Begin by graphing f(x)=log_{2}x Then use transformations of this graph to graph the given function. What is the vertical asymptote? Use the graphs to determine each function s domain and range. g(x)=frac{1}{2}log_{2}x

Step 1
For the following Logarithmic function defined by
${\displaystyle f\left(x\right)={\log }_{2}x}$
Graphing Logarithmic function given in equation requires setting up table of coordinates, so that
$\begin{array}{|ccc|}\hline & \text{Table of coordinates of the Logarithmic function}f(x)={\log }_{2}x& \\ x& f(x)=y={\log }_{2}x& (x,y)\\ 1& f(1)=y={\log }_2(1)\Rightarrow 2^y=1=2^0\Rightarrow y=0& (1,0)\\ 2& f(2)=y={\log }_2(2)\Rightarrow 2^y=2=2^1\Rightarrow y=1& (2,1)\\ 4& f(4)=y={\log }_2(4)\Rightarrow 2^y=4=2^2\Rightarrow y=2& (4,2)\\ 8& f(8)=y={\log }_2(8)\Rightarrow 2^y=8=2^3\Rightarrow y=3& (8,3)\\ 16& f(16)=y={\log }_2(16)\Rightarrow 2^y=16=2^4\Rightarrow y=4& (16,4)\\ 32& f(32)=y={\log }_2(32)\Rightarrow 2^y=32=2^5\Rightarrow y=5& (32,5)\\ 64& f(64)=y={\log }_2(64)\Rightarrow 2^y=64=2^6\Rightarrow y=6& (64,6)\\ \hline\end{array}$
Step 2
We plot the following points between ${\displaystyle (x,y)}$ determined from the table of coordinates and connect them with the continuous curve which represent the Logarithmic function ${\displaystyle f\left(x\right)={\log }_{2}x}$ as shown in Figure (1)
Figure (1) represent the graph of Logarithmic function ${\displaystyle f\left(x\right)={\log }_{2}x}$

Step 3
Note that:
The y-axis which represented by the equation ${\displaystyle x=0}$ is the vertical asymptote, so that the curve approches, but never touches the positive portion of the y-axis as shown in figure (1).
The domain of ${\displaystyle f\left(x\right)={\log }_2\left(x\right)}$ is all positive real numbers ${\displaystyle x\in \left(0\mathrm{\infty }\right)}$ and the range is all real numbers ${\displaystyle y\in (-\mathrm{\infty },\mathrm{\infty })}$
Step 4
To graph the Logarithmic function ${\displaystyle g\left(x\right)=\frac{1}{2}{\log }_2\left(x\right)}$, we vertically shri
the graph of the function ${\displaystyle f\left(x\right)={\log }_2\left(x\right)}$ because ${\displaystyle 0<\frac{1}{2}<1(f\left(x\right)={\log }_2\left(x\right)}$ Vertically shri
to ${\displaystyle g\left(x\right)=\frac{1}{2}{\log }_2\left(x\right))}$, then the Logarithmic function