Dte: 3/1/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: 1. f x = 3x 6, find the inverse, f 1 x., Using your grphing clcultor, Grph 1. f x,f 1 x nd y =x pg 308 #48, 50, 5 HW Requests: ening of Logs worksheet Homework: Properties of Logs Worksheet complete Red section 3.3, 3.4 Announcements:
Dte: 3/13/13 Objective: SWBAT pply properties of exponentil functions nd will pply properties of rithms. Bell Ringer: pg 308 #0 pg 316 #8-10, Write n eqution for the following dt using Power regression. x 1 4 9 16 5 y - -4-6 -8-10 HW Requests: pg 308 #1-15 odds Homework: Properties of Logs Worksheet complete Red section 3.3, 3.4 Announcements:
Exponentil Functions nd the Bse e x kx Any exponentil function f ( x) b cn be rewritten s f ( x) e, for ny ppropritely chosen rel number constnt k. kx If 0 nd k 0, f ( x) e is n exponentil growth function. kx If 0 nd k 0, f ( x) e is n exponentil decy function. Slide 3-3
Common Logrithms nd Nturl Logrithms Becuse of their frequent use in rel-world nd pplied problems, rithms with bse 10 re referred to s common rithms. Common s written without n indicted bse re ssumed to be bse 10. 10 x = x The stndrd clcultor button ssumes bse 10. Becuse e is frequently used s n exponentil bse, bse e is defined s the nturl rithm e x = ln x
Exponentil Functions nd the Bse e Slide 3-5
Nturl Log is the inverse of exponentil bse e =.718 Properties of rithms hold the sme for ln x Here s the question then: Why is ln e = 1? Why is e ln x = x? Now wht does the grph of the common nd nturl look like? Cn we nlyze this function? Does it hve similr properties to other prent functions? Pg 305 Grph e x, ln x nd y = x
Properties of Logrithms
Properties of Logrithms 1. CONDENSED N = = EXPANDED N. 3. N r = N = r (Properties bsed on rules of exponents since s = exponents) Definition: b n p if nd only if b p n
Using the properties, write the expression s sum nd/or difference of s (expnd). 6 b c 4 3 When working with s, re-write ny rdicls s rtionl exponents. using the second property: using the first property: N N using the third property: r r N N b 6 c 3 4 4 6 b 6 c 3 4 6 6 b 6 6 4 6 b 6 3 c 3 c
Using the properties, write the expression s single rithm (condense). 1 3 x 3 y using the third property: r r this direction 3 x 3 y 1 using the second property: N this direction N 3 x y 1 We re good!
ore Properties of Logrithms This one sys if you hve n eqution, you cn tke the of both sides nd the equlity still holds. If N, then N If N, then N This one sys if you hve n eqution nd ech side hs of the sme bse, you know the "stuff" you re tking the s of re equl.
Chnging bses when needed There is n nswer to this nd it must 8 3 be more thn 3 but less thn 4, but we cn't do this one in our hed. ( to the wht is 8?) 16 4 ( to the wht is 16?) 10 3.3 ( to the wht is 10?) Check by putting 3.3 in your clcultor (we rounded so it won't be exct) Let's put it equl to x nd we'll solve for x. Chnge to exponentil form. use property & tke of both sides (we'll use common ) If N, then N use 3rd property r r solve for x by dividing by use clcultor to pproximte 10 x x x x 10 10 10 10 x 3. 3
If we generlize the process we just did we come up with the: Chnge-of-Bse Formul Exmple for TI-83 The bse you chnge to cn be ny bse so generlly we ll wnt to chnge to bse so we cn use our clcultor. Tht would be either bse 10 or bse e. b b ln ln common bse 10 LOG LN nturl bse e
Use the Chnge-of-Bse Formul nd clcultor to pproximte the rithm. Round your nswer to three deciml plces. 3 16 Since 3 = 9 nd 3 3 = 7, our nswer of wht exponent to put on 3 to get it to equl 16 will be something between nd 3. ln16 3 16.54 ln 3 put in clcultor
Since s nd exponentils of the sme bse re inverse functions of ech other they undo ech other. f x 1 x f x x Remember tht: f f 1 x nd f 1 f x This mens tht: f f 1 x x inverses undo ech ech other f 1 f x x 5 = 5 7 3 3 = 7
Introduction To Logrithms cont.
Exmple 4 Evlute: 7 7 Solution: 7 7 y 7 y 7 y First, we write the problem with vrible. Now tke it out of the rithmic form nd write it in exponentil form.
Exmple 5 Evlute: 4 4 16 b n p if nd only if b p n Solution: 4 4 16 y 4 y 4 16 y 16 First, we write the problem with vrible. Now tke it out of the exponentil form nd write it in rithmic form. Just like 3 8 converts to 8 3
Exmple 1 Solve: 3 (4x 10) 3 (x 1) Solution: Since the bses re both 3 we simply set the rguments equl. 4x 10 x 1 3x 10 1 3x 9 x 3
Exmple Solve: 8 (x 14) 8 (5x) Solution: Since the bses re both 8 we simply set the rguments equl. x 14 5x x 5x 14 0 (x 7)(x ) 0 Fctor (x 7) 0 or (x ) 0 x 7 or x continued on the next pge
Exmple continued Solve: 8 (x 14) 8 (5x) Solution: x 7 or x It ppers we hve solutions. Look t the definition of rithm, not only must we use positive bses, but lso the rguments must be positive. Therefore - is not solution.
( 8) undefined WHY? One esy explntion is to simply rewrite this rithm in exponentil form. We ll then see why negtive vlue is not permitted. ( 8) y First, we write the problem with vrible. y 8 Now tke it out of the rithmic form nd write it in exponentil form. Wht power of would gives us -8? 3 8 nd 3 1 8 Hence expressions of this type re undefined.