11.2 Logarithmic Functions
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1 .2 Logrithmic Functions In the lst section we delt with the eponentil function. One thing tht we notice from tht discussion is tht ll eponentil functions pss the horizontl line test. Tht mens tht the eponentil function must hve n inverse function. If we try to find this inverse function using our stndrd methods we would get the following: f y y We see tht we re stuck. For this reson we simply define the inverse of n eponentil function in such wy tht it stisfies the properties of inverse functions. We cll this function the rithmic function with bse. Definition: Logrithmic function- Let nd be positive rel numbers such tht. The rithm of with bse is written nd is defined s follows: f y if nd only if is the rithmic function with bse. y For emple, the following is true sttement 5. The wy we cn verify this is by 24 using the bove definition to chnge the sttement into eponentil form (since we cn del 5 esily with eponentils). If we chnge it over we get which is clerly true sttement. 24 There re two primry techniques we use to chnge equtions from rithmic to eponentil form. The first is to just use the definition bove by lbeling ech of the vlues, y, nd. Then just substituting them into the definition in the proper spot. The other wy of chnging between rithmic nd eponentil forms is by using much more descriptive sttement to decipher wht ech piece of rithmic epression is. Tht sttement is the following: bse nswer = eponent, where bse is the bse of the eponentil, nswer is the opposite side of the equl sign of the eponentil, nd eponent is the eponent of the eponentil. Mny people prefer this method becuse it is much more descriptive thn trying to remember two epressions tht both contin, y, nd. Emple : Evlute the following.. 27 b. 5 c. 2 0 d Since we wnt to know the vlue of 27, lets sy tht its. Then we hve 27 = Now using bse nswer = eponent we cn chnge the epression over to eponentil form. We cn see tht the bse is, nswer is 27 nd eponent is. This gives us = 27 Clerly, = since = 27. So 27 =.
2 b. Similrly, we will set 5 = nd find by using bse nswer = eponent to convert to eponentil form. This gives us 5 =. So = 0 since 5 0 =. So 5 = 0. c. Agin set 2 0 = nd convert to eponentil form. This gives us 2 = 0. But no such will eists since ny power of 2 is greter thn zero. There for we sy 2 0 is not rel number. This lso tells us tht 0 is not in the domin of y = 2 ( fct tht we will need lter). d. Lstly, set 5 5 = nd convert to eponentil form. We get 5 = 5. Clerly =. So therefore 5 5 =. Emple 2: Solve the following for.. 5 = b. 4 = 0. Agin using bse nswer = eponent we will convert the eqution to eponentil form. So 5 = becomes 5 =. So clerly = 5. b. Likewise we convert 4 = 0 to 4 0 =. So =. These previous emples give us the following properties. Bsic Properties of Logrithms. = 0 becuse 0 = 2. = becuse =. = becuse = We will need these properties lter. Net we wnt to tlk bout two rithms tht hve specil bses. First, The Common Log 0 is clled the common. We usully write just to men the common. Also, common is on every scientific clcultor. It s the button Emple : Use your clcultor to evlute 9. Round to three deciml plces. Using your clcultor input of 9. Depending on the type you hve you might hve to input first or 9 first. Consult your instructions or sk you instructor for help. You should get 9 = So rounding off we get 9 = The Nturl Log Recll from the lst section we defined the nturl bse, e. Therefore, the function defined by f() = e = ln is clled the nturl rithmic function. This is lso on your clcultor. It s the button ln So ln is just rithm with bse of e. We never write e. It is considered improper nottion.
3 Emple 4: Use your clcultor to evlute ln 6. Round to three deciml plces. Using your clcultor input ln of 6. Agin depending on the type you hve you might hve to input ln first or 6 first. Consult your instructions or sk you instructor for help. You should get ln 6 = So rounding off we get ln 6 =.792. Emple 5: Solve the following for. Round to three deciml plces.. = -2. b. ln =.4 Since this time we re not looking for the vlue of the rithm we cnnot do the problem like emples or 4. So, since this time we hve the output nd we wnt to know wht will give ech of these vlues we just use the shift, 2 nd, or inv buttons on our clcultor.. So since we hve the output of 2. we hit the 2 nd button before the button. Consult your instructions or sk you instructor for help. We should get = (with rounding). b. Similrly we use the 2 nd button before the ln button. We should get = (with rounding). Lets sy we wnted to evlute 5 7. The problem is if we chnge it to eponentil form we get 5 = 7, which does not seem esy, nd we don t hve 5 button on our clcultor to solve it. In order to rectify tht sitution we hve formul for chnging from ny bse to nother bse tht we cn use (usully bse 0 or e). This formul is clled the chnge of bse formul. Chnge of Bse Formul b or specificlly b ln ln We cn use this to evlute ny with ny bse. Emple 6: Evlute. Round your nswer to three deciml plces b c. (+e ) d. / We use the chnge of bse formul to chnge from bse 5 to ny bse we wnt. Since we wnt to use our clcultors to find the vlue, we will use either bse e or bse 0. Lets use bse e, tht is ln. So we hve 5 7 ln 7 ln 5.209
4 b. Agin we will use the chnge of bse formul. This time lets chnge to bse 0. This gives c. Since it doesn t mtter wht bse we choose for the chnge of bse formul we will use ln for simplicity. This gives d. Finlly, chnge of bse formul gives us ln e e ln ln ln.82.2 Eercises Evlute without using clcultor ln e ln e ln e ln e /. ln 4. y y ln e Solve the following for without using clcultor = = 44. = = = = = = = / = /2 52. = = / 54. /2 = / = 56. 2/ = 57. /4 = = = = = /2 = / = / = /5 = - Evlute. Round your nswer to three deciml plces ln 68. ln ln.4 7. ln ln ln.000 Solve the following for. Round your nswer to three deciml plces. 79. = = 4 8. ln = ln = = ln = ln = = Evlute. Round your nswer to three deciml plces / / / /7 08. /7 4/5 09. / e 0. 6/ (6.87) (2.24) (.657) 4. 4 (+e 4 ) 5. 6 (/+e) 6. 7 (-e 2 )
5 7. (ln +6) 8. 5 ( 5 e) 9. (ln( 4 5) Let f() = 2, g() = (-) nd h() = ln (+). Find the following. 20. f() 2. f(4) 22. g(4) 2. g(2) 24. h(0) 25. h(e-) 26. f() 27. g(-b) 28. h(+) 29. f(+h) 0. g(+h). h(+h) 2. (f o g)(). (g o f)()
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