Logarithmic Functions
The function ex is the unique exponential function whose tangent at (0, 1) has slope 1.
Write in log form. Either (method 1: easy problems) use th
Lecture
The number e1 =Math
e ≈ 2.7 140
and hence
2 < e < 3 12
⇒ the graph of ex lies between the graphs of
definition
of log or
( method
2: hard
problems)
takeuse
thethe
Write
in log form.
Either
(method
1: easy
problems)
xExam 2xcovers Lectures
Math
140
Lecture
12
7
-12.
Study
the
recommended
exercises.
2 and 3 .
definition
of
log
or
(
method
2:
hard
problems)
take
the
appropriate log of both sides and cancel inverses.
Review
area, circumference,
formulas
- inside front
cover.
Exam
2 covers
Lectures 7 -12.volume
Study the
recommended
exercises.
3
x
x
appropriate
Review
area,
circumference,
volume
formulas
inside
front
cover.
`
2
8 log of both sides and cancel inverses.
−x e .
RECALL
Thegraphs
graphs
e and
Below
are.the
of of
ex and
e
.
3
x
x
`
2
8
RECALL. The graphs of e and e .
Method 1. Since 2x and log2(x) are inverse,
Method
3 log1. Since
8 2x and log2(x) are inverse,
2
`5 2x 31 log
6 28
2x
1
x
`5 Method
6 2. Take the log of both sides.
y=e
x
2x Take
1
Method
the
of both sides. log 6
log 5 52.
loglog
y=e
5 6 so 2x 1
5
slope=1
2x 1
log 5 5
log 5 6 so 2x 1 log 5 6
slope=1
Solve for x, write the answer using logarithms.
1
2
Solve
1
`5x2 for4x, write the answer using logarithms.
`5x x 24 log 4
5
y=0 h.a.
xx2 loglog
Leave answers in exact form, no d
5 45 4
y=0 h.a.
x-intercept
none
y-intercept
1
x
log
4
Leave
answers in exact form, no d
3t
1
5
`e
8
x-intercept
none
y-intercept
1y = 0
vert. asym.
none
hor. asym.
3t
1
`e 3t 81 ln 8
vert.
asym.
none
hor.
asym.
y(0,5),
=0
domain
(-5,5)
range
3t
3t 1ln 8ln 8 1
domain
range
(0,5),
`Graph y (-5,5)
e x 1 1.
x
1
3t
t 13lnln8 8 1 1
`Graph
.
y domain,
e
1range,
Give the
intercepts and asymptotes.
t 1 ln
1
1 8
Give theGraph
domain,
and
Example:
f (x) range,
= ex−1 intercepts
− 1 and give
theasymptotes.
domain, range, intercepts,
`3 2x and53xasymptotes.
Use natural logarithms and method 2
1 natural logarithms and method 2.
`3 2x ln53x 2x1 ln 5xUse
2x
x 1
ln
2x3ln 3 lnx 5 1 ln 5
2x
x 2lnln3 3 xln 51 lnln55
1
x 2 ln 3 ln 5 ln=5 ln 5 See Practice Exam 2, pr
x ln 5/ 2 ln 3 ln 5 ln
ln 9/5
5
1
-1
y=-1
x ln 5/ 2 ln 3 ln 5 = ln 9/5 See Practice Exam 2, pr
Since they are inverses, the graph of log b x is the
h.a.
-1
y=-1
Since
they areofinverses,
is theln
log b x and
h.a.
reflection
bx acrossthe
thegraph
majorofdiagonal
x-intercept x = 1
x
reflection
of b ofacross
1
1
the reflection
ex. the major diagonal and ln
x-intercept
y-intercept xy==1e 1 2.7 1
.63
1
1
the reflection of ex.
y-intercept
.63
hor. asym. yy==e-1 1 2.7 1
-x
y= e
-x
y= e
hor.
asym.
domain
domain
y(-5,5)
= -1
(-5,5)
range
range
(-1,5).
(-1,5).
x
y=e
x
y=e
Logarithms
Logarithms
Assume
> 0,
b =a1.
bx is
and
it has
has an
an inverse.
inverse.
Assume
a >b0,
and
6= Thus
1. Then
ax 1-1
is 1-1
and
y=ln(x)
1
x
Assume
b >.0,
b = 1.
Thus
b of
is 1-1
and
it
hasb,anisinverse.
y=ln(x)
1
DEFINITION
,
the
log
x
to
the
base
the
log
x
b
DEFINITION
.oflog
, the log of function
x to the base
the
xexponential
1 y=0 h.a.
b
inverse
the
bx. a,b,is isthe
Definition: loga (x),
the log of x to the base
inverse of the exponential function
x
y=0 h.a.
1
x inverse of the exponential function b .
x
a . ln(x), the natural logarithm, = log x = the inverse of e .
e
x
x
The
natural
logarithm,
ln(x) =
=log
loge (x),
is the inverse of the function
x=0f (x)
v.a. = e .
ln(x),
the isnatural
logarithm,
Note, “ln”
“el-n” not
“one-n”, not
“eye-n”.
e x = the inverse of e .
x=0 v.a.
Note, “ln” is “el-n” not “one-n”, not “eye-n”.
For the graph of ln(x)
Remember
is in
theopposite
inverse ofdirections
the function
if and cancel.
only if f t
(y) = x. So,
Inverses yact
andf (x)
inverses
x-intercept
1
For
the graph of ln(x)
Inverses
opposite
t x-intercept
y = and
ln(x)inverses
iff e y cancel.
b y x.directions
x.
y logactxiniff
vert. asym.
x = 0.
1
• y = logba (x) iff ayy = x
y
x
yln(e
= ln(x)
x.
ylog log
xb x iff b
vert.
asym.
x(0,5)
= 0.
domain
) = x.iff e x.
x
b b x
x
x )=x
• log
abx(a
domain
(0,5)
ln(x)
ln(e
)
=
x.
log
x
log
`Graph
y
ln
x 1
bb
e = x.
b
x
b x (x) =
•b log
alog
eln(x) = x.
xx x
If we have ln(b ) instead of ln(ex), then ln and bx don’t
x
x
x
x ), then ln and b don’t
If we
have ln(bcancel
) instead
ln(e
completely
andof
ln(e
) = x becomes:
completely
cancel
andexponent
ln(ex) =comes
x becomes:
.
The
down to the outside.
ln bx xlnb
x
ln b 0 xlnb. The exponent comes
1 down to the outside.
FACT. e = 1 ˆ 0 = ln(1).
e = e ˆ 1 = ln(e)
1 1 = ln(e)
FACT. e0 = 1 ˆ 0 = ln(1).
e1 = e ˆ
`Simplify to a rational.
`Simplify
log 5 5 5 to a rational.
log 5 5 5 log 5 5 1 5 1/2 log 5 5 3/2 32 .
log 51 5
log 51 5 log1 5 1 5 1/2 log 5 3/2 32 .
a
`Graph y ln x 1
1.
1.
y-intercept
hor. asym.
y-intercept
hor.
asym.
range
range
x=-1
vertx=-1
asym
vert asym
none
none
none
none
(-5,5).
(-5,5).
1
1
-..63
-..63
y-intercept: ln(0+1)+1 = ln1 +1 = 0+1 = 1
x-intercept:ln(0+1)+1
ln(x+1)+1==ln1
0 iff
y-intercept:
+1 ln(x+1)
= 0+1 ==1-1 iff x +1 = e-1
x-intercept:
= 0 iff -.63
ln(x+1) = -1 iff x +1 = e-1 i
x = 1/e ln(x+1)+1
- 1 1/(2.7)-1
vert.
hor. asym: none
x =asym:
1/e -x1= -11/(2.7)-1 -.63
Thus for the natural logarithm we have,
• y = ln(x) iff ey = x
• ln(ex ) = x
• eln(x) = x
Example: Simplify to a rational function:
√
1. log5 (5 5) √
1
3
⇒
log5 (5 5) = log5 (51 · 5 2 ) = log5 (5 2 ) =
3
2
2. log2 ( 18 )
⇒
log2 ( 18 ) = log2 ( 213 ) = log2 (2−3 ) = −3
In order to simplify a logarithmic function, the logarithm and exponential function need to
have the same base.
Example: To simplify log8 (2), first we need to write 2 in terms of an exponential function
with base 8:
8x = 2
⇐⇒
1
Thus, log8 (2) = log8 (8 3 ) =
(23 )x = 21
⇐⇒
3x = 1
⇐⇒
1
3
Solving functions using logarithms:
2
• 5x = 4
2
⇒
log5 (5x ) = log5 (4)
⇒
x2 = log
p5 (4)
⇒
x = ± log5 (4)
• e3t+1
⇒
⇒
⇒
⇒
=8
ln(e3t+1 ) = ln(8)
3t + 1 = ln(8)
3t = ln(8) − 1
t = ln(8)−1
3
Solving logarithmic functions:
• log5 (3x + 7) = 2
⇒
5log5 (3x+7) = 52
⇒
3x + 7 = 52
⇒
3x = 25 − 7
⇒
x=6
• logx (49) = 2
⇒
xlogx (49) = x2
⇒
49 = x2
⇒
x = 7 since the base of a logarithm is always positive.
2
x=
1
3
x 2 ln 3 ln 5 ln 5
2x ln 3 x 1 ln 5 ln 5
x xln2 5/
3 5 ln 5ln=5ln 9/5 See Practice Exam 2, prob. 11b.
ln 32 lnln
ln 5
Since
graph of log x is the
x lnthey
5/ 2are
ln 3inverses,
ln 5 = lnthe
9/5 See PracticebExam 2, prob. 11b.
x
reflection
b acrossthe
thegraph
majorofdiagonal
ln(x) is
Since
they logarithmic
areofinverses,
isare
theinverses,
log bthey
x and
Graphing
functions:
Since
the
x
x
the reflection
. major diagonal.
xof ethe
reflection
of
b
across
reflection of b across the major diagonal and ln(x) is
the reflectiony=e
of xex.
x
y=e
an inverse.
is the
n inverse.
is the
inverse of ex.
x
nverse of e .
es cancel. t
e y x.
es cancel. t
y
x.
d bx don’t
bx don’t
the outside.
3
2.
3
.
533/2
2.
3x 1, x
3.
3x 1, x
1 1
x=0 v.a.
1
y=0 y=ln(x)
h.a.
y=0 h.a.
Forx=0
the graph
v.a. of ln(x)
x-intercept
1
For
graph of ln(x)
vert.theasym.
x = 0.
x-intercept
1(0,5)
domain
vert.
asym.
x
`Graph y ln=x0. 1
domain
(0,5)
y-intercept
hor. asym.
y-intercept
range
hor.
asym.
1.
range
x=-1
`Graph
. + 1) +
ln x f (x)
1 =1ln(x
Example:y Graph
1. asym
vert
x=-1
vert asym
none
none
none
(-5,5).
none
(-5,5).
1
-..63
1
= ln(e)
he
outside.
= ln(e)
3/2
55
y=ln(x)
1
1
3
1
3
y-intercept: ln(0+1)+1 = ln1 +1 = 0+1 = 1
-..63
x-intercept: ln(x+1)+1 = 0 iff ln(x+1) = -1 iff x +1 = e-1 iff
y-intercept:
= ln1 +1-.63
= 0+1 = 1
x = 1/e ln(0+1)+1
- 1 1/(2.7)-1
x-intercept:
= 0 iff ln(x+1)
= -1none
iff x +1 = e-1 iff
vert. asym: ln(x+1)+1
x = -1
hor. asym:
x = 1/e(-1,
- 15) 1/(2.7)-1 -.63
domain:
range: (-5,5).
vert.
asym:
x
=
-1
hor.
asym:
none
`Graph y 1 ln 1 x . refl. across
y, right 1, refl. across x, up 1.
domain: (-1, 5)
range: (-5,5).
The 5 log properties of Lecture 13 are needed for Exam 2.
`Graph y 1 ln 1 x . refl. across y, right 1, refl. across x, up 1.
The 5 log properties of Lecture 13 are needed for Exam 2.
Properties of Logarithms
Let a > 0, a 6= 1, and x, y > 0.
• loga (xy) = loga (x) + loga (y)
• loga ( xy ) = loga (x) − loga (y)
• loga (xr ) = r · loga (x)
Write as a single logarithm:
• 2 · log10 (x) + log10 (y)
= log10 (x2 ) + log10 (y)
= log10 (x2 y)
• log2 (x) − 4 log2 (y)
= log2 (x) − log2 (y 4 )
= log2 ( yx4 )
• ln(x2 + y 2 ) − ln(y 3 + a)
2
2
= ln( xy3+y
+a )
***Note: loga (xy) = loga (x) + loga (y) 6= loga (x + y)
3
graph of logb (x) is the
Write as a sum/ difference/ multiple of the simplest possible logarithms:
14 • logb y32y
+a
= 41 · logb ( y32y
+a )
1
= 4 [logb (2y) − logb (y 3 + a)]
= 14 [logb (2) + logb (y) − logb (y 3 + a)
• ln √ 21 2
x +y
1
= ln[(x2 + y 2 )− 2 ]
= − 12 · ln(x2 + y 2 )
Theorem:
loga M = loga N iff M = N
Solve for x: (Remember to check that the solutions don’t give undefined logarithms.)
• log2 (4x) − log2 (3) = log2 (x + 2)
⇒ log2 (4x) − log2 (x + 2) = log2 (3)
4x
⇒ log2 ( x+2
) = log2 (3)
4x
⇒ x+2 = 3
⇒ 4x = 3(x + 2) = 3x + 6
⇒ 4x − 3x = 6
⇒ x=6
Check: log2 (24) and log2 (8) are both defined.
• log5 (x + 6) + log5 (x + 2) = 1
⇒ log5 [(x + 6)(x + 2)] = 1
⇒ (x + 6)(x + 2) = 5
⇒ x2 + 8x + 12 = 5
⇒ x2 + 8x + 7 = 0
⇒ (x + 1)(x + 7) = 0
⇒ x = −1 or −7
Check: log5 (−7 + 6) is undefined. log5 (−1 + 6) and log5 (−1 + 2) are both defined.
Thus, the solution is x = −1.
• ln x = ln(x + 6) − ln(x − 4)
⇒ ln x = ln[(x + 6)/(x − 4)]
⇒ x = (x + 6)/(x − 4)
⇒ x(x − 4) = x + 6
⇒ x2 − 4x = x + 6
⇒ x2 − 5x − 6 = 0
⇒ (x − 6)(x + 1) = 0
⇒ x = −1 or 6
Check: ln(−1) is undefined. ln 6, ln 12 and ln 2 are all defined. Thus, the solution is
x = 6.
4
Solve for x using natural logarithm.
5x−2 = 33x+2
ln 5x−2 = ln 33x+2
(x − 2) ln 5 = (3x + 2) ln 3
x ln 5 − 2 ln 5 = 3x ln 3 + 2 ln 3
x ln 5 − 3x ln 3 = 2 ln 3 + 2 ln 5
x(ln 5 − 3 ln 3) = 2 ln 3 + 2 ln 5
3+2 ln 5
x = 2lnln5−3
ln 3
Change of Base Formula: loga (x) =
Proof:
logb (x)
logb (a)
=
logb (aloga (x) )
logb (a)
=
logb (x)
logb (a)
loga (x)·logb (a)
logb (a)
= loga (x)
Examples:
• Express log4 (2) in terms of log3 :
3 (2)
⇒ log4 (2) = log
log (4)
3
• Express ln(2) in terms of log5 :
5 (2)
⇒ ln(2) = loge (2) = log
log (e)
5
• Express log5 (11) in terms of natural logarithms:
⇒ log5 (11) = ln(11)
ln(5)
5