The function g is defined by $$g(x)=\frac{3\ln(x)-7}{\ln(x)-2}, \quad x > 0, \quad x \neq k$$ where k is a constant - Edexcel - A-Level Maths Pure - Question 14 - 2020 - Paper 2

To prove that $g'(x) > 0$ for all values of x in the domain, we first differentiate g(x) using the quotient rule:

$g'(x) = \frac{(\ln(x) - 2)(3(\frac{1}{x}) - 0) - (3\ln(x) - 7)(\frac{1}{x})}{(\ln(x) - 2)^2}$

After simplifying, we find:

$g'(x) = \frac{(3 - 3\ln(x) + 7)(1/x)}{(\ln(x) - 2)^2}$

To show that $g'(x) > 0$, we need:

$3 - 3\ln(x) + 7 > 0$

which simplifies to:

$10 > 3\ln(x) \Rightarrow \ln(x) < \frac{10}{3}$

Since x > 0, the inequality holds for all x in the domain of g, thereby proving $g'(x) > 0$.

To find the range of values of a for which $g(a) > 0$, we first set up the inequality:

$\frac{3\ln(a) - 7}{\ln(a) - 2} > 0$

This implies that:

1. The numerator must be positive: $3\ln(a) - 7 > 0 \Rightarrow \ln(a) > \frac{7}{3} \Rightarrow a > e^{7/3}$

2. The denominator must be positive: $\ln(a) - 2 > 0 \Rightarrow \ln(a) > 2 \Rightarrow a > e^2$

Combining these inequalities gives:

$a > e^{7/3}$

Thus, the range of values is:

$a \in (e^{7/3}, \infty)$.