Problem 22 - Algebra and Trigonometry

Question

Find the solution of the exponential equation, rounded to four decimal places. \(10^{1-x}=6^{x}\)

Step-by-Step Solution

Answer

The solution of the equation is approximately \(x \approx 0.5623\).

Step 1: Take the logarithm of both sides

To solve the equation \(10^{1-x} = 6^x\), take the logarithm of both sides. Using the property \( \log(a^b) = b\log(a) \), we get: \(\log(10^{1-x}) = \log(6^x)\).

Step 2: Apply the power rule for logarithms

Using the property \( \log(a^b) = b\log(a) \), rewrite the equation: \((1-x)\log(10) = x\log(6)\).

Step 3: Simplify the equation

Distribute the logarithms: \(\log(10) - x\log(10) = x\log(6)\). Rearrange the terms to isolate \(x\): \(\log(10) = x\log(6) + x\log(10)\).

Step 4: Factor out x

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

Step 5: Solve for x

Divide both sides by \((\log(6) + \log(10))\) to find \(x\): \[ x = \frac{\log(10)}{\log(6) + \log(10)} \].

Step 6: Calculate the value of x

Using a calculator, find the logarithm values and compute: \(\log(10) \approx 1\), \(\log(6) \approx 0.7782\). Plug these into the equation: \[ x \approx \frac{1}{0.7782 + 1} \approx 0.5623 \].

Key Concepts

Logarithms

Logarithms are fundamental mathematical tools that help us understand and solve various types of equations, especially exponential ones. In simple terms, a logarithm tells you the power to which a number must be raised to obtain another number. For example, in the equation \(10^{1-x} = 6^x\), taking logarithms on both sides helps us break down the equation into more manageable parts.

• The relationship is defined as \( \log_b(a) = c \), which means \( b^c = a \).
• Logarithms turn multiplication into addition, which simplifies handling exponential terms.

By applying logarithms to both sides, we can use properties like the power rule: \(\log(a^b) = b\log(a)\). This helps in separating the variables and constants, ultimately aiding in solving the equation.

Properties of Exponents

In solving exponential equations, understanding the properties of exponents is key. Exponents indicate how many times a number (the base) is multiplied by itself. For instance, \(10^{1-x}\) represents 10 raised to the power of \(1-x\), and \(6^x\) encapsulates 6 raised to the power x.

• Multiplying Exponents: \((a^b) \times (a^c) = a^{b+c}\)
• Dividing Exponents: \((a^b) \div (a^c) = a^{b-c}\)
• Power Rule: \((a^b)^c = a^{bc}\)

When encountering the given equation, we leverage these rules to manipulate terms. Specifically, by taking the logarithm, we translate these exponential terms into linear terms, making it easier to isolate variables and constants.

Solving Equations

Solving equations, especially exponential ones, involves a structured approach. The central task is to isolate the variable to find its value. With equations like \(10^{1-x} = 6^x\), follow these general steps:

• First, take the logarithm of both sides to transform the equation: \(\log(10^{1-x}) = \log(6^x)\).
• Apply properties of logarithms and exponents, such as the power rule, to simplify: \((1-x)\log(10) = x\log(6)\).
• Re-organize terms to isolate the variable: \(\log(10) = x\log(6) + x\log(10)\).
• Factor out the common variable, leading to a solvable form: \(\log(10) = x(\log(6) + \log(10))\).
• Divide to solve for the variable: \(x = \frac{\log(10)}{\log(6) + \log(10)}\).

Each step reduces complexity, enhancing understanding of the problem's structure and solution path.

Mathematical Problem Solving

Mathematical problem solving is both an art and a science, combining systematic methods with creative thinking. It requires patience and practice. Here's a general strategy using the context of solving \(10^{1-x} = 6^x\):

• Understand the Problem: Recognize that it is an exponential equation, and identify knowns and unknowns.
• Devise a Plan: Decide to use logarithms and properties of exponents to simplify the equation.
• Execute the Plan: Follow through the sequence of algebraic manipulations.
• Check Your Work: Substitute back calculated values to verify accuracy.

By integrating concept understanding with a solid methodology, you can approach such problems confidently. Problem solving ultimately enhances logical thinking and mathematical fluency.