Find the exact solutions to the equations (a) ln(3x - 7) = 5 (b) 3^y e^{2y} = 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 2 - 2009 - Paper 2

First, we apply the natural logarithm to both sides:

$ext{ln}(3^y e^{2y}) = ext{ln}(15)$

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

$ext{ln}(3^y) + ext{ln}(e^{2y}) = ext{ln}(15)$

This can be expressed as:

$y ext{ln}(3) + 2y = ext{ln}(15)$

Combining the terms yields:

$y ( ext{ln}(3) + 2) = ext{ln}(15)$

Thus,

$y = \frac{ ext{ln}(15)}{ ext{ln}(3) + 2}$

The exact solution is:

$y \approx -2 + ext{ln}(15)$.

To find the inverse of the function $f(x) = e^x + 3$, we start by setting $y = e^x + 3$. Rearranging gives:

$y - 3 = e^x$

Taking the natural logarithm:

$x = ext{ln}(y - 3)$

Hence, the inverse function is:

$f^{-1}(y) = ext{ln}(y - 3)$.

For the domain of $f^{-1}$, since $y - 3 > 0$, we have:

$y > 3$.

To find the composition of the functions $f$ and $g$, we evaluate:

$fg(x) = f(g(x)) = f( ext{ln}(x - 1))$

Substituting into $f$ gives:

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

Next, we consider the range of $fg(x)$. Since $g(x)$ is defined for $x > 1$, the output of $fg(x)$ is:

ightarrow y > 3.$$ Thus, the range is: $$y > 3$$.