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

To prove this, we first differentiate g(x) using the quotient rule:

[ g'(x) = \frac{(\ln(x) - 2)(3 \cdot \frac{1}{x}) - (3 \ln(x) - 7)(\frac{1}{x})}{(\ln(x) - 2)^2} ]

Simplifying the numerator:

[ g'(x) = \frac{3(\ln(x) - 2) \cdot 1 - (3 \ln(x) - 7) \cdot 1}{x(\ln(x) - 2)^2} ]

This further simplifies to:

[ g'(x) = \frac{(7 - 2) - 3\ln(x) + 3 \cdot 2}{x(\ln(x) - 2)^2} = \frac{7 - 3\ln(x)}{x(\ln(x) - 2)^2} ]

Now, for g'(x) > 0, we require:

[ 7 - 3\ln(x) > 0 ]
[ \Rightarrow \ln(x) < \frac{7}{3} ]

As ( x > 0 ), this condition holds true for all values in the domain, confirming that g'(x) > 0 for all x in the domain of g.

To find the range of values of a such that g(a) > 0, we need to set up the inequality:

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

This implies that both the numerator and denominator must have the same sign. We solve:

1. Numerator:
   [ 3 \ln(a) - 7 > 0 ]
   [ \Rightarrow \ln(a) > \frac{7}{3} ]
   [ \Rightarrow a > e^{7/3} ]

2. Denominator:
   [ \ln(a) - 2 > 0 ]
   [ \Rightarrow \ln(a) > 2 ]
   [ \Rightarrow a > e^2 ]

Thus, the range of values of a for which g(a) > 0 is:

[ a > e^{\max(2, 7/3)} ]

Since ( 7/3 \approx 2.33 ), we have:

[ a > e^{7/3} ]