Question

read explain 1b part a and complete your turn #1 and 2 (adapted from lesson 15.2).
from the parent function ( f(x)=log _{b} x ) with base ( b ), the domain is ( (0, infty) ) and the range is all real numbers, also written as ( (-infty, infty) ). when determining the domain and range for the transformed function ( g(x)=a log _{b}(x - h)+k ), the horizontal translations will change the domain from the parent functions domain.
to determine the domain of ( g(x) ), find the vertical asymptote. since the vertical asymptote of ( g(x) ) is ( x = h ), the domain is ( (h, infty) ). if there is a reflection across the ( y )-axis, ( log _{b}(-x) ), then the domain is ( (-infty, h) ).
the range of any logarithm function will be all real numbers; any vertical translation or reflection across the ( x )-axis does not change the range.

example 1b what is the domain and range of the following equations?
a. ( y=log _{6}(x + 4) )
b. ( y=log _{3}(x)-7 )
comparing to ( a log _{b}(x - h)+k ), observe that
comparing to ( a log _{b}(x - h)+k ), there is no
( h=-4 ). since the vertical asymptote is ( x=-4 ), the
horizontal translation. thus, the domain is ( (0, infty) ).
the domain is ( (-4, infty) ). the range is ( (-infty, infty) ).
the range is ( (-infty, infty) ).

your turn

1. what is the domain and range of
2. what is the domain and range of

( y=log _{3}(x - 5)+2 )?
( y=-log (x + 2)-8 )?
domain:
domain:
range:
range:

1. what is the domain of ( y=log (7 - x) )?

compare the work of two students. determine who found the domain correctly. explain why.
student a
there is a vertical asymptote at ( x = h ). this means
( 7 - x=0 ). then ( -x=-7 ). so, ( x = 7 ).
since ( log (7 - x) ) can be seen as ( log (-x + 7), x<0 ).
the domain is ( (-infty, 7) ).
student b
there is a vertical asymptote at ( x = h ). this means
( 7 - x=0 ). then ( -x=-7 ). so, ( x = 7 ).
since ( log (7 - x) ) can be seen as ( log (x - 7), x>0 ).
the domain is ( (7, infty) ).

Explanation:

1. For \(y = \log_3(x - 5)+2\)

Step1: Find the domain

For the function \(y=\log_b(x - h)+k\), the domain is determined by \(x - h>0\). Here \(b = 3\), \(h = 5\), \(k=2\).
We set \(x-5>0\).
\(x-5>0\Rightarrow x>5\). So the domain is \((5,\infty)\).

Step2: Find the range

The range of any logarithmic function \(y = a\log_b(x - h)+k\) (where \(b>1\)) is all real numbers. Because vertical translations (\(k = 2\)) do not change the range of the parent - function \(y=\log_bx\) (range of \(y=\log_bx\) is \((-\infty,\infty)\)). So the range is \((-\infty,\infty)\).

2. For \(y=-\log(x + 2)-8\)

Step1: Find the domain

For the function \(y = a\log_b(x - h)+k\) (here \(a=-1\), \(b = 10\), \(h=-2\), \(k =-8\)), we set \(x+2>0\).
\(x + 2>0\Rightarrow x>-2\). So the domain is \((-2,\infty)\).

Step2: Find the range

The range of \(y=\log(x)\) is \((-\infty,\infty)\). Vertical reflections (\(a=-1\)) and vertical translations (\(k=-8\)) do not change the range of the logarithmic function. So the range is \((-\infty,\infty)\).

3. For \(y=\log(7 - x)\)

Analyze Student A and Student B

For the function \(y=\log(7 - x)=\log(-(x - 7))\), the argument of the logarithm must be positive.
We set \(7 - x>0\).
\(7 - x>0\Rightarrow -x>-7\Rightarrow x<7\).
Student A:
They correctly set the argument of the logarithm \(7 - x>0\). Solving \(7 - x>0\) gives \(x<7\), so the domain is \((-\infty,7)\).
Student B:
They made a mistake in rewriting \(\log(7 - x)\) as \(\log(x - 7)\). The correct way to find the domain is by setting the argument of the given function \(7 - x>0\), not by an incorrect rewrite that changes the sign - condition for the domain.

Answer:

1. Domain: \((5,\infty)\), Range: \((-\infty,\infty)\)
2. Domain: \((-2,\infty)\), Range: \((-\infty,\infty)\)
3. Student A found the domain correctly. Because for \(y = \log(7 - x)\), we need \(7 - x>0\) (argument of the logarithm must be positive). Solving \(7 - x>0\) gives \(x<7\), so the domain is \((-\infty,7)\). Student B incorrectly rewrote \(\log(7 - x)\) as \(\log(x - 7)\) which led to an incorrect domain condition.