Problem 38 - BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Question

\(y=\ln 3 x-6\)

Step-by-Step Solution

Answer

The solution to the equation \(y=\ln 3 x-6\) is \(x = \frac{e^{(y + 6)}}{3}\)

Step 1: Isolate the Logarithmic Term

First, let's isolate the logarithmic term by sending the -6 on the left side of the equation to the other side, by doing basic algebra. Therefore, we have the following equation: \(y + 6 = \ln 3x\)

Step 2: Convert to Exponential Form

Next, we have to convert the logarithm to its equivalent exponential form. Recall that if \(y = \ln x\), then \(e^y = x\). Here, we get: \(e^{(y + 6)} = 3x\)

Step 3: Solve for x

Now let's isolate x. To do this, divide each side of the equation by 3. That gives us our final solution: \(x = \frac{e^{(y + 6)}}{3}\).

Key Concepts

Convert Logarithm to Exponential Form

Understanding the relationship between logarithmic and exponential forms is foundational to solving logarithmic equations. The logarithm, denoted as \( \ln x \) when referring to the natural logarithm, represents the power to which a given base (in the case of the natural logarithm, the base is \( e \)) must be raised to obtain the number \( x \).

For instance, converting the logarithmic equation \( y = \ln x \) to its exponential counterpart involves raising the base \( e \) to the power of \( y \) to get \( x \) as the result. Thus, \( e^y = x \) symbolizes the same relationship. This conversion from logarithmic to exponential form is a vital skill and enables students to move on to the next steps in solving logarithmic equations.

Isolating the Variable

Solving any algebraic equation typically involves isolating the variable of interest on one side of the equation. When working with logarithmic equations, this step is crucial for clarity and simplifies the process of conversion to exponential form.

In our original exercise, \( y = \ln (3x) - 6 \), we intend to solve for \( x \) by first isolating the logarithmic term. We do so by adding 6 to both sides of the equation, resulting in \( y + 6 = \ln (3x) \). Now, the logarithm is on one side and everything else is on the other side, neatly setting the stage for conversion to exponential form. It's a strategic move that streamlines the solving process and reduces potential errors.

Solving Exponential Equations

Once we've converted the logarithmic equation to its exponential form, we're ready to tackle solving an exponential equation. The goal is to find the value of the variable that satisfies the exponential relationship. Referring back to our problem, after converting to the exponential form, we found \( e^{(y + 6)} = 3x \).

The next step involves algebraically manipulating the equation to isolate \( x \). Since \( x \) is being multiplied by 3, we divide both sides of the equation by 3, yielding a final solution of \( x = \frac{e^{(y + 6)}}{3} \). This illustrates how an understanding of exponential rules simplifies the resolution of complex equations and allows students to find solutions with confidence.