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 = log₃x - 3

Explanation:

Step1: Recall log - rule

Use the property $\log_a a^k=k$.

Step2: Calculate $\log_3\frac{1}{9}$

Since $\frac{1}{9}=3^{-2}$, then $\log_3\frac{1}{9}=-2$.

Step3: Calculate $y$ for $x = \frac{1}{9}$

Substitute $\log_3\frac{1}{9}=-2$ into $y = \log_3x - 3$. So $y=-2 - 3=-5$.

Step4: Calculate $\log_3\frac{1}{3}$

Since $\frac{1}{3}=3^{-1}$, then $\log_3\frac{1}{3}=-1$.

Step5: Calculate $y$ for $x=\frac{1}{3}$

Substitute $\log_3\frac{1}{3}=-1$ into $y = \log_3x - 3$. So $y=-1 - 3=-4$.

Step6: Calculate $\log_31$

Since $1 = 3^0$, then $\log_31=0$.

Step7: Calculate $y$ for $x = 1$

Substitute $\log_31=0$ into $y=\log_3x - 3$. So $y=0 - 3=-3$.

Step8: Calculate $\log_33$

Since $3 = 3^1$, then $\log_33=1$.

Step9: Calculate $y$ for $x = 3$

Substitute $\log_33=1$ into $y=\log_3x - 3$. So $y=1 - 3=-2$.

Step10: Calculate $\log_39$

Since $9 = 3^2$, then $\log_39=2$.

Step11: Calculate $y$ for $x = 9$

Substitute $\log_39=2$ into $y=\log_3x - 3$. So $y=2 - 3=-1$.

$x$	$\log_3x$	$y=\log_3x - 3$
$\frac{1}{3}$	$-1$	$-4$
$1$	$0$	$-3$
$3$	$1$	$-2$
$9$	$2$	$-1$

For the domain of $y = \log_3x-3$, the argument of the logarithm must be positive. So the domain is $(0,\infty)$.
For the range, since the range of the basic logarithmic function $y=\log_3x$ is $(-\infty,\infty)$ and we are just shifting it down by 3 units, the range of $y=\log_3x - 3$ is also $(-\infty,\infty)$.

Answer:

$x$	$\log_3x$	$y=\log_3x - 3$
$\frac{1}{3}$	$-1$	$-4$
$1$	$0$	$-3$
$3$	$1$	$-2$
$9$	$2$	$-1$

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