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 \(z - 1\) for \(z=\frac{17}{16}\)

When \(z = \frac{17}{16}\), \(z-1=\frac{17}{16}-1=\frac{17 - 16}{16}=\frac{1}{16}\)

Step2: Calculate \(y=\log_4(z - 1)\) for \(z=\frac{17}{16}\)

Using the property \(\log_a b=c\) means \(a^c = b\). Let \(a = 4\), \(b=\frac{1}{16}\), then \(4^y=\frac{1}{16}\). Since \(4=\ 2^2\) and \(\frac{1}{16}=2^{-4}\), \((2^2)^y=2^{-4}\), \(2^{2y}=2^{-4}\), so \(2y=-4\), \(y = - 2\)

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

When \(z = 2\), \(z - 1=2 - 1=1\)

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

If \(z-1 = 1\), then \(y=\log_41\). Since \(4^0=1\), \(y = 0\)

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

When \(z = 5\), \(z - 1=5 - 1=4\)

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

If \(z - 1=4\), then \(y=\log_44\). Since \(4^1=4\), \(y = 1\)

The completed table:

\(z\)	\(z - 1\)	\(y=\log_4(z - 1)\)
\(2\)	\(1\)	\(0\)
\(5\)	\(4\)	\(1\)

For the domain of \(y=\log_4(z - 1)\), the argument of the logarithm \(z-1>0\), so the domain is \((1,\infty)\).
For the range, since the logarithm function can take on all real - values, the range is \((-\infty,\infty)\)

Answer:

\(z\)	\(z - 1\)	\(y=\log_4(z - 1)\)
\(2\)	\(1\)	\(0\)
\(5\)	\(4\)	\(1\)

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