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. $y = \log_4(x - 1)$

Explanation:

Step1: Calculate \(x - 1\) for \(x=\frac{5}{4}\)

\(\frac{5}{4}-1=\frac{5 - 4}{4}=\frac{1}{4}\)

Step2: Calculate \(y=\log_4(x - 1)\) for \(x=\frac{5}{4}\)

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

Step3: Calculate \(x - 1\) for \(x = 2\)

\(2-1 = 1\)

Step4: Calculate \(y=\log_4(x - 1)\) for \(x = 2\)

Since \(4^0=1\), \(y = 0\)

Step5: Calculate \(x - 1\) for \(x = 5\)

\(5 - 1=4\)

Step6: Calculate \(y=\log_4(x - 1)\) for \(x = 5\)

Since \(4^1 = 4\), \(y = 1\)

Step7: Calculate \(x - 1\) for \(x = 17\)

\(17-1 = 16\)

Step8: Calculate \(y=\log_4(x - 1)\) for \(x = 17\)

Since \(4^2=16\), \(y = 2\)

The completed table:

\(x\)	\(x - 1\)	\(y=\log_4(x - 1)\)
\(\frac{5}{4}\)	\(\frac{1}{4}\)	\(-1\)
\(2\)	\(1\)	\(0\)
\(5\)	\(4\)	\(1\)
\(17\)	\(16\)	\(2\)

For the domain of \(y=\log_4(x - 1)\), the argument of the logarithm must be positive. So \(x-1>0\), which gives \(x > 1\). The domain is \((1,\infty)\).
For the range, the range of any non - vertical logarithmic function \(y = \log_a(u)\) is \((-\infty,\infty)\). So the range of \(y=\log_4(x - 1)\) is \((-\infty,\infty)\)

Answer:

Completed table:

\(x\)	\(x - 1\)	\(y=\log_4(x - 1)\)
\(\frac{5}{4}\)	\(\frac{1}{4}\)	\(-1\)
\(2\)	\(1\)	\(0\)
\(5\)	\(4\)	\(1\)
\(17\)	\(16\)	\(2\)

Domain: \((1,\infty)\)
Range: \((-\infty,\infty)\)