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Let's break down the problem step by step:

### Given:
- \( f(x) = e^{2x} \)
- \( g(x) = x + 1 \)

### Part (a): Calculate \( f^{-1}(3) \times g^{-1}(3) \).

#### 1. Find \( f^{-1}(x) \):
To find the inverse of \( f(x) = e^{2x} \), we express \( y = f(x) = e^{2x} \), and solve for \( x \) in terms of \( y \).

\[
y = e^{2x} \quad \Rightarrow \quad \ln(y) = 2x \quad \Rightarrow \quad x = \frac{1}{2} \ln(y)
\]

Thus, \( f^{-1}(x) = \frac{1}{2} \ln(x) \).

#### 2. Find \( f^{-1}(3) \):
Using the inverse function:
\[
f^{-1}(3) = \frac{1}{2} \ln(3)
\]

#### 3. Find \( g^{-1}(x) \):
For \( g(x) = x + 1 \), the inverse is straightforward. Set \( y = g(x) = x + 1 \), and solve for \( x \):

\[
x = y - 1
\]

Thus, \( g^{-1}(x) = x - 1 \).

#### 4. Find \( g^{-1}(3) \):
\[
g^{-1}(3) = 3 - 1 = 2
\]

#### 5. Multiply \( f^{-1}(3) \times g^{-1}(3) \):
Now, multiply the two values:
\[
f^{-1}(3) \times g^{-1}(3) = \frac{1}{2} \ln(3) \times 2 = \ln(3)
\]

So, the answer to part (a) is \( \ln(3) \).

---

### Part (b): Show that \( (f \circ g)^{-1}(3) = \ln(\sqrt{3}) - 1 \).

First, we need to calculate the composition \( f \circ g \), find its inverse, and then evaluate it at 3.

#### 1. Find \( f \circ g \):
The composition \( (f \circ g)(x) \) means \( f(g(x)) \), so:
\[
f(g(x)) = f(x + 1) = e^{2(x+1)} = e^{2x+2} = e^2 \cdot e^{2x}
\]

Thus, \( f \circ g(x) = e^2 \cdot e^{2x} \).

#### 2. Find the inverse of \( f \circ g(x) \):
Set \( y = f \circ g(x) = e^2 \cdot e^{2x} \), and solve for \( x \):

\[
y = e^2 \cdot e^{2x} \quad \Rightarrow \quad \frac{y}{e^2} = e^{2x} \quad \Rightarrow \quad \ln\left(\frac{y}{e^2}\right) = 2x
\]

\[
\ln(y) - 2 = 2x \quad \Rightarrow \quad x = \frac{1}{2} \ln(y) - 1
\]

Thus, \( (f \circ g)^{-1}(x) = \frac{1}{2} \ln(x) - 1 \).

#### 3. Evaluate \( (f \circ g)^{-1}(3) \):
Now substitute \( x = 3 \) into the inverse function:
\[
(f \circ g)^{-1}(3) = \frac{1}{2} \ln(3) - 1
\]

Notice that \( \frac{1}{2} \ln(3) \) is the same as \( \ln(\sqrt{3}) \), so:

\[
(f \circ g)^{-1}(3) = \ln(\sqrt{3}) - 1
\]

This proves the desired result.

---

### Final Answers:
- **Part (a):** \( \ln(3) \)
- **Part (b):** \( \ln(\sqrt{3}) - 1 \)

Would you like further details or clarifications on any part of the solution?

Here are 5 related questions:
1. How do you generally find the inverse of an exponential function?
2. What are the steps for composing two functions?
3. Can you explain why \( f^{-1}(x) \times g^{-1}(x) \) worked out this way?
4. How would you find the inverse of a more complex composite function?
5. What is the significance of natural logarithms in inverse calculations?

**Tip:** Always check your composition when dealing with inverse functions to ensure that the result makes sense when plugging back into the original!

The functions f and g are defined by: f(x) = e^(2x), g(x) = x + 1. (a) Calculate f^(-1)(3) × g^(-1)(3). (b) Show that (f ∘ g)^(-1)(3) = ln(√3) - 1.

This problem involves finding the inverse of exponential and linear functions, calculating their products, and showing the inverse of their composition. It covers key concepts such as function inverses, composition of functions, and logarithmic calculations. Suitable for advanced high school students (Grades 11-12).

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