9. (i) Find the exact solutions to the equations (a) ln(3x – 7) = 5 (b) 3^2 e^{2 + x} = 15 (ii) The functions f and g are defined by f(x) = e^x + 3, x ∈ ℝ g(x) = ln(x – 1), x ∈ ℝ, x > 1 (a) Find f^{-1} and state its domain - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 2

First, we rewrite the equation to isolate the exponential term:

$9 e^{2 + x} = 15$ Now, divide both sides by 9:

$e^{2 + x} = \frac{15}{9} = \frac{5}{3}$ Next, we take the natural logarithm of both sides:

$2 + x = ext{ln}(\frac{5}{3})$ To isolate $x$, we rearrange:

$x = ext{ln}(\frac{5}{3}) - 2$ The exact solution for $x$ is:

$x \approx -0.0874$

To find the inverse function $f^{-1}(x)$ of the function defined as:

$f(x) = e^x + 3$ We start by replacing $f(x)$ with $y$:

$y = e^x + 3$ To find the inverse, we exchange $x$ and $y$ and solve for $y$:

$x = e^y + 3$ Now we isolate $e^y$:

$e^y = x - 3$ Next, we take the natural logarithm of both sides:

$y = ext{ln}(x - 3)$ Therefore, the inverse function is:

$f^{-1}(x) = ext{ln}(x - 3)$ The domain of $f^{-1}$ is the range of $f$, which is:

$x > 3$

To find the composite function $fg(x)$, we substitute $g(x)$ into $f(x)$:

$g(x) = ext{ln}(x - 1)$ So, $fg(x) = f(g(x)) = f( ext{ln}(x - 1))$ Now substituting into $f(x)$ gives us:

$fg(x) = e^{ ext{ln}(x - 1)} + 3$ This simplifies to:

$fg(x) = (x - 1) + 3 = x + 2$ The range of $fg(x)$ is therefore:

$x + 2 \text{ where } x \in ℝ$ Thus, the range is:

$y > 2$