Logarithms & Logarithmic
Functions
for _ f ( x) = b , where_ b > 1
x
One-to-One function
Includes (0,1)
Domain: all Real Numbers
We have seen that the exponential function is one-to-one, so
it has an inverse.
It s inverse is the logarithmic function.
A LOGARITHM IS AN EXPONENT!
Exponential _ form:
f ( x ) = b x → example → 52 = 25
In logarithm terms, 2 is the LOGARITHM of 25 for a
base of 5.
Exponential _ form:
example → 33 = 27
In logarithm terms, 3 is the LOGARITHM of 27 for a base of 3.
example → 4 = 8
1.5
In logarithm terms, 1.5 is the LOGARITHM of 8 for a base of 4.
Definition of a LOGARITHM:
The _ LOGARITHM _ of _ x _ with _ base _ b,
( which _ equals_ y )_ is _ defined _ by:
y = logb x _ if _ and _ only _ if _ b y = x
for _ every _ x > 0_ and _ for _ every _ real _ number _ y
The logarithm y is the exponent to which b is raised to get x.
Notice the difference between exponential
form and logarithmic form.
Logarithmic _ Form:
log 2 16 = 4
1
log5
= −2
25
Exponential _ Form:
2 4 = 16
5−2 =
1
25
Exponential _ Form:
Logarithmic _ Form:
log10 17 = t
10 = 17
log 3 (2 x − 5) = y
3y = 2 x − 5
t
Since, exponential functions are one-to-one with
b > 0 and b not equal to one it follows that . . . .
b = b _ if _ and _ only _ if _ u = v
u
v
For instance . . . .
2 x −1
if _ 3 = 3
x
_ then.....
x = 2 x − 1, so _ x = 1
Example Evaluating Logarithm problems:
log 4 64 = ?
In other words, 4 to what power equals 64?
4 = 64, _ so _ log4 64 = 3
3
log e e = ?
4
In other words, e to what power equals e to the fourth power?
e4 = e4 , _ so_ loge e4 = e4
Since_ b = 1_ for _ b ≠ 0_ and _ b = b......
0
logb 1 = 0
logb b = 1
1
Examples of solving logarithmic equations ……..
Often it is easiest to change the equation to
exponential form to solve.
log 3 x = 4, find _ x
3 = x, so_ x = 81
4
1
log 4 (2 x − 1) = , find _ x
2
1
2
4 = 2x − 1
2 = 2x − 1
3
3 = 2 x , so _ x =
2
One more ……..
Find _ y _ for:
y = 5 + 8 log e (3x )
1. First for the log term.
y −5
log e (3x) =
8
2. Second, put in exponential form and solve for x.
y −5
log e (3x ) =
8
e
y −5
8
1
= 3x , so _ x = e
3
y −5
8
Base 10 is the common logarithm
and can be determined using the LOG
button on your calculator
1.653212514
10
−.3010299957
10
≈.5
≈ 45
Base e is the natural logarithm and
can be determined using the LN
button on your calculator
e
e
−.6931471806
≈.5
3.80666249
≈ 45
Find _ the_ inverse_ of _ f ( x) = 2 x
1. Put in logarithmic form
y = 2 → log2 y = x
x
2. Exchange the x and y variables
log 2 x = y → this_ is_ the _ inverse _ of _ f ( x )
In General Terms …….
The _ inverse _ of _ the _ exp onential _ function _ f ( x ) = b x
is_ the _ log arithmic _ function _ defined _ by _ g ( x ) = logb x
Exponential _ Function :
Inverse( LOG )_ Function:
f (x) = 4 x
f −1 ( x ) = log 4 x
! 1$
f (x) = # &
" 3%
f (x) = 10
f (x) = e x
x
x
f −1 ( x ) = log 1 x
3
f −1 ( x ) = log x
f −1 ( x ) = ln x
Exponential _ Function :
Inverse( LOG )_ Function:
f (x) = 4 x
f −1 ( x ) = log 4 x
! 1$
f (x) = # &
" 3%
f (x) = 10
f (x) = e
x
x
x
f −1 ( x ) = log 1 x
3
f −1 ( x ) = log x
f −1 ( x ) = ln x
Domain exponential function = Range logarithmic
function
Range exponential function = Domain logarithmic
function
so, Logarithmic _ function _ Domain ! x _ in _ int erval _(0, ")
Range ! y _ in _ int erval _ ( #", " )
That does not mean the domain is
always 0 to infinity, though.
Consider..... f ( x ) = log(2 x − 1)
2x − 1 > 0
sin ce_ there_ is_ NO_ x _ such _ that _10 x = 0_ or _10x < 0
1
#1 &
2x !1 > 0, so _ x > , Domain _ is _ thus _ % , "(
$2 '
2
y = ex
y = ln( x)
Symmetric about y = x line, property of inverse functions
Log graphs contain the point (1,0)
General shape of LOG graph where the base is greater
than 1
g ( x ) = log 7 ( x )
f ( x) = log 3 ( x )
The log with
the smaller
base will be
on top.
Think Why.
More accurately,
treats the y-axis as
a vertical
asymptote
General shape of LOG graph where the base is between 0
than 1
f ( x ) = log 1 ( x )
The log with
the larger
base will be
on top above
the x-axis and
below
beneath the xaxis.
Think Why.
3
g ( x ) = log 1 ( x )
2
Transformations of LOG functions:
Transformations of LOG functions:
From the definition of a logarithm we know . ….
b = x _ equals_ logb x = y
y
Substituting the second equation s value for y
into the first equation, we get …….
b
logb x
= x_ x > 0
So for the typical bases used you get …………
log x
10
= x _ and _ e
ln x
= x_ x > 0
So you have the following:
10
e
e
=6
ln 6
10
= 100
log 100
log 4 x
2
ln x +5
= 4x
= x +5
2
Basic Properties of Logarithms:
Suppose _ that _ M , N , and _ b _ are _ positive
real _ numbers, where _ b ≠ 1, and _ r _ is_ any
real _ number. Then _ we _ have.......
Pr oduct _ Rule:
logb MN = logb M + logb N
example _ of _ product _ rule:
log 2 16 = log 2 8 + log 2 2
and _ log 2 16 = log 2 4 + log 2 4
Basic Properties of Logarithms:
Suppose _ that _ M , N , and _ b _ are _ positive
real _ numbers, where _ b ≠ 1, and _ r _ is_ any
real _ number. Then _ we _ have.......
Quotient _ Rule:
M
logb
= logb M − logb N
N
example _ of _ quotient _ rule:
log 2 16 = log 2 32 − log 2 2
and _ log 2 8 = log 2 32 − log 2 4
Basic Properties of Logarithms:
Suppose _ that _ M , N , and _ b _ are _ positive
real _ numbers, where _ b ≠ 1, and _ r _ is_ any
real _ number. Then _ we _ have.......
Power _ Rule:
logb N = r logb N
r
example _ of _ power _ rule:
log 2 16 = 4
log 2 16 = log 2 4 2 = 2 log 2 4 = 4
The Power Rule is crucial because it
lets us directly solve exponential
functions.
Power _ Rule:
logb N r = r logb N
17 x = 367
one − to − one _ function, so _ take _ the _ log_ of _ both _ sides
log 17 x = log 367
x log 17 = log 367
log 367
x=
≈ 2.084333628
log 17
Also from the Power Rule we get the following results …..
logb b = r
r
so,log 10 = r _ and _ ln e = r
r
r
Problems using the basic properties of Logs:
Writing expressions as sums & differences of logs
x
log8 = log8 x − log8 5
5
x
log8 =
5
(
7 11
log x y
)=
! (y + 7)3 $
ln #
&=
y %
"
(
)
log x 7 y11 = log x 7 + log y11
(
7 11
log x y
) = 7log x +11(log y)
1
! (y + 7)3 $
1
3
2
ln #
& = ln(y + 7) ' ln y = 3ln(y + 7) ' ln y
2
y %
"
Problems using the basic properties of Logs:
Combining sums and differences of logs
2 ln x − 3 ln y =
2 ln x − 3 ln y = ln x 2 − ln y 3
x2
2 ln x − 3 ln y = ln( 3 )
y
log(c − cd ) − log(2c − 2d ) =
2
2
"
(c
! cd) %
log(c 2 ! cd) ! log(2c ! 2d) = log $
'
(2c
!
2d)
#
&
" (c 2 ! cd) %
" c(c ! d) %
( c+
= log $
= log $
= log * '
'
) 2,
# 2(c ! d) &
# (2c ! 2d) &
Because Logs are one-to-one functions, we can use the
following formulas to help solve equations…….
1)_ _ If _ M = N , then _ logb M = logb N
Applying it …..
10 = 257, then _ log 10 = log 257
x
x
so, x log 10 = log 257 → x = log 257
2)_ _ If _ logb M = logb N , then _ M = N
log5 (2 x + 8) = log5 125, then _ 2 x + 8 = 125
More advanced equations to solve …….
log3 ( x + 1) + log 3 ( x + 3) = 1, solve_ for _ x,det er min e_ legitimate_ answers
1. Rewrite the equation so you have one log expression on
one side of the equation.
log3 ( x + 1)( x + 3) = 1
2. Convert the equation to exponential form.
3 = ( x + 1)( x + 3)
1
3. Solve the equation.
x2 + 4x + 3 = 3
x2 + 4x = 0
x ( x + 4) = 0, so _ x = 0_ and _ x = −4
−4 _ not _ possible _ as_ log 3 ( −3)_ is_ not _ a _ real _ number
Change of Base Formula: Very helpful to change to common or natural logs from
other bases
log a x
logb x =
log a b
log 12 ln 12
log 3 12 =
=
log 3
ln 3
If you forget the formula it is easily derived.
1. Start with the definition of the log, then change to
exponential form.
log b x = y
b =x
y
2. Now, take the log of each side of the equation.
log a b = log a x
y log a b = log a x
y
3. Solve for y.
log a x
y=
log a b
Solving equations with one exponential expression:
42 x −1 = 3, solve_ for _ x
1. Take the Log of each side
log 4
2 x−1
= log 3
2. Use the power rule to get rid of the exponent.
Power _ Rule:_ log N r = r log N
(2 x − 1) log 4 = log 3
3. Solve for x
(2x !1)log 4 = log3
log3
log3
2x !1 =
" 2x =
+1
log 4
log 4
1 # log3 &
x= %
+1(
2 $ log 4 '
Another equations with one exponential expression:
e
−3t
=.5, solve_ for _ t
1. Take the Natural Log of each side
ln e
−3t
= ln.5
2. Use the power rule to get rid of the exponent.
Power _ Rule:_ log N r = r log N
−3t ln e = ln.5
3. Remember, ln e = 1. Now, Solve for x.
− ln.5
−3t (1) = ln.5 → t =
3
Solve equations with TWO exponential expression:
3 x −2
5
= 3 , solve_ for _ x
x
1. Take the Log of each side, and
remove the exponent
log 53 x −2 = log 3x
(3x − 2) log 5 = x log 3
2. Distribute terms and get all terms with an x to one side.
(3x − 2) log 5 = x log 3
3x log 5 − x log 3 = 2 log 5
3x log 5 − 2 log 5 = x log 3
x (3 log 5 − log 3) = 2 log 5
3x log 5 − x log 3 = 2 log 5
2 log 5
x=
3 log 5 − log 3
3. Factor out x, divide by remaining term.