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

$x$	$\log_{3} x$	$y = \log_{3} x - 4$
$\frac{1}{3}$
$1$
$3$
$9$

Explanation:

Step1: Calculate $\log_3 \frac{1}{9}$

Recall that $\log_a a^b = b$. We know that $3^{-2}=\frac{1}{9}$, so $\log_3 \frac{1}{9}=\log_3 3^{-2}=-2$.

Step2: Calculate $y = \log_3 \frac{1}{9}-4$

Substitute the value of $\log_3 \frac{1}{9}$ into the equation for $y$. So $y=-2 - 4=-6$.

Now we can fill in the table for other values of $x$:

• For $x = \frac{1}{3}$:

Step3: Calculate $\log_3 \frac{1}{3}$

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

Step4: Calculate $y=\log_3 \frac{1}{3}-4$

Substitute: $y=-1 - 4=-5$.

• For $x = 1$:

Step5: Calculate $\log_3 1$

Since $3^0 = 1$, $\log_3 1 = 0$.

Step6: Calculate $y=\log_3 1-4$

Substitute: $y=0 - 4=-4$.

• For $x = 3$:

Step7: Calculate $\log_3 3$

Since $3^1 = 3$, $\log_3 3 = 1$.

Step8: Calculate $y=\log_3 3-4$

Substitute: $y=1 - 4=-3$.

• For $x = 9$:

Step9: Calculate $\log_3 9$

Since $3^2 = 9$, $\log_3 9 = 2$.

Step10: Calculate $y=\log_3 9-4$

Substitute: $y=2 - 4=-2$.

Now, to find the domain and range:

• Domain: The function $y = \log_3 x-4$ is a logarithmic function. The argument of a logarithm must be positive, so $x>0$. Thus, the domain is $(0, \infty)$.
• Range: The range of a logarithmic function $\log_a x$ is all real numbers, and subtracting 4 does not change the range. So the range is $(-\infty, \infty)$ or $\mathbb{R}$.

Filled Table:

$x$	$\log_3 x$	$y = \log_3 x - 4$
$\frac{1}{3}$	$-1$	$-5$
$1$	$0$	$-4$
$3$	$1$	$-3$
$9$	$2$	$-2$

Domain and Range:

• Domain: $x \in (0, \infty)$
• Range: $y \in (-\infty, \infty)$

Answer:

Step1: Calculate $\log_3 \frac{1}{9}$

Recall that $\log_a a^b = b$. We know that $3^{-2}=\frac{1}{9}$, so $\log_3 \frac{1}{9}=\log_3 3^{-2}=-2$.

Step2: Calculate $y = \log_3 \frac{1}{9}-4$

Substitute the value of $\log_3 \frac{1}{9}$ into the equation for $y$. So $y=-2 - 4=-6$.

Now we can fill in the table for other values of $x$:

• For $x = \frac{1}{3}$:

Step3: Calculate $\log_3 \frac{1}{3}$

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

Step4: Calculate $y=\log_3 \frac{1}{3}-4$

Substitute: $y=-1 - 4=-5$.

• For $x = 1$:

Step5: Calculate $\log_3 1$

Since $3^0 = 1$, $\log_3 1 = 0$.

Step6: Calculate $y=\log_3 1-4$

Substitute: $y=0 - 4=-4$.

• For $x = 3$:

Step7: Calculate $\log_3 3$

Since $3^1 = 3$, $\log_3 3 = 1$.

Step8: Calculate $y=\log_3 3-4$

Substitute: $y=1 - 4=-3$.

• For $x = 9$:

Step9: Calculate $\log_3 9$

Since $3^2 = 9$, $\log_3 9 = 2$.

Step10: Calculate $y=\log_3 9-4$

Substitute: $y=2 - 4=-2$.

Now, to find the domain and range:

• Domain: The function $y = \log_3 x-4$ is a logarithmic function. The argument of a logarithm must be positive, so $x>0$. Thus, the domain is $(0, \infty)$.
• Range: The range of a logarithmic function $\log_a x$ is all real numbers, and subtracting 4 does not change the range. So the range is $(-\infty, \infty)$ or $\mathbb{R}$.

Filled Table:

$x$	$\log_3 x$	$y = \log_3 x - 4$
$\frac{1}{3}$	$-1$	$-5$
$1$	$0$	$-4$
$3$	$1$	$-3$
$9$	$2$	$-2$

Domain and Range:

• Domain: $x \in (0, \infty)$
• Range: $y \in (-\infty, \infty)$