The problem from the image can be broken down as follows:

Problem 12

We are given the function: $f(x) = \ln(x-1) + \ln 3, \text{ for } x > 1.$

Part (a)

• Find $f(x)$.
• Let $g(x) = e^x \text{ for } x \in \mathbb{R}$.

Part (b)

• Find $g \circ f(x)$, giving your answer in the form $ax + b$, where $a, b \in \mathbb{Z}$.
• Find the domain and range of $g \circ f$.

Step-by-Step Solution

Part (a)

The given function is: $f(x) = \ln(x-1) + \ln 3.$ Using the properties of logarithms, particularly the product rule $\ln a + \ln b = \ln(ab)$, we can rewrite this as: $f(x) = \ln(3(x-1)).$ Thus, $f(x) = \ln(3(x-1))$.

Part (b)

The function $g(x) = e^x$. Now, to find $g \circ f(x)$, we calculate: $g(f(x)) = g(\ln(3(x-1))) = e^{\ln(3(x-1))}.$ Since $e^{\ln y} = y$, this simplifies to: $g(f(x)) = 3(x-1).$ Expanding this, we get: $g(f(x)) = 3x - 3.$ Thus, $g(f(x)) = 3x - 3$, which is in the form $ax + b$ where $a = 3$ and $b = -3$.

Domain and Range of $g \circ f$

• The domain of $f(x) = \ln(3(x-1))$ requires $x > 1$, so the domain of $g \circ f(x)$ is $x > 1$.
• The range of $g(f(x)) = 3x - 3$ is all real numbers since it is a linear function, which can take any real value as $x$ increases.

Final Answers:

• $f(x) = \ln(3(x-1))$
• $g(f(x)) = 3x - 3$, with $a = 3$ and $b = -3$
• The domain of $g \circ f$ is $x > 1$, and the range is $\mathbb{R}$.

Let me know if you want further clarifications or details. Here are 5 related questions for further practice:

1. Can you simplify the expression $\ln(a) + \ln(b) - \ln(c)$?
2. How do you find the inverse of a function like $f(x) = \ln(3x - 2)$?
3. What is the derivative of $f(x) = \ln(2x + 5)$?
4. If $g(x) = e^x$, what is the derivative of $g(f(x))$ where $f(x) = \ln(x-1)$?
5. How do you find the domain of $f(x) = \ln(x^2 - 4)$?

Tip: Always remember the logarithmic identities like the product rule, quotient rule, and power rule when simplifying or transforming logarithmic functions.