|
|
|
|
|
|
||||||||||
|
|
||||||||||||||
The following properties of logarithms are important
and used frequently in our study of logarithms.
They correspond closely to
our rules for exponents studied earlier.
|
Exponential Form |
Logarithmic Form |
|
34 = 81 |
log381 = 4 |
|
52 = 25 |
log525 = 2 |
|
25 = 32 |
log232 = 5 |
|
103 = 1000 |
log101000 = 3 |
|
|
|
|
|
|
UNIT 4 :
EXPONENTIAL & LOGARITHMIC FUNCTIONS REVIEW QUESTIONS
Solutions
|
Interval |
|
|
|
|
Test Number |
- 4 |
- 1 |
4 |
|
Sign
of (x – 3 ) |
( - ) |
( - ) |
( + ) |
|
Sign
of (x + 3) |
( - ) |
( + ) |
( + ) |
|
Sign
of (x – 3 )(x + 3) |
( + ) |
( - ) |
( + ) |
|
Interval |
|
|
|
|
Test Number |
- 3 |
- 1 |
5 |
|
Sign
of (x + 2) |
( - ) |
( + ) |
( + ) |
|
Sign
of (x – 3) |
( - ) |
( - ) |
( + ) |
|
Sign
of (x + 2)(x – 3) |
( + ) |
( - ) |
( + ) |
|
Interval |
|
|
|
|
Test Number |
- 5 |
0 |
3 |
|
Sign
of (2x – 5 ) |
( - ) |
( - ) |
( + ) |
|
Sign
of (x + 4) |
( - ) |
( + ) |
( + ) |
|
Sign
of (2x – 5)/(x + 4) |
( + ) |
( - ) |
( + ) |
11. Solve for
x.
Solutions:
12. Solve for
x.
Solutions:
13. Solve for
x.
Solutions:
14. Solve for
x.
Solutions:
15. A biologist grows a colony of bacteria in a
Petri dish. Under ideal conditions, the
doubling period is 3 h. How long will it
take for the colony to grow to 32 times its original size?
Recall:
A
= ? A0 = ? t = ? d = 3 h
, where
·
A is the number of bacteria after the
given time frame
·
A0 is the starting
number of bacteria
·
2 is the growth factor
·
t is the total time elapsed in the
experiment
·
d is the doubling period
Solution: In
this problem, we are asked to find the time it takes to grow to 32 times its original
size. We are not told the exact starting
and final numbers of bacteria. The key
here is that whatever the starting amount is, the final amount will be exactly
32 times larger. Therefore we let the
starting amount be 1 g . Hence the final amount will be 32 g.
We now have the following information in the box at right.
A
= 32 g A0 = 1 g t = ? d = 3 h
Therefore it will take 15 h to grow to 32 times its
original size.
16. The half-life of radium is 1600 years. What fraction of radium remains from a sample
after 12 800 years?
Solution:
Again let the starting mass be 1 g.
A
= ? A0 = 1 g t = 12 800
years h = 1 600 years
Recall:
Therefore 
of the original amount
remains after 12 800 years.
A
= 3.125 g A0 = 100g t = 2 000 years
h = ?
