# Mathematics Number: Logarithms

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1 plce of mind F A C U L T Y O F E D U C A T I O N Deprtment of Curriculum nd Pedgogy Mthemtics Numer: Logrithms Science nd Mthemtics Eduction Reserch Group Supported y UBC Teching nd Lerning Enhncement Fund

2 Logrithms

3 Introduction to Logrithms Find the vlue of where = 8. Press for hint A. B. C. D. E Write 8 s power with se. Eponent lw: y if nd only if y

4 Solution Answer: B Justifiction: Try writing the right hnd side of the eqution s power of two. 3 8 Then use the eponent lw: y if nd only if y 8 3 3

5 Introduction to Logrithms II Find the vlue of where = 18. A. B. C. D. E None of the ove

6 Solution Answer: E Justifiction: We cnnot write 18 s power of. This mens we cnnot solve this question the sme wy s the previous one Although we cn t find the ect nswer, we cn guess the nswer is etween 4 nd 5. The nswer cnnot e written s n integer, so the correct nswer is E, None of the ove. In order to nswer this question, we will hve to lern out rithms.

7 Introduction to Logrithms III If y = then y =. Notice how the rithm ( y) returns the eponent of. Using the ove definition, wht is the vlue of 8? A. B. C. D. E None of the ove

8 Solution Answer: B Justifiction: This question is ectly the sme s question 1. We must find the vlue of where = 8. In question 1, we were sked to find the eponent such tht = 8. This is ectly the sme s finding the vlue of since 8 In generl, we cn convert etween s nd eponents s follows: y = y = A helpful reminder: equls y y = to the

9 Introduction to Logrithms IV Find the vlue of where = 18. Epress the nswer using rithms. Hint: Try epressing the question in terms of rithms. A. B. C. D. E y None of the ove If y = then y =. y = y = Press for hint

10 Solution Answer: A Justifiction: Recll how equtions in the form y =, cn e rewritten with rithms, nd vice-vers: y = y = 18 Notice how pplying this to = 18 llows us to solve for, which we were not le to do without rithms: This is the ect nswer to the eqution ove, so. We will see lter tht this result cn e generlized to property of rithms:

11 Logrithm Lws The following is summry of importnt rithm properties: n n c c Other useful properties include: y 1 if nd only if Note: When you see, it is ssumed tht the se of the rithm is. When you see ln, the se is ssumed to e e.. ln e y

12 Properties of Logrithms I Wht is c written s single rithm? A. B. C. D. E. c c c c c

13 Solution Answer: D Justifiction: We will use the two following two properties to simplify the epression: 1. First comine the terms property 1 shown ove: c. using the first Net comine the term tht is sutrcted using property : c c Does order mtter? Would simplifying the sutrcted term first chnge the nswer? c

14 Properties of Logrithms II Wht is the vlue of 0 8. A. B. C. D. E

15 Solution Answer: D Justifiction: Use eponent nd rithm lws to simplify the prolem. We cn strt y pulling the eponent the outside of the rithm: since since n n 0 The solution tells us tht 16 = 0 8.

16 Alterntive Solution Answer: D Justifiction: Since the rithm is in se, nother strtegy is to write ll terms with se of since since since ( ) y n 1 y n

17 Properties of Logrithms III Which of the following is not equivlent to the following eqution, where > 0, 1? c A. c B. C. D. E. 1 c c c c

18 Solution Answer: D Justifiction: Answer A writes the rithm in eponentil form: c c We cn lso use the property y = + y to rek up the : 1 When we let 1, we get nswer C. When we let, we get nswer B. Answer continues on the net slide

19 Solution Continued Answer E uses the chnge of se property to rewrite the rithm with se : Answer D is very similr to nswers B nd C, lthough. Therefore nswer D is the only non-equivlent epression:

20 Properties of Logrithms IV Which of the following is equivlent to the epression shown: k k A. B. C. D. E. k k 1 k Chnge of se property: y c c y Press for hint

21 Solution Answer: C Justifiction: The most common mistke tht my rise when nswering this question is incorrectly stting: k k k This sttement is not correct Mking this error will led to nswer A or B. Insted, we cn use the chnge of se property to write the epression s single rithm: Chnge of se property: k c k y y c

22 Properties of Logrithms V Which of the following re equivlent to the following epression ( ) n A. B. C. D. E. ( None of the ove ) n n n( ) n n n

23 Solution Answer: E Justifiction: Epressions in the form ( ) nd ( ) n cnnot e simplified. Therefore, ( ) cnnot e simplified. n The two most common errors re: ( ) n n Compre the incorrect sttements ove with the correct rithm lws: ( ) ( n ) n( )

24 Properties of Logrithms VI If > > 0, how do () nd () compre? A. B. C. D. E.

25 Solution Answer: A Justifiction: It my e esier to compre rithms y writing them s eponents insted. Let: A B Converting to eponents gives: A nd B Given in the question A B A B Sustitute = A, = B The lrger the eponent on the se, the lrger our finl vlue. This is true whenever the ses re the sme nd greter thn 1. For emple, 4 > 3. The eponent A must e lrger thn the eponent B Sustitute A =, B =

26 Properties of Logrithms VII Wht is the vlue of the following epression? A. B. C. D. E. Cnnot e simplified

27 Solution Answer: A Justifiction: This is one of the rithm lws: If =, is the eponent to which must e rised in order to equl (in order words, = ) y the definition of rithms. equls = to the equls = to the It is importnt to rememer tht rithms return the vlue of n eponent. When the eponent returned is the originl rithm, we get the property:

28 Alterntive Solution Answer: A Justifiction: We cn lso solve this prolem using other properties of rithms. Let. The vlue of will e our nswer. If we write this eqution using rithms: = y = y In this question, We cn now use the property = if nd only if =. y

29 Properties of Logrithms VIII If > > 1 nd c > 1 how do (c) nd (c) compre? A. B. C. D. E. ( c) ( c) ( c) ( c) ( c) ( c) ( c) ( c) ( c) ( c) Use the chnge of se property: ( c) ( c) ( ) ( c) c ( ) Press for hint

30 Solution Answer: C Justifiction: We cn use the chnge of se property to mke the two epressions esier to compre: ( c) ( c) ( ) Since,, nd c re greter thn 1, (), (), nd (c) re ll positive. From the previous question we lerned tht () > () if >. Since () is in the denomintor, it will produce smller frction thn (). ( c) ( ) ( c) ( ) c ( c) ( ) ( c) ( c)

31 Alterntive Solution Answer: C Justifiction: You cn lso chnge the question to eponents nd compre. Let: A ( c) c A B If we equte these two equtions we get: A B ( c) c B Since >, the eponent A must e smller thn B. If this were not the cse, the LHS will hve lrger se nd eponent thn the RHS, so they cnnot e equl. A B ( c) ( c)

32 Properties of Logrithms IX Given 3 1.1, wht is the pproimte vlue of 3 49? A. B. C. D. E. (1.1) (1.1) (1.1) (1.1) Write 3 49 in terms of 3 using the following property: n n Press for hint

33 Solution Answer: C Justifiction: Without knowing tht 3 1.1, we cn still determine n pproimte vlue for Since 3 = 3 nd 3 81 = 4, we should epect 3 49 is etween 3 nd 4. We cn lredy rule out nswers A, D, nd E. To get etter estimte, we cn use the properties of rithms: (1.1) since since n 3 n 1.1

34 Properties of Logrithms X If 1 wht does equl? m n 0 m n A. B. C. D. E Write m n 1 s m n Press for hint

35 Solution Answer: A Justifiction: If we simplify the eqution reltionship etween m nd n: m n 1 m m n m n 1 n We cn now plug this vlue into the other epression: m n 0 m n since 1 m n 1, we cn find the

36 Alterntive Solution Answer: A Justifiction: The solution cn lso e found y simplifying first, lthough much more work is required to get finl nswer: 0 m n 0 0 m m 0 0 n n 0 m n m m Chnge to se 0 = Since m = n 0 m n

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