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1 . Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve. The grph of n eponentil function depends f ( ) on the vlue of. > 1 0 < < 1 y 5 (-1, 1/) (1,) y 5 (-1, 1/) 1 (1,) Points on the grph: (-1, 1/), (0,1), (1, ) Properties of eponentil functions 1. The domin is the set of ll rel numbers: Df = R. The rnge is the set of positive numbers: Rf = (0, +). (This mens tht is lwys positive, tht is > 0 for ll. The eqution = negtive number hs no solution). There re no -intercepts. The y-intercept is (0, 1) 5. The -is (line y = 0) is horizontl symptote 6. An eponentil function is incresing when > 1 nd decresing when 0 < < 1 7. An eponentil function is one to one, nd therefore hs the inverse. The inverse of the eponentil function f() = is rithmic function g() = () 8. Since n eponentil function is one to one we hve the following property: If u = v, then u = v. (This property is used when solving eponentil equtions tht could be rewritten in the form u = v.) Nturl eponentil function is the function f() = e, where e is n irrtionl number, e n The number e is defined s the number to which the epression ( 1 1 ) n pproches s n becomes lrger nd lrger. Since e > 1, the grph of the nturl eponentil function is s below

2 5 y (1,e) (-1, 1/e) Emple: Use trnsformtions to grph f() = - -. Strt with bsic function nd use one trnsformtion t time. Show ll intermedite grphs. This function is obtined from the grph of y = by first reflecting it bout y-is (obtining y = - ) nd then shifting the grph down by units. Mke sure to plot the three points on the grph of the bsic function! Remrk: Function y = hs horizontl symptote, so remember to shift it too when performing shift up/down y = y = y = - Emple: Use trnsformtions to grph f() = e -1. Strt with bsic function nd use one trnsformtion t time. Show ll intermedite grphs. Bsic function: y = e y = e -1 (shift to the right by1)

3 y= e -1 (horizontl compression times) y = e -1 ( verticl stretch times) Emple: Solve (i) (ii) (iii) Rewrite the eqution in the form u = v Since =, we cn rewrite the eqution s Using properties of eponents we get. Use property 8 of eponentil functions to conclude tht u = v Since we hve =. Solve the eqution u = v ( 1) / Solution set = {0, ½ }. Logrithmic functions A rithmic function f() = (), > 0, 1, > 0 (rithm to the bse of ) is the inverse of the eponentil function y =. Therefore, we hve the following properties for this function (s the inverse function) (I) y = () if nd only if y = Emple: This reltionship gives the definition of (): () is n eponent to which the bse must be rised to obtin ) (8) is n eponent to which must be rised to obtin 8 (we cn write this s = 8) Clerly this eponent is, thus (8) = b) 1/ (9) is n eponent to which 1/ must be rised to obtin 9: ( 1/ ) = 9. Solving this eqution for, we get =, nd = or = -. Thus 1/ (9) = -.

4 c) () is n eponent to which must be rised to obtin : =. We know tht such number eists, since is in the rnge of the eponentil function y = (there is point with y-coordinte on the grph of this function) but we re not ble to find it using trditionl methods. If we wnt to refer to this number, we use (). The reltionship in (I) llows us to move from eponent to rithm nd vice vers Emple: - Chnge the given rithmic epression into eponentil form: = The eponentil form is: =. Notice tht this process llowed us to find vlue of, or to solve the eqution () = - Chnge the given eponentil form to the rithmic one: =. Since is the eponent to which is rised to get, we hve = (). Note tht the bse of the eponent is lwys the sme s the bse of the rithm. Common rithm is the rithm with the bse 10. Customrily, the bse 10 is omitted when writing this rithm: 10 () = () Nturl rithm is the rithm with the bse e (the inverse of y = e ): ln() = e () (II) Domin of rithmic function = (0, ) (We cn tke rithm of positive number only.) Rnge of rithmic function = (-, + ) (III) ( ) =, for ll rel numbers ( ), for ll > 0 Emple 5 = 5, lne () =,, e ln7 = 7 (IV) Grph of f() = () is symmetric to the grph of y = bout the line y= > 1 0 < < 1 y 5 y = y= (1,) (-1, 1/) 1 (,1) y = () (1/, -1) y y = 5 y= (-1, 1/) (,1) 1 (1,) (1/, -1) - y = () Points on the grph of y = () : (1/, -1), (1,0), (, 1) (V) The -intercept is (1, 0). (VI) There is no y-intercept (VII) The y-is (the line = 0) is the verticl symptote (VIII) A rithmic function is incresing when > 1 nd decresing when 0 < < 1 (IX) A rithmic function is one to one. Its inverse is the eponentil function

5 (X) Becuse rithmic function is one to one we hve the following property: If (u) = (v), then u = v (This property is used to solve rithmic equtions tht cn be rewritten in the form (u) = (v).) Emple: Use trnsformtions to grph f() = - (-1) +. Strt with bsic function nd use one trnsformtion t time. Show ll intermedite grphs. Plot the three points on the grph of the bsic function ) y = () b) y = (-1) c) y = (-1) d) y =- (-1) e) y = - (-1) + Remrk: Since rithmic function hs verticl symptote, do not forget to shift it when shifting left/right Emple: Find the domin of the following functions (A rithm is defined only for positive (> 0) vlues) ) f() = 1/ ( ) Df: > 0 = 0 = =

6 Df = (, ) (, ) b) g()= ln 9 Dg: = 0 9 = 0 = - = 9 = -/ = use the test points to determine the sign in ech intervl Dg= (, / ) (, ) Emple: Solve the following equtions ) 5 ( + + ) = (i) Find the domin of the rithm(s) + + > = 0 = 1 1 (1)() 1 15 not rel number (ii) Since y = + + hs no -intercepts nd the grph is prbol tht opens up, the grph must lwys sty bove -is. Therefore, + + > 0 for ll Chnge the eqution to the eponentil form nd solve + + = = = 0 = 1 1 (1)( 1) 1 85 since there re no restrictions on, bove numbers re solutions of the eqution.

7 b) e -+1 = 1 This is n eponentil eqution tht cn be solved by chnging it to the rithmic form -+ 1 = e (1) -+1 = ln(1) -= -1 + ln1 = 1 ln1 1 ln1 Since this is n eponentil equtions, there re no restrictions on. Solution is = Properties of rithms: Suppose > 0, 1 nd M, N > 0. Properties of rithms 1 ln1 (i) (1) = 0 () = 1 Emple: (1) = 0 15 (15)= 1 ln(1) = 0 ln(e) = 1 (ii) M M Emple: (7) e ln() = (iii) ( r ) = r Emple: ( ) = ln(e ) = (iv) (MN) = (M) + (N) Emple : 5 (10)= 5 (5)+ 5 () (M) + (N) = (MN) ln(+1) + ln(-1)= ln[(+1)(-1)] (v) Emple: M ( M ) ( N) N 15 (15) () M 1 ( M ) ( N) (1) () N (vi) (M r )= r (M) Emple: ( ) = () r (M) = (M r ) 5 (+1)= [(+1) 5 ] (vii) If M = N, then (M) = (N) Emple: if =, then () = () If (M) = (N), then M = N if (-1) = (-5), then -1 = -5 (viii) Chnge of the bse formul b M ( M ) ( ), where b is ny positive number different thn 1 b In prticulr, M lnm ( M ) nd ( M ) ( ) ln( ) This formul is used to find vlues of rithms using clcultor.

8 Emple: Evlute () ln () ln() ( ) Emple : Write s sum/difference of rithms. Epress powers s product. 1 ( ) [ ( ) ] 1 1 ( ) [( ) ] ( ) ( ) 1 ( Emple: Write s single rithm 1 1/ 1) (+1) (-1)- () = = [(+1) ] [(-1) ] ()= ( 1) = ( 1) ( 1) ( 1) ( ) ( 1) ( 1). Eponentil nd rithmic equtions A rithmic eqution is n eqution tht contins vrible inside rithm. Since rithm is defined only for positive numbers, before solving rithmic eqution you must find its domin ( lterntively, you cn check the pprent solutions by plugging them into the originl eqution nd checking whether ll rithms re well defined). There re two types of rithmic equtions: (A) Equtions reducible to the form (u) = r, where u is n epression tht contins vrible nd r is rel number To solve such eqution chnge it to the eponentil form r = u nd solve. Emple: Solve (-1)+ () = 5 (i) (ii) (iii) Determine the domin of the eqution. (Wht is inside of ny rithm must be positive) -1 > 0 > 1 (Only numbers greter thn 1 cn be solutions of this eqution) Use properties of rithms to write the left hnd side s single rithm (-1) + () = 5 ((-1) ) = 5 Chnge to the eponentil form 5 = (-1)

9 (iv) Solve = (-1) / = (-1) -1 = / = 1 + / (v) Since = 1 + / is greter thn 1, it is the solution (B) Equtions reducible to the form (u) = (v). To solve such eqution use the (vii) property of rithms to get the eqution u = v. Solve the eqution. Emple: Solve 5 () + 5 (-)= 5 (+ ). (i) Determine the domin of the eqution. (Wht is inside of ny rithm must be positive) > 0 nd > 0 nd + > 0 >0 nd > nd > - If is to stisfy ll these inequlities, then > (Only numbers greter thn cn be solutions of this eqution) (ii) Use properties of rithms to write ech side of the eqution s single rithm 5 ((-)) = 5 ( + ) (iii) Since the rithms re equl ( (M) = (N), we must hve (M = N) (-) = + (iv) Solve (-) = + = + - = 0 (-)(+1) = 0 = or = -1 (v) Since ny solution must be greter thn, only = is the solution Eponentil equtions These re equtions in which vrible ppers in the eponent. Since eponentil functions re defined for ll rel numbers, there re no restrictions on vrible nd we do not hve to check the solutions. There re three types of eponentil equtions: (A) Equtions tht cn be reduced to the form u = r, where u is n epression tht contins vrible nd r is positive rel number. If r is negtive or 0, the eqution hs no solution.

10 To solve such eqution, chnge into rithmic form nd solve Emple: Solve -1 = 5 (i) Write the eqution in the desired form (eponent = number) -1 = 5/ (ii) Chnge to the rithmic form -1 = (5/) (iii) Solve = 1 + (5/) = 1 (5 / ) To find n pproimte vlue, use the chnge of the bse formul to rewrite (5/) s (5/)/ (B) Equtions tht cn be reduced to the form u = v. To solve such n eqution use the property of eponentil functions tht sys tht if u = v, then u = v nd solve it. 6 Emple Solve (i) 16 Use the properties of eponents to write the eqution in the desired form. Notice tht ll bses (16,, ) re powers of, 16 =, = 1, = (ii) Use the property (7) + = 1 6 (iii) Solve + 1 = 0 (+6)(-) = 0 = - 6 or = Solutions: -6, (C) Equtions tht cn be reduced to the form u = b v To solve such eqution pply the (or ln ) to both sides of the eqution (property (vii) of rithms), use the property of rithms to bring the u nd v outside of the rithms nd solve for the vrible. Keep in mind tht () nd (b) re just numbers ( like 1. or ) Emple: Solve +1 = 5 1- (i) Apply to both sides ( +1 ) = (5 1- ) (ii) Use properties of rithms. (Enclose the powers into the prentheses) (+1)() = (1-)(5) (iii) Solve

11 Eliminte prentheses ()+ () = (5) -(5) Bring the terms with to the left hnd side () + (5) = (5) () Fctor out (()+(5)) = (5) () ( 5) ( ) Divide, to find = ( ) (5) You could use properties of rithms to write the solution s = (5 / ) ( 5 ) (5 / ) ( 50) If n eponentil eqution cnnot be trnsformed to one of the types bove, try to substitute by u n eponentil epression within the eqution. This might reduce the eqution to n lgebric one, like qudrtic or rtionl. Emple: Solve = 0 (i) Rewrite the eqution so tht ppers eplicitly ( ) + 1 = 0 ( ) + ( ) 1 = 0 (ii) Substitute u = u + u 1 = 0 (iii) Solve the eqution for u (u+6)(u-) = 0 u = -6 or u = (iv) Bck- substitute nd solve for = -6 or = No solution = 1 Solution: = 1

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