Question

given the function below, fill in the table of values, use the table of values to graph the function, and then identify the functions domain and range. y = log₂x - 4 x | log₂x | y = log₂x - 4 1/4 | | 1/2 | | 1 | | 2 | | 4 | | 8 |

Explanation:

Step1: Recall log - rule

Use the rule $\log_a b=c$ means $a^c = b$. For $x=\frac{1}{4}$, we have $\log_2\frac{1}{4}=\log_22^{-2}=-2$.

Step2: Calculate $y$ - value

Substitute $\log_2\frac{1}{4}=-2$ into $y = \log_2x - 4$. Then $y=-2 - 4=-6$.

$x$	$\log_2x$	$y=\log_2x - 4$
$\frac{1}{2}$	$\log_2\frac{1}{2}=\log_22^{-1}=-1$	$-1 - 4=-5$
$1$	$\log_21 = 0$	$0 - 4=-4$
$2$	$\log_22 = 1$	$1 - 4=-3$
$4$	$\log_24=\log_22^{2}=2$	$2 - 4=-2$
$8$	$\log_28=\log_22^{3}=3$	$3 - 4=-1$

The domain of the function $y = \log_2x-4$ is $(0,+\infty)$ since the argument of the logarithm function $x>0$. The range is $(-\infty,+\infty)$ because the range of the basic logarithmic function $y = \log_2x$ is $(-\infty,+\infty)$ and subtracting 4 from it does not change the range.

Answer:

$x$	$\log_2x$	$y=\log_2x - 4$
$\frac{1}{2}$	$-1$	$-5$
$1$	$0$	$-4$
$2$	$1$	$-3$
$4$	$2$	$-2$
$8$	$3$	$-1$

Domain: $(0,+\infty)$; Range: $(-\infty,+\infty)$