Problem 22 - Precalculus Mathematics for Calculus

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 is approximately 0.5620.

Step 1: Take Logarithm on Both Sides

To solve the equation \(10^{1-x} = 6^x\), we first take the natural logarithm on both sides to simplify the equation. This gives us: \[ \ln(10^{1-x}) = \ln(6^x) \] Applying the logarithm power rule \(\ln(a^b) = b \ln(a)\), the equation becomes: \[ (1-x) \ln(10) = x \ln(6) \]

Step 2: Distribute and Rearrange

Next, we distribute the logarithms to get rid of the parentheses: \[ \ln(10) - x \ln(10) = x \ln(6) \] Rearrange terms to bring all terms involving \(x\) to one side: \[ \ln(10) = x \ln(6) + x \ln(10) \] Factor out \(x\) from the right-hand side: \[ \ln(10) = x(\ln(6) + \ln(10)) \]

Step 3: Solve for x

Isolate \(x\) by dividing both sides of the equation by \(\ln(6) + \ln(10)\): \[ x = \frac{\ln(10)}{\ln(6) + \ln(10)} \]

Step 4: Compute the Numerical Answer

Calculate the numerical value using a calculator:First, find \(\ln(10)\) and \(\ln(6)\): \[ \ln(10) \approx 2.3026 \] \[ \ln(6) \approx 1.7918 \] Substitute back: \[ x = \frac{2.3026}{1.7918 + 2.3026} \approx 0.5620 \] Thus, the solution rounded to four decimal places is approximately \(x = 0.5620\).

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a fundamental concept in mathematics, especially when dealing with exponential equations. It operates with the base \( e \), which is an irrational constant approximately equal to 2.71828. The natural logarithm provides a way to transform exponential relationships into linear ones, making them easier to manipulate and solve.

In the context of solving exponential equations, such as \( 10^{1-x} = 6^x \), taking the natural logarithm of both sides helps us convert the exponential forms into a form we can handle algebraically. This is a powerful tool because it simplifies the equation significantly. By applying the logarithm power rule, \( \ln(a^b) = b \ln(a) \), we can transform our complex exponential terms into linear terms like \( (1-x) \ln(10) \) and \( x \ln(6) \).

This transformation is essential for making further algebraic manipulations possible, setting the stage to eventually solve for the variable \( x \).

Logarithmic Properties

Logarithmic properties are key to simplifying and solving equations with logarithms. One crucial property is the power rule: \( \ln(a^b) = b \ln(a) \). This rule allows us to take an exponent out of a logarithm, turning it into a multiplier, as seen in our example equation \( (1-x) \ln(10) = x \ln(6) \).

Another useful property is the product rule: \( \ln(ab) = \ln(a) + \ln(b) \). This is often applied when you need to combine or separate terms involving products inside a logarithm, although it is not directly used in this specific example. Additionally, the quotient rule: \( \ln\left( \frac{a}{b} \right) = \ln(a) - \ln(b) \), can be useful when working with divisions inside a logarithmic function.

In algebraic manipulation, these properties help in rearranging and simplifying complex expressions, so the variable of interest can be isolated and evaluated. In solving for \( x \), recognizing and applying these logarithmic properties is crucial to simplify the equation to a solvable form.

Rounding Numbers

Rounding numbers is a common practice when dealing with solutions that require decimal approximations. It involves adjusting a number to a specified level of precision, often to make it easier to read or work with.

In our step-by-step solution, when calculating \( x = \frac{\ln(10)}{\ln(6) + \ln(10)} \), we arrived at a numeric expression that is not an integer. By computing, we found \( x \approx 0.5619863449 \). Rounding this to four decimal places, as instructed, gives us \( x \approx 0.5620 \).

• To round to four decimal places, consider the fifth decimal digit. If this digit is 5 or greater, increase the fourth decimal digit by one.
• If it's less than 5, leave the fourth decimal digit as it is.

This rounding rule helps ensure consistency and provides an easier-to-use result when an exact number is not needed.