Example of Log and exponetial problems for test Review
1.
Given the following graph, find the log equation:
( 43, 5 )
Log 1 = 0
12
so log ( x – 11 ) = 0 when x = 12
Now find the base a
5  loga (43  11)
y  log a ( x  11)
5  loga (32)
use exponential form a 5  32
THUS
y  log2 ( x  11)
a2
The VA is x = 11
2. Given the following graph, find the log equation:
y  a x equation goes thru y  1 , when x  0
( 3, 22 )
normally, so y  a x  b , and b  14 since
a 0  1 for all values of a
15
Now 22  a 3  14
equation is
3. Find the domain, range , asymptote , and sketch for
Normal exponetial
y intercept is y = 1.
a3  8
so
y  2 x  14 , with
a 2
HA y  14
y  (5) x  3
y = 5 x has domain ( - ∞, ∞ ), range ( 0, ∞ ), and HA y = 0, and the
Our graph has a vertical shift of 3 down, so domain is the same, but range ( -3, ∞ ), and HA y = -3
and the y int is y = -2
-2
-3
4.
Find the domain, range , asymptote , and sketch for
The regular equation
and VA x  0
y  log x
has domain (0, )
y   log(x  13)
range (- , )
x int x  1
This new equation has a horiz shift of 13 rt , so the VA is now x  13 , the x int is x  14
[ why ? log(14 - 13)  log 1  0 ] ... and the domain is ( 13, ). The graph is reflectedacross the
x axis, but the range stays the same.
14
VA
X=13
5. .
Find the domain, range , asymptote , and sketch for
y  log(x  22)
The regular equation y  log x has domain (0, ) range (- , )
x int x  1
and VA x  0
This new equation has a horiz shift of 22 left , so the VA is now x  -22 ,
the x int is x  -21
and the domain is ( - 22, ). range stays the same.
VA
X=-22
-21