Answer: (a) \(a=\frac43\); (b) \(f(x)\in\mathbb R\); (d)(i) \(g(g(x))=4x-9\); (d)(ii) \(x=\frac{e^2+31}{12}\).

Use the laws of logarithms and exponentials first, then check that any logarithm arguments are positive where required.

(a) For \(f(x)=2\ln(3x-4)\), the logarithm requires

\(3x-4\gt0.\)

So

\(x\gt\frac43.\)

Hence the least possible value of \(a\) is

\(a=\frac43.\)

(b) As \(x\) runs over \(x\gt \frac43\), the expression \(3x-4\) runs over all positive values. Therefore \(\ln(3x-4)\) can take any real value.

So the range of \(f\) is

\(f(x)\in\mathbb R.\)

(c) The graph \(y=f(x)\) has vertical asymptote

\(x=\frac43.\)

Its \(x\)-intercept is found from

\(2\ln(3x-4)=0.\)

Thus \(3x-4=1\), so

\(x=\frac53.\)

So \(y=f(x)\) meets the \(x\)-axis at \(\left(\frac53,0\right)\).

The inverse function is

\(y=f^{-1}(x)=\frac{e^{x/2}+4}{3}.\)

This graph has horizontal asymptote

\(y=\frac43.\)

Its \(y\)-intercept is

\(y=\frac{1+4}{3}=\frac53.\)

So \(y=f^{-1}(x)\) meets the \(y\)-axis at \(\left(0,\frac53\right)\). The two graphs are reflections in the line \(y=x\).

(d)(i) Since \(g(x)=2x-3\),

\(g(g(x))=g(2x-3)=2(2x-3)-3=4x-9.\)

(d)(ii) Solve

\(f(g(g(x)))=4.\)

Using \(g(g(x))=4x-9\),

\(f(4x-9)=2\ln(3(4x-9)-4).\)

So

\(2\ln(12x-31)=4.\)

Hence

\(\ln(12x-31)=2.\)

Therefore

\(12x-31=e^2,\)

and

\(x=\frac{e^2+31}{12}.\)