Answer: (a) \((\ln(3x+2))^2+1\); (b) \(x=\frac{e^2-2}{3}\); (c) \(x=-0.243\) approximately.

Answer: (a) \((\ln(3x+2))^2+1\); (b) \(x=\frac{e^2-2}{3}\); (c) \(x=-0.243\) approximately.

(a) Since \(f(x)=x^2+1\) and \(g(x)=\ln(3x+2)\),

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

Therefore

\(fg(x)=(\ln(3x+2))^2+1.\)

(b) Solve

\((\ln(3x+2))^2+1=5.\)

Then

\((\ln(3x+2))^2=4,\)

so

\(\ln(3x+2)=\pm2.\)

Because the composite function \(fg(x)\) requires \(g(x)\) to be in the domain of \(f\), we need \(g(x)\gt -\frac23\). The value \(\ln(3x+2)=-2\) is not allowed.

Hence

\(\ln(3x+2)=2.\)

So

\(3x+2=e^2, \qquad x=\frac{e^2-2}{3}.\)

(c) Now

\(gg(x)=g(g(x))=\ln(3\ln(3x+2)+2).\)

Set this equal to \(1\):

\(\ln(3\ln(3x+2)+2)=1.\)

Therefore

\(3\ln(3x+2)+2=e.\)

So

\(\ln(3x+2)=\frac{e-2}{3}.\)

Exponentiating gives

\(3x+2=e^{(e-2)/3}.\)

Hence

\(x=\frac{e^{(e-2)/3}-2}{3}.\)

Numerically,

\(x=-0.243\)

to 3 significant figures.