IB Mathematics AHL 1.9 Laws of logarithms AI HL Paper 1- Exam Style Questions- New Syllabus

(a)
Using the laws of logarithms:
$\ln(2-x) + \ln(-2-x) = \ln((2-x)(-2-x)) = \ln(x^2 – 4)$.
Apply the power law: $-2\ln(-x) = \ln((-x)^{-2}) = \ln\left(\frac{1}{x^2}\right)$.
Combine the terms: $f(x) = \ln\left( \frac{x^2 – 4}{x^2} \right)$.
Therefore, $g(x) = \frac{x^2 – 4}{x^2}$ (or $1 – \frac{4}{x^2}$).
$\boxed{g(x) = \frac{x^2-4}{x^2}}$

(b)
The range of the inverse function $f^{-1}$ is equal to the domain of the original function $f$.
Given the domain of $f$ is $x < -2$:
$\boxed{f^{-1}(x) < -2}$ (or $(-\infty, -2)$)

(c)
Set $y = \ln\left( \frac{x^2 – 4}{x^2} \right)$. To find the inverse, swap the variables $x$ and $y$:
$x = \ln\left( \frac{y^2 – 4}{y^2} \right)$
Exponentiate both sides: $e^x = \frac{y^2 – 4}{y^2} = 1 – \frac{4}{y^2}$.
Rearrange to solve for $y$:
$\frac{4}{y^2} = 1 – e^x$
$y^2 = \frac{4}{1 – e^x}$
$y = \pm \frac{2}{\sqrt{1 – e^x}}$
Since the range from part (b) requires $y < -2$, we take the negative root:
$y = -\frac{2}{\sqrt{1 – e^x}}$.
Comparing this to the form $\frac{a}{\sqrt{1 – e^x}}$ gives $a = -2$.
$\boxed{-2}$