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THE LOGARITHMIC FUNCTION WITH BASE b is thefunction

 y   = logb x

b is no!"#ll$ # nu"%e! g!e#te! th#n & '#lthough it nee(onl$ %e g!e#te! th#n ) #n( not e*u#l to &+ The function is(e,ne( fo! #ll x - ) He!e is its g!#.h fo! #n$ %#se b

Note the follo/ing0

1 Fo! #n$ %#se2 the x3inte!ce.t is & Wh$4To see the answer, pass your mouse over thecolored area.To cover the answer again, click "Refresh"("Reload").

The logarithm of 1 is 0.  y  = logb1= 0.

1 The g!#.h .#sses th!ough the .oint 'b2 &+ Wh$4

The logarithm of the base is 1. logbb = 1.

1The g!#.h is %elo/ the x3#5is 33 the log#!ith" isneg#ti6e 33 fo!

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) 7 x 7 &

 

Which nu"%e!s #!e those th#t h#6e neg#ti6elog#!ith"s4

Proper fractions.

1The function is (e,ne( onl$ fo! .ositi6e 6#lues of x

 

logb'89+2 fo! e5#".le2 "#:es no sense Since b is #l/#$s.ositi6e2 no .o/e! of b c#n .!o(uce # neg#ti6e nu"%e!

1 The !#nge of the function is #ll !e#l nu"%e!s

1 The neg#ti6e y 3#5is is # 6e!tic#l #s$".tote 'To.ic&;+

Example 1. Translation of axes.  He!e is the g!#.h ofthe n#tu!#l log#!ith"2  y  = ln x  'To.ic <)+

 An( he!e is the g!#.h of  y  = ln 'x 8 <+ 33 /hich isits translation < units to the !ight

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The x3inte!ce.t h#s "o6e( f!o" & to An( the 6e!tic#l #s$".tote h#s "o6e( f!o" ) to <

Problem 1.  S:etch the g!#.h of y  = ln 'x > +

This is a translation 3 units to the left. The x -intercept has moved from 1 to −. !nd the verticalas"mptote has moved from 0 to −3.

Exponential functions

B$ (e,nition0

logb y  = x  "e#ns bx = y 

In othe! /o!(s2 co!!es.on(ing to e6e!$ log#!ith"function /ith %#se bthe!e is #n e5.onenti#l function /ith%#se b0

 y   = bx

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It is (e,ne( fo! e6e!$ !e#l nu"%e! x He!e is itsg!#.h0

The!e #!e t/o i".o!t#nt things to note01 The y 3inte!ce.t is #t ')2 &+ Fo!2 b) = &

1 The neg#ti6e x3#5is is # ho!i?ont#l #s$".tote Fo!2/hen x is # l#!ge neg#ti6e nu"%e! 33 eg b8&)2))) 33 then y  is # 6e!$ s"#ll .ositi6e nu"%e!

Problem .

#+ Let f 'x+ = ex W!ite the function f '8x+

 f #−x $ = e− x 

The argument  x  is replaced b" − x .

%+ Wh#t is the !el#tionshi. %et/een the g!#.h of  y  = ex #n( the g!#.h%+ of  y  = e8x 4

 y  = e−x 

 is the re%ection about the y-axis  of y  = e x .

c+ S:etch the g!#.h of y  = e8x

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Inverse relations

E5.onenti#l functions #n( log#!ith"ic functions /ith%#se b #!e in6e!ses

The functions logbx  #n( bx  #!e in6e!ses

Fo! in #n$ %#se b0

i+ blogbx = x2

#n(

ii+ logbbx = x

Rule i+ e"%o(ies the (e,nition of # log#!ith"0 logbx isthe exponentto /hich b "ust %e !#ise( to .!o(uce x

Rule ii+ /e h#6e seen in the .!e6ious To.ic

No/2 let

 f 'x+ = bx  #n( g'x+ = logbx

Then Rule i+ is  f 'g'x++ = x

 An( Rule ii + is g' f 'x++ = x

These !ules s#tisf$ the (e,nition of # .#i!of in6e!se functions 'To.ic &@+ The!efo!e fo! #n$ %#se b2the functions

 f 'x+ = bx  #n( g'x+ = logbx

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#!e in6e!ses

Problem 3.  E6#lu#te the follo/ing

  #+

 log<< = &

%+ log

&).<

 = '.

c+ ln ex >

&

 

= x  (

1

(+ <log<

 = & e+ &)log &)) = 100

f+ eln 'x 8

+= x  −&

Problem ). 

#+ Wh#t function is the in6e!se of  y = ln x4

 y  =

e

 x 

.%+ Let  f 'x+ = ln x  #n( g'x+ = ex2 #n( sho/th#t f  #n( g s#tisf$ the%+ in6e!se !el#tions

 f #g# x $$ = lne x  = x *

g# f # x $$ = eln

 x 

 = x .

He!e #!e the g!#.hs of  y  = ex  #n(  y  = ln x 0

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 As /ith #ll .#i!s of in6e!se functions2 thei! g!#.hs#!e s$""et!ic#l /ith !es.ect to the line  y  = x  'See To.ic&@+

Problem &.  E6#lu#te ln e#!ccos '8&+

ln earccos #−1$ = arccos #−1$ = +.

,The angle hose cosine is −1

is +.,

See To.ic &@ of T!igono"et!$

Exponential and logarithmic equations

Example .  Sol6e this e*u#tion fo! x 0

x > & = <

 olution To !ele#se x > & f!o" the e5.onent2 t#:e thein6e!se function 33 the log#!ith" /ith %#se  33 of %othsi(es E*ui6#lentl$2 /!ite the log#!ith"ic fo!" 'To.ic <)+

log-x > &  = log< 

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x > & = log< 

x > & = 9

 

x  =

Example 3.  Sol6e fo! x 0

<x 8 9  = x

 olution We "#$ t#:e the log of %oth si(es eithe! /iththe %#se < o! the %#se Let us use %#se <0

log<<x 8 9  =

 log<x

 

x 8 9  = 

x log<2 #cco!(ing to the !(L#/

 

x 8 x log<  = 

9

 

x'& 8 log<+  = 

9

 

x  = 

9& 8 log<

log< is so"e nu"%e! The e*u#tion is sol6e(

Problem '.  Sol6e fo! x 0

<x 8   = < 

log x  − &  = log3 

 x  − & = & 

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 x   = 10

Problem /.  Sol6e fo! x The solution "#$ %e e5.!esse(#s # log#!ith"

&)x 8 &  = <<x > &

log 103 x  − 1  = log  x  ( 1

 

3 x  − 1 = # x  ( 1$ log  

3 x  − 1 =  x  log ( log  

3 x  −  x  log = 1 ( log  

 x #3 − log $ = 1 ( log  

 x   =1 ( log 3 − log

Problem .  Sol6e fo! x 0

esin x  = &

 

ln esin x   = ln 1 

sin x   = 0 

 x  is the radian angle hose sine is 0 

 x   = 0.

Example ).  Sol6e fo! x0

log'<x > + = olution.  To f!ee the #!gu"ent of the log#!ith"2 t#:ethe in6e!se function 33 x 33 of %oth si(es Th#t is2 let e#chsi(e %e the e5.onent /ith %#se E*ui6#lentl$2 /!itethe e5.onenti#l fo!"

<x > =

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<x  = &< 8  

<x  = &<< 

x  = &

Problem 2.  Sol6e fo! x 0

log9'x 8 + = ) 

f e let each side be the exponent ith base )*then 

3 x  − &  =

 )0  = 1

 

3 x   = 

' 

 x   = 

Problem 10.  Sol6e fo! x 0

log<'xD > + = 9 

 x 4 ( / = )  = 1' 

 x 4 = 1' − / = 2 

 x   = 53

Example &.  Sol6e fo! x0

log '<x > &+ = log &&olution If /e let e#ch si(e %e the e5.onent /ith &) #sthe %#se2 then #cco!(ing to the in6e!se !el#tions0

<x > &= &&

Th#t i".lies

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x=

Problem 11.  Sol6e fo! x0

ln 'x 8 &+ = ln '<x > ;+

f e let each side be the exponent ith base e*then

& x  − 1  =  x  (  

3 x   = 2 

 x   = 3.

S:ill in Alge%!#2 Lesson @

Creating one logarithm from a sum

Example '.  Use the l#/s of log#!ith"s 'To.ic <)+ to/!ite the follo/ing #s one log#!ith"

log x  > log y   8 < log  

  olution log x  > log y   8 < log     = log xy   8 log  D

 

= logxy   D

Problem 1.  W!ite #s one log#!ith"0

k log x  > m log y   8 n log  

Problem 13.  W!ite #s one log#!ith"0

log '<x 8 ;+ 8 log 'xD 8 &+

log '<x 8 ;+ 8 log 'xD 8 &+ = log x −  x 4 − 1'

 

= log # x − )$

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# x  − )$# x  ( )$ 

= log

 x  ( )

Example /.  B$ "e#ns of Rule i #%o6e 33

n  = logbbn2

33 /e c#n /!ite #n$ nu"%e! #s # log#!ith" in #n$ %#se

Fo! e5#".le2

= log<<

 

@ = log.@

 

t  = ln et

 

= log &)))

Problem 1). 

#+ < = ln e4   %+ & = ln e

Example .  W!ite the follo/ing #s one log#!ith"0

logbx  > n

  olution. logbx  > n   = logbx  > logbbn

 

= logbxbn

Problem 1&.  W!ite #s one log#!ith"0

log < >

log ( 3 = log ( log 103

 

= log 6 103

 

= log 000

Problem 1'.  W!ite #s one log#!ith"0

ln A 8 t

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ln ! − t   = ln ! − ln et

 

= ln ! ( ln e−t

 

= ln ! e−t

Problem 1/.  Sol6e fo! x0

log<x  > log<'x > <+  = 

 

log7 x # x  ( $8  =

 

3. 

f e no let each side be the exponent ith base *then 

 x # x  ( $  = 

3 = .

 

 x 4 (  x  −   =

 

0

 

# x  − $# x  ( )$  = 

0

 

 x   = 

or −).

See S:ill in Alge%!#2 Lesson

We "ust !eect the solution x = 8 92 ho/e6e!2 %ec#usethe neg#ti6e nu"%e! 89 is not in the (o"#in of log<x

Problem 1.  Sol6e fo! x

ln '& > x+ 8 ln '& 8 x+  = 

&

 

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  =  1.

 

f e no let each side be the exponent ith base e*then 

=  e

 

1 ( x   =  e − e x 

 

e x  ( x 

  =

  e − 1 

#e ( 1$ x   =  e − 1

 

 x   =