Problem 74 - College Algebra with Corequisite Support

Question

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth. \(\ln (3)+\ln (4.4 x+6.8)=2\)

Step-by-Step Solution

Answer

The solution is approximately \( x = -0.9857 \).

Step 1: Use the Property of Logarithms

The equation given is \( \ln(3) + \ln(4.4x + 6.8) = 2 \). We can use the property of logarithms that states \( \ln(a) + \ln(b) = \ln(a \cdot b) \). Therefore, the equation becomes \( \ln(3 \cdot (4.4x + 6.8)) = 2 \).

Step 2: Simplify the Logarithmic Expression

Multiply inside the logarithm: \( \ln(3 \cdot (4.4x + 6.8))\) becomes \( \ln(13.2x + 20.4) \). The equation is now \( \ln(13.2x + 20.4) = 2 \).

Step 3: Exponentiate Both Sides

To remove the logarithm, exponentiate both sides of the equation using base \( e \), giving \( 13.2x + 20.4 = e^2 \). The value of \( e^2 \) is approximately 7.3891.

Step 4: Solve for x

Now, solve the equation \( 13.2x + 20.4 = 7.3891 \). Subtract 20.4 from both sides to get \( 13.2x = 7.3891 - 20.4 = -13.0109 \). Then divide by 13.2: \( x = \frac{-13.0109}{13.2} \approx -0.9857 \).

Step 5: Round the Solution

Since we rounded each step to appropriate decimal places, check the result: \( x = -0.9857 \) is rounded to the nearest ten-thousandth.

Key Concepts

Logarithmic Equations

Logarithmic equations are equations that involve logarithms of unknown variables. They can look quite intimidating, but with the right approach, they become manageable. In our example, the equation is \( \ln(3) + \ln(4.4x + 6.8) = 2 \). The goal in solving logarithmic equations is typically to isolate the variable, making it easier to solve.

• Look for opportunities to use logarithmic identities to simplify the equation.
• After simplifying, often transforming the logarithmic equation into an exponential equation is the best next step.
• This transformation helps clear the log, revealing a path to solve for the unknown.

Understanding logarithmic equations and their properties allow us to tackle these seemingly complex problems systematically.

Properties of Logarithms

Logarithms come with their own set of properties that make them much easier to deal with. One of these properties is how to combine two logs into one. The equation given initially as \( \ln(3) + \ln(4.4x + 6.8) \) can be combined using the property \( \ln(a) + \ln(b) = \ln(a \cdot b) \), leading to \( \ln(3 \cdot (4.4x + 6.8)) \).

• Sum of logs becomes a product inside a single log.
• This makes the equation simpler and easier to handle.
• Understanding these properties allows quick simplification.

By using the properties of logarithms, we can simplify the expression, which is crucial for progressing to the next steps, such as exponentiation.

Exponentiation

Exponentiation is the opposite operation of logarithms. To solve for the variable, we can exponentiate both sides of the equation. This method manipulates the equation into a more solvable form. In our example, we took \( \ln(13.2x + 20.4) = 2 \) and transformed it into \( 13.2x + 20.4 = e^2 \).

• Exponentiation helps "remove" the logarithm from an equation.
• It does this by using the base of the logarithm (often \(e\) for natural logs).
• After this step, the variable is in a linear equation, which is easier to solve.

Effective use of exponentiation is key in transitioning from logarithmic to algebraic terms, paving the way to find the variable's value.

Rounding Numbers

Rounding numbers is a crucial step in maintaining precision in answers while making them easy to interpret. Throughout solving this logarithmic equation, you may need to round:

• When testing or calculating approximate values (e.g., for base \(e\) calculations).
• In the solution, to ensure the answer is rounded to the specified decimal, as in our example where the result was \(x \approx -0.9857\).
• Rounding helps in focusing on significant figures crucial for reporting results accurately.

Always be mindful of the level of precision needed, and ensure you round to the appropriate number of decimal places to accurately reflect this in your final solution.