Problem 51 - Precalculus with Limits

Question

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. \(2-6 \ln x=10\)

Step-by-Step Solution

Answer

The solution to the logarithmic equation is \(x \approx 0.263\).

Step 1: Isolate the logarithmic term

First step is to isolate the term with the logarithm. In this case, add 6 to both sides of the equation \(2-6 \ln x = 10\) to isolate \(-6 \ln x\) on one side. This gives \(-6 \ln x = 10-2 = 8\).

Step 2: Divide by coefficient of logarithm

Next, divide both sides by -6 to remove the coefficients of the logarithm. This gives us \(\ln x = -\frac{8}{6} = -\frac{4}{3}\).

Step 3: Apply inverse of logarithm function

The inverse of the logarithm function is the exponential function. By applying the inverse, we get \(x = e^{-4/3}\).

Step 4: Evaluate the expression

Lastly, evaluate \(e^{-4/3}\) using a calculator to obtain \(x = e^{-4/3} \approx 0.263\).

Step 5: Check the solution

Substitute the solution \(x \approx 0.263\) back into the original equation to confirm. Since \(2-6 \ln (0.263) \approx 10\), the solution is correct.

Key Concepts

Isolating the Logarithmic Term

When you're faced with a logarithmic equation like the one in the exercise, your first goal is to isolate the term that contains the logarithm. This is a crucial step because it sets the stage for solving the equation. Think of it like peeling an onion, where your aim is to get to the core of the problem.

In the provided equation, we have something like this: \( 2 - 6 \ln x = 10 \). The term we want to isolate is \( -6 \ln x \).

Here's a simple trick to do so:

• Look at the equation and identify the term with the logarithm. Here, that's \( \ln x \).
• Remove or "move" everything else over to the other side. For our equation, you add \(6\) to both sides: \( 2 - 6 \ln x + 6 = 10 + 6\).
• Simplify to get \( -6 \ln x = 8 \).

By isolating the logarithmic term, you're making it easier to perform operations that involve only the logarithm and its corresponding variable.

Exponential Function

Once we've isolated the logarithmic term, the next step is to eliminate the logarithm itself. This is where the exponential function comes into play. The exponential function is essentially the opposite or inverse of a logarithm.

You might wonder how this works. Mathematically, if you have \( \ln x = a \), then applying the exponential function "undoes" the logarithm. In mathematical terms, it changes to \( x = e^a \).

For our exercise:

• We have \( \ln x = -\frac{4}{3} \). By applying the exponential function, you convert it to \( x = e^{-4/3} \).
• This transformation is powerful because it allows us to solve for \( x \) directly.

Always remember, to "cancel" a logarithm, just think exponentially! This concept is key to solving many equations involving logs.

Solve Algebraically

Now that we've managed the logarithms with the help of exponential functions, let's talk about solving the equation algebraically, which essentially involves evaluating expressions and checking solutions.

After everything is neatly set up like in our equation \( x = e^{-4/3} \), you move on to find a numerical result. Using a calculator or technological tools:

• Calculate \( e^{-4/3} \) to approximate \( x \). In this problem, you get \( x \approx 0.263 \).

Once you have this value, it's imperative to check your work. Substitute back into the original equation:

• Verify that \( 2 - 6 \ln(0.263) \approx 10 \). If it holds true, your solution is confirmed.

This careful approach ensures accuracy in your solution and builds confidence in solving logarithmic equations algebraically. By understanding each step, you're better prepared to tackle similar problems.