Answer
(a) $(-\frac{1}{2},\infty)$.
(b) $2$, $(40,2)$.
(c) $121$, $(121,3)$.
(d) $ 4$.
Work Step by Step
Given $G(x)=log_3(2x+1)-2$, we have:
(a) the domain $2x+1\gt0$ or $(-\frac{1}{2},\infty)$.
(b) $G(40)=log_3(2(40)+1)-2=2$ giving point $(40,2)$.
(c) $G(x)=log_3(2x+1)-2=3 \Longrightarrow 2x+1=3^5 \Longrightarrow x=121$ giving point $(121,3)$.
(d) $G(x)=log_3(2x+1)-2=0 \Longrightarrow 2x+1=3^2 \Longrightarrow x=4$.
