Logarithmic Functions

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Logrithmic Functions Definition: Let > 0,. Then log is the number to which you rise to get. Logrithms re in essence eponents. Their domins re powers of the bse nd their rnges re the eponents needed to produce those prticulr powers. Emple: Demonstrte tht log 4 6. Here the bse is 4 nd 6. To wht number do you hve to rise 4 in order to get 6? Answer:, so log 4 6. Emple: Show tht log 0 0,000 4. Here the bse is 0, nd 0,000. Wht number do you hve to rise 0 to, in order to get 0,000 (4 zeros)? Answer: 4, so log 0 0,000 4. MATH 80 Lecture of Ronld Brent 08 All rights reserved.

Emple: log 0.00? Here the bse is 0, nd.00. Write.00 s power of 0. Since.00 0, log 0.00. The function log 0 is known s the common logrithm. Sometimes you will see the epression log. Theorem: Let > 0,. Then log. log. MATH 80 Lecture of Ronld Brent 08 All rights reserved.

Grphs of Logrithmic Functions: If you wish to grph the function f ( ) log, you need only grph the function g( ), nd flip it round the line y y y y y log MATH 80 Lecture of Ronld Brent 08 All rights reserved.

y y 0 y y log MATH 80 Lecture 4 of Ronld Brent 08 All rights reserved.

Solving Eponentil Equtions We re now ble to solve eponentil equtions by getting t the eponent of term. Emple: Solve for : 4 6 Since you cn write 6 s power of 4, this problem cn be solved without logrithms. To wht do you rise 4, to get 6? Answer: Emple: Solve for : 5 5 Agin, since you cn write 5 s power of 5, this problem cn be solved without logrithms. To wht do you rise 5, to get 5? Answer: MATH 80 Lecture 5 of Ronld Brent 08 All rights reserved.

Emple: Solve for : Since one cn t write s power of, this problem cn t be solved without logrithms. But now, knowing tht log is the inverse of one simply Log both sides with bse of. ( ) log log or log You see tht logrithms isolte the eponent. Emple: Solve of : 5 4 First log both sides using bse of 5: ( 5 ) log 4 log 5 5 or log 5 4 And now tke the squre root of both sides so tht ± log 5 4 MATH 80 Lecture 6 of Ronld Brent 08 All rights reserved.

Lws of Logrithms Log Lws: Let > 0,, nd let > 0 nd y > 0. Then: ) log y log + log y ) log log log y y r ) log r log 4) log 0 for ll (but we knew tht lredy!) logb 5) For ny > 0, with, log, for ny convenient b for logb emple, 0 or e. This chnge of bse lw cn be lifesver if you cn t compute log. MATH 80 Lecture 7 of Ronld Brent 08 All rights reserved.

Emple: Solve log + log ( ). You cn combine the left side using rule. So log ( ), using the first lw. Now you cn pply : log ( ) You should check this solution:. log + log () log 9 + log + MATH 80 Lecture 8 of Ronld Brent 08 All rights reserved.

4 Emple: Solve log ( 6) 8. 4 Using the third lw, you obtin 4log4 ( 6) 8, or log4 ( 6). Now pply 4 : 6 4. So 6 6 0 Either fctoring or using the qudrtic formul, nd 8 or. MATH 80 Lecture 9 of Ronld Brent 08 All rights reserved.

Emple: Evlute log 6 7. You cn t evlute this directly. You could try to use clcultor, but most of them don t hve log. But using the chnge-of-bse formul, log0 7 log6 7, log 6 0.845 so log 6 7. 0860. 0.778 You could hve used the nturl logrithm, 6 0 log 6 7 ln 7 ln 6.9459.798.0860 MATH 80 Lecture 0 of Ronld Brent 08 All rights reserved.

The Nturl Logrithm In the lst lecture, we introduced the eponentil function e. Its inverse, log e, is clled the nturl logrithm. For simplicity, log e is denoted by the symbol ln. To get its grph, we flip e bout the line y. y y e y 4 (,e) (e,) 4 4 y ln 4 MATH 80 Lecture of Ronld Brent 08 All rights reserved.

Emple: Solve ln( ) 8. Apply e to get ( e 8 ), or ( ± e 4 ), we pick + sign since is rel so 4 ( ) e which cn be solved to give + 4 ± e. For most purposes you cn leve your nswer in this form. If its deciml pproimtion is needed, you cn use your clcultor to determine one. MATH 80 Lecture of Ronld Brent 08 All rights reserved.

Also mentioned in the lst lecture; ny eponentil k e for some constnt k. First write s e ln. So cn be written in the form (ln) ( e ln ) e (Recll: So k ln Tht s it! Now consider the following: m n mn ( ).) Emple: Write 4 in the form k e. 4 (ln4) 4 ( e ln ) e, nd since 4. 86 ln, 4 e. 86 Emple: Write ( 0.455) in the form k e. ln0.455 (ln0.455) ( 0.455) ( e ) e, nd since ln 0.455 0. 787 e 0. 787 (0.455), MATH 80 Lecture of Ronld Brent 08 All rights reserved.

Derivtives of logrithmic functions: To compute the derivtive of ln, recll tht y ln mens derivtive of both sides: y e. Tking the d d d e d y or e y dy d or nd using the chin rule: dy d y e (ln ) (ln u( )) u ( ) u( ) MATH 80 Lecture 4 of Ronld Brent 08 All rights reserved.

Emples: ) ( ) ln( ) f. f ( ) 4 b) f ( ) ln( + ). f 4 + 4 ( ) 4 + + + cos f. f ( ) cot sin c) ( ) ln(sin ) d) ( ) sin(ln ) f. f ( ) cos(ln ) ( f ( ) ()ln + ln + e) f ) ln MATH 80 Lecture 5 of Ronld Brent 08 All rights reserved.

f) f ( ). ln + f ( ) (ln + )() (/ (ln + ) ln + (ln + ) ) ln (ln + ) g) f ( ) ln ln + (ln f ( ) )( ln (ln ) + ) (ln (ln ) + )( ) MATH 80 Lecture 6 of Ronld Brent 08 All rights reserved.

Using the chnge of bse formul we cn write log ln ln And so d d d ln d log ln d ln ln d (ln ) (log u( )) u ( ) (ln ) u( ) f ( ) log, f ( ) (ln ) f ( ) log ( + 4), f ( ) (ln )( + 4) MATH 80 Lecture 7 of Ronld Brent 08 All rights reserved.

Let f ( ) ln. Find out everything bout this function nd then grph it. ) Domin: > 0 ) Intercepts: Note: lim ln 0. + 0 ) Symmetry: No Symmetry intercepts come from solving ln 0, so. 4) Asymptotes: Since lim f ( ), there re no symptotes. 5,6) Intervls of Increse nd Decrese. Rel. Etrem Since f ( ) ()ln + ln +, f ( ) < 0 for < e, nd f ( ) > 0 for > e decreses on < e. The is reltive min t ( ) e is criticl point. Also,. So f increses on e, e. > e, nd MATH 80 Lecture 8 of Ronld Brent 08 All rights reserved.

7,8) Since f ( ), nd since > 0, f ( ) > 0, nd hence the function is concve up everywhere nd there re no inflection points. ln() 0 8 6 4 0 0 4 5 - MATH 80 Lecture 9 of Ronld Brent 08 All rights reserved.

Let f ( ) ln. Find out everything bout this function nd then grph it. ) Domin: > 0 ) Intercepts: Note: lim ln 0. 0 ) Symmetry: No Symmetry + intercepts come from solving ln 0, so. 4) Asymptotes: Since lim f ( ), there re no symptotes. 5,6) Intervls of Increse nd Decrese. Rel. Etrem Since f ( ) ln + (ln + ), f ( ) < 0 for < e /, nd f ( ) > 0 for nd decreses on / e is criticl point. Also, / > e. So f increses on / < e. The is reltive min t / e, e. / > e, MATH 80 Lecture 0 of Ronld Brent 08 All rights reserved.

7,8) Since f ( ) (ln + ) + ln +, e / is n inflection point, becuse f ( ) > 0 for / / >e, nd f ( ) < 0 for <e. So f is / / concve up on >e, nd concve down on <e. The point of inflection is ^*ln() e /, e..0 The function looks like:.5.0.5.0 0.5 0.0 0.0 0.5.0.5.0-0.5 MATH 80 Lecture of Ronld Brent 08 All rights reserved.

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