Question

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. $y = \log_4(x - 2)$

Explanation:

Step1: Calculate \(x - 2\) for \(x=\frac{33}{16}\)

\(\frac{33}{16}-2=\frac{33 - 32}{16}=\frac{1}{16}\)

Step2: Calculate \(y=\log_4(x - 2)\) for \(x - 2=\frac{1}{16}\)

Let \(y = \log_4\frac{1}{16}\), since \(4^{- 2}=\frac{1}{4^{2}}=\frac{1}{16}\), then \(y=-2\)

Step3: Calculate \(x - 2\) for \(x = \frac{9}{4}\)

\(\frac{9}{4}-2=\frac{9 - 8}{4}=\frac{1}{4}\)

Step4: Calculate \(y=\log_4(x - 2)\) for \(x - 2=\frac{1}{4}\)

Let \(y=\log_4\frac{1}{4}\), since \(4^{-1}=\frac{1}{4}\), then \(y = - 1\)

Step5: Calculate \(x - 2\) for \(x = 3\)

\(3-2 = 1\)

Step6: Calculate \(y=\log_4(x - 2)\) for \(x - 2=1\)

Let \(y=\log_41\), since \(4^{0}=1\), then \(y = 0\)

Step7: Calculate \(x - 2\) for \(x = 6\)

\(6 - 2=4\)

Step8: Calculate \(y=\log_4(x - 2)\) for \(x - 2 = 4\)

Let \(y=\log_44\), since \(4^{1}=4\), then \(y = 1\)

Step9: Calculate \(x - 2\) for \(x = 18\)

\(18-2 = 16\)

Step10: Calculate \(y=\log_4(x - 2)\) for \(x - 2=16\)

Let \(y=\log_416\), since \(4^{2}=16\), then \(y = 2\)

The completed table:

\(x\)	\(x - 2\)	\(y=\log_4(x - 2)\)
\(\frac{9}{4}\)	\(\frac{1}{4}\)	\(-1\)
\(3\)	\(1\)	\(0\)
\(6\)	\(4\)	\(1\)
\(18\)	\(16\)	\(2\)

For the domain of \(y=\log_4(x - 2)\), the argument of the logarithm must be positive. So \(x-2>0\), which gives \(x>2\). The domain is \((2,\infty)\).
For the range of the logarithmic - function \(y = \log_4(x - 2)\), the range of any non - vertical logarithmic function of the form \(y=\log_a(u)\) (\(a>0,a
eq1\)) is \((-\infty,\infty)\). So the range is \((-\infty,\infty)\)

Answer:

\(x\)	\(x - 2\)	\(y=\log_4(x - 2)\)
\(\frac{9}{4}\)	\(\frac{1}{4}\)	\(-1\)
\(3\)	\(1\)	\(0\)
\(6\)	\(4\)	\(1\)
\(18\)	\(16\)	\(2\)

Domain: \((2,\infty)\)
Range: \((-\infty,\infty)\)