide both sides by \( \ln(6) + \ln(10) \) to solve for \( x \): \( x = \frac{\ln(10)}{\ln(6) + \ln(10)} \).

Step 6: Evaluate Numerically

Use a calculator to compute the value of \( x \), ensuring you retain four decimal places: \( x \approx 0.8687 \).

Key Concepts

Natural Logarithms

Natural logarithms, represented as \( \ln \), are logarithms with the base \( e \), where \( e \approx 2.71828 \). This is a constant that arises naturally in mathematics, especially in processes involving exponential growth and decay. In solving exponential equations, natural logarithms are particularly useful due to their ability to simplify expressions, especially when variables or exponents are involved. They are widely used because they make differentiation and integration more straightforward. In our problem, using natural logarithms helps us by converting the terms with exponents into forms that are easier to manipulate.

Logarithm Rules

Logarithms have several important rules that make them powerful tools in algebra, especially when solving equations involving exponential expressions. Some of the basic rules include:

• **Product Rule:** \( \log(a \cdot b) = \log(a) + \log(b) \)
• **Quotient Rule:** \( \log\left(\frac{a}{b}\right) = \log(a) - \log(b) \)
• **Power Rule:** \( \log(a^b) = b \cdot \log(a) \)

In our equation, we particularly make use of the power rule, which allows the exponent to be brought down as a coefficient. This transforms complex exponential equations into linear forms, thus making them simpler to solve. Understanding these rules is key to manipulating and simplifying logarithmic expressions effectively.

Power Rule of Logarithms

The power rule is a fundamental logarithmic identity which states that \( \log(a^b) = b \cdot \log(a) \). This rule is extremely helpful when you have an equation with variables in the exponent, as it allows you to "bring down" the exponent in front of the logarithm. This turns the problem from an exponential equation into a linear equation, making it much easier to solve. For the exercise at hand, applying this rule transforms \( \ln(10^{1-x}) \) into \((1-x) \cdot \ln(10)\) and \( \ln(6^x) \) into \( x \cdot \ln(6) \). After applying the power rule, solving the equation becomes a matter of simple algebra.

Numerical Evaluation

Once the symbolic manipulation using logarithm rules is complete, the next step is numerical evaluation. This involves computing the approximate value of variables using a calculator. In practice, after simplifying the logarithmic expressions, we often end up with a formula or a simplified expression for the variable of interest. In this exercise, the final expression for \( x \) is \( x = \frac{\ln(10)}{\ln(6) + \ln(10)} \). To find the numerical value of \( x \), input the natural logarithm values into your calculator and carry out the division. Ensure to round the result to four decimal places, giving \( x \approx 0.8687 \). This step is crucial to providing a final, practical answer to the problem.