Find the exact solutions to the equations (a) ln(3x - 7) = 5 (b) 3^{2x+2} = 15 (ii) The functions f and g are defined by f(x) = e^{x} + 3, \, x \in \mathbb{R} g(x) = ln(x - 1), \, x \in \mathbb{R}, \, x > 1 (a) Find f^{-1} and state its domain - Edexcel - A-Level Maths Pure - Question 3 - 2010 - Paper 2

First, we can take the logarithm (base 3) of both sides:

$\log_3(3^{2x+2}) = \log_3(15)$

Using the properties of logarithms, we rewrite the left-hand side:

$2x + 2 = \log_3(15)$

Solving for x, we rearrange:

$2x = \log_3(15) - 2$ $x = \frac{\log_3(15) - 2}{2}$

Using the change of base formula:

$\log_3(15) = \frac{\ln(15)}{\ln(3)}$

The final result for x is:

$x = \frac{\frac{\ln(15)}{\ln(3)} - 2}{2}$

To find the inverse function f^{-1}(x), start with:

$y = f(x) = e^x + 3$

Swap x and y:

$x = e^y + 3$

Isolating y gives:

$e^y = x - 3$ $y = \ln(x - 3)$

Thus, the inverse function is:

$f^{-1}(x) = \ln(x - 3)$

The domain of f^{-1} requires that:

$x - 3 > 0 \\ \Rightarrow x > 3$

So, the domain is:

$x > 3$

To find the composite function fg, we have:

$fg(x) = f(g(x)) = f(\ln(x - 1))$

This becomes:

$f(\ln(x - 1)) = e^{\ln(x - 1)} + 3 = (x - 1) + 3 = x + 2$

Next, we determine the range of fg. The range of g(x) is:

$x > 1 \Rightarrow g(x) \in \mathbb{R}$

So the range of fg, which is the same as f's behavior as x increases, is:

$x + 2, \text{ which is } x + 2 \geq 3$

Therefore, the range is:

$y \geq 3$