Q. 111 - Precalculus Enhanced with Graphing Utilities

Question

Suppose that $G\left(x\right)={\log }_3(2x+1)-2$

(a) What is the domain of G?

(b) What is G(40)? What point is on the graph of G?

(c) If G(x)=3, what is x? What point is on the graph of G?

(d) What is the zero of G?

Step-by-Step Solution

Answer

(a) The domain of g is$\left(-\frac{1}{2},\infty \right)$.

(b) The value of g(40) is 2.

(c) The value of x is 121.

(d) The zero of g is 4.

Step 1. Given information.

The following function is given $G\left(x\right)={\log }_3(2x+1)-2$.

Step 2. The domain of G.

(a) Here, x+1=0

$x>-1 / 2$

So, the domain is (-1 / 2,0).

Step 3. The value of G(40)

$$\begin{gathered} \left(b\right) G\left(x\right)={\log }_3(2 x+1)-2 \\ G\left(40\right)={\log }_3(2* 40+1)-2 \\ =2 \end{gathered}$$

Step 4. The value of x if G(x)=3.

$$\begin{gathered} \left(c\right) {\log }_3(2 x+1)-2=3 \\ {\log }_3(2 x+1)=5 \\ 2 x+1 =3^5 \\ x=121 \end{gathered}$$

Step 5. The zero of G.

$$\begin{gathered} \left(d\right) {\log }_3(2 x+1)-2 =0 \\ {\log }_3(2 x+1) =2 \\ 2x+1 =3^{2 } \\ x =4 \end{gathered}$$