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 - 1

Explanation:

Step1: Recall log - rule for $x=\frac{1}{9}$

If $y = \log_{a}x$, then $a^{y}=x$. For $y=\log_{3}\frac{1}{9}$, we know $3^{y}=\frac{1}{9}=3^{- 2}$, so $\log_{3}\frac{1}{9}=-2$.

Step2: Calculate $y$ for $x = \frac{1}{9}$

Substitute $\log_{3}\frac{1}{9}=-2$ into $y=\log_{3}x - 1$. Then $y=-2 - 1=-3$.

Step3: Recall log - rule for $x=\frac{1}{3}$

For $y = \log_{3}\frac{1}{3}$, since $3^{y}=\frac{1}{3}=3^{-1}$, so $\log_{3}\frac{1}{3}=-1$.

Step4: Calculate $y$ for $x=\frac{1}{3}$

Substitute $\log_{3}\frac{1}{3}=-1$ into $y=\log_{3}x - 1$. Then $y=-1 - 1=-2$.

Step5: Recall log - rule for $x = 1$

For $y=\log_{3}1$, since $3^{y}=1 = 3^{0}$, so $\log_{3}1=0$.

Step6: Calculate $y$ for $x = 1$

Substitute $\log_{3}1=0$ into $y=\log_{3}x - 1$. Then $y=0 - 1=-1$.

Step7: Recall log - rule for $x = 3$

For $y=\log_{3}3$, since $3^{y}=3 = 3^{1}$, so $\log_{3}3=1$.

Step8: Calculate $y$ for $x = 3$

Substitute $\log_{3}3=1$ into $y=\log_{3}x - 1$. Then $y=1 - 1=0$.

Step9: Recall log - rule for $x = 9$

For $y=\log_{3}9$, since $3^{y}=9 = 3^{2}$, so $\log_{3}9=2$.

Step10: Calculate $y$ for $x = 9$

Substitute $\log_{3}9=2$ into $y=\log_{3}x - 1$. Then $y=2 - 1=1$.

The completed table:

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

The domain of the function $y = \log_{3}x-1$ is $(0,\infty)$ because the argument of the logarithm function $\log_{3}x$ must be positive.
The range of the function $y=\log_{3}x - 1$ is $(-\infty,\infty)$ since the range of the basic logarithmic function $y = \log_{3}x$ is $(-\infty,\infty)$ and subtracting 1 from it does not change the range.

Answer:

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

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