Math 1090
Logarithms, Day 1
27 March, 2012
This packet will be integrated with today’s lesson. Throughout class, you will gain
some familiarity with logarithmic functions and their graphs, and learn to solve
some simple equations involving logarithms and exponentials. Once this packet is
complete, you should have a good set of notes on Section 4.3 in your textbook.
Name:
Warm Up: Fill in the blanks.
1. If 3x “ 81, then x “
2. If 2x “ 16, then x “
3. If ex “ e2 , then x “
4. If 5x “ 1, then x “
5. If 4x “
1
,
16
then x “
6. If 10x “ 100, then x “
Part I:
7. If 2x “ 6, then x “
.
Solution: Think about the function f pxq “ 2x . We know that f pxq is one-to-one
and onto, so it has an inverse function f ´1 . There is a unique solution to the equation
2x “ 6, namely f ´1 p6q.
8. Let gpxq “ 3x . If 3x “ 7, what is x?
Now that we have some idea of how to solve these problems, let’s introduce the standard
notation. For the rest of this packet, a, b P R, a ‰ 1 and b ‰ 1.
Definition: The exponential function f pxq “ ax , is invertible, and its inverse is the
logarithmic function loga pxq.
Let’s think about what this means. First of all,
“ logapbq “ x” means “ax “ b.”
Also, loga pxq and ax are inverse functions, so
logapaxq “ alogapxq “ x
Here are some logarithmic and exponential equations. Translate each one into the equivalent
logarithmic or exponential equation. The first one is done for you.
9. log4 p16q “ 2 ùñ 42 “ 16
13. log4 p2q “
1
2
ùñ
10. log5 p125q “ 3 ùñ
14. 103 “ 1000 ùñ
11. 43 “ 64 ùñ
15. loga a2 “ 2 ùñ
12. log2 p 18 q “ ´3 ùñ
16. 5´a “
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1
5a
ùñ
We can use this translation to evaluate logarithmic expressions. For each expression below,
evaluate by translating into an exponential equation and solving.
17. log7 49
Solution: Let x “ log7 49. We can rewrite this as 7x “ 49. Since 7x “ 72 , x “ 2.
18. log4 256
22. log17 1
19. log3 27
23. loge e4
20. log3
1
9
21. log4 2
24. log3 ´3
25. log4 0
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Part II: Graphing Logarithmic Functions
“To know the graph of a function is to
understand the function.” - Socrates
26. Using your knowledge that loga pxq is the inverse function of ax , graph log3 pxq on the
axes below.
27. Using your knowledge that loga pxq is the inverse function of ax , graph log10 pxq on the
axes below.
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28. Using your knowledge that loga pxq is the inverse function of ax , graph log 1 pxq on the
2
axes below.
29. Using your knowledge that loga pxq is the inverse function of ax , graph log 1 pxq on the
4
axes below.
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Part III: Using your knowledge of logarithms, solve for x. The first problem is done for
you.
30. 6x “ 10
Solution: We can convert the exponential expression to the equivalent logarithmic
expression:
6x “ 10 ùñ log6 p10q “ x
So x “ log6 p10q. Sometimes we’ll be able to simplify this and sometimes we won’t,
but we’ll talk more about that in the next lesson.
31. 7x “ 48
32. 2.5x “ 5
33. 36x “ 62
34. π x “ 25
35. 5x`1 “ 125
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36. 72x´1 “ 49
37. 12x`4 “ 144
38. 3´x “ 27
39. 2x`5 “ 15
40. log4 pxq “ 16
41. log14 px ` 1q “ 16
42. log2 p2x ` 1q “ 5
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