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. $y = \log_{4}x - 3$

Explanation:

Step1: Recall log - rule

Use the rule $\log_a b = c$ means $a^c=b$. For $x = \frac{1}{16}$, we want to find $\log_4\frac{1}{16}$. Let $\log_4\frac{1}{16}=y$, then $4^y=\frac{1}{16}=4^{- 2}$, so $\log_4\frac{1}{16}=-2$.

Step2: Calculate $y$ - value

Substitute $\log_4\frac{1}{16}=-2$ into $y = \log_4x - 3$. Then $y=-2 - 3=-5$.

Step3: Repeat for other $x$ - values

For $x=\frac{1}{4}$, let $\log_4\frac{1}{4}=y$, then $4^y=\frac{1}{4}=4^{-1}$, so $\log_4\frac{1}{4}=-1$ and $y=-1 - 3=-4$.
For $x = 1$, $\log_41 = 0$ (since $4^0 = 1$), and $y=0 - 3=-3$.
For $x = 4$, $\log_44 = 1$ (since $4^1 = 4$), and $y=1 - 3=-2$.
For $x = 16$, $\log_416 = 2$ (since $4^2 = 16$), and $y=2 - 3=-1$.

$x$	$\log_4x$	$y=\log_4x - 3$
$\frac{1}{4}$	$-1$	$-4$
$1$	$0$	$-3$
$4$	$1$	$-2$
$16$	$2$	$-1$

The domain of the function $y=\log_4x - 3$ is $(0,\infty)$ because the argument of the logarithm function $x>0$. The range of the function is $(-\infty,\infty)$ since the range of the basic logarithmic function $y = \log_4x$ is $(-\infty,\infty)$ and subtracting 3 does not change the range.

Answer:

$x$	$\log_4x$	$y=\log_4x - 3$
$\frac{1}{4}$	$-1$	$-4$
$1$	$0$	$-3$
$4$	$1$	$-2$
$16$	$2$	$-1$

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