Question

Solve for \(y\) in terms of \(x\).\(\ln y=2 \ln x+\ln (x-3)\)

Short answer

\(y = x^3-3x^2\)

Step-by-step solution

Applying the Power Rule of Logarithms

The equation \(\ln y=2 \ln x+\ln (x-3)\) requires applying the power rule of logarithms for environment \(2 \ln x\), resulting in \( \ln y= \ln x^2 + \ln (x-3)\)

Applying the Product Rule of Logarithms

Combine the two terms on the right side of the equation using the product rule which states that the \(\ln a + \ln b = \ln (a*b)\). This gives us \(\ln y = \ln(x^2(x-3))\) or \(\ln y = \ln(x^3-3x^2)\)

Removing the Logarithm

Since the logarithm base on both sides is the same, you can remove the log and equate the arguments resulting in \(y = x^3-3x^2\)

Key concepts

Power Rule of Logarithms

The power rule of logarithms is a handy tool that simplifies expressions involving logarithms. It states that for any logarithm, when you have a constant multiplied by the logarithm of a number, you can move that constant to the exponent of the number inside the log. This is expressed as:

• \( n \ln a = \ln(a^n) \)

This rule helps to condense complicated logarithmic expressions and makes them easier to work with.
In the context of solving the given exercise, when we have the expression \(2 \ln x\), we utilize the power rule by recognizing that the 2 can be moved to the exponent of \(x\). This transforms the expression into \(\ln(x^2)\). This simplification is crucial for further deriving the solution and reduces the complexity of logarithmic equations, paving the way for using other rules like the product rule that follows.
Understanding the power rule is essential as it often is the first step in breaking down multi-element logarithmic expressions, thus streamlining the process of solving equations.

Product Rule of Logarithms

The product rule of logarithms is another fundamental principle that is very useful when working with equations involving logarithms. It has a simple premise: the logarithm of a product is the sum of the logarithms of the factors. Mathematically, this is represented as:

• \( \ln(a) + \ln(b) = \ln(ab) \)

This transformation allows you to combine multiple logarithmic terms into a single term, making the expression simpler to handle.
In the exercise, after using the power rule, we deal with \(\ln(x^2) + \ln(x-3)\). According to the product rule, we can combine this into one logarithmic term, \(\ln(x^2(x-3))\), simplifying it further to \(\ln(x^3-3x^2)\). This step is critical as it creates a direct relationship between the variables which aids in efficiently solving for \(y\). Understanding how to utilize the product rule is key to manipulating and solving logarithmic equations, making it an invaluable skill in algebra.

Exponential Functions

Exponential functions involve exponents and are often encountered when we're dealing with growth and decay phenomena, but they also play a crucial role in simplifying equations involving logarithms. In the context of the given problem, once we have simplified the logarithmic expression using both the power and product rules, we next need to remove the logarithm to solve for \(y\).

• The base of the natural logarithm \(\ln\) is \(e\), thus if \(\ln y = \ln f(x)\), then \(y = f(x)\).

This is achieved by equating the arguments of the logs, assuming both sides have the same base. This step leverages the understanding of exponential functions because the natural logarithm is the opposite operation of raising \(e\) to a power.
By isolating \(y\), as seen in this exercise, we find that \(y = x^3 - 3x^2\). This final expression is no longer logarithmic or exponential but is derived using our knowledge of these functions. Applying the concept of exponential functions here is key to interpreting and solving logarithmic equations effectively.