Exponentials and Logs
Prepared by: Sa’diyya Hendrickson
Name:
Date:
Consider the following definitions:
1 Exponential function (base b): Let b be a positive real number such that b ∕= 1.
Then, the function f defined by f (x) = bx is called an exponential function (base b)
for all x ∈ R.
E.g. f (x) = 2x is the exponential, base 2.
2 Logarithmic function (base b): Let b be a positive real number such that b ∕= 1.
Then, the function g defined by g(x) = logb x is a logarithmic function (base b),
where y = log b x is called a logarithm and has the following meaning:
y = log b x is the exponent that when placed on the base b, the result is x.
Symbolically:
y = log b x
⇐⇒
by = b
log b x
=x
! "
E.g. g(x) = log 2 x is the logarithm, base 2. By definition, log 2 12
is the! exponent
that when placed on the base 2, the result is 12 . Thus,
"
1
log 2 2 = −1 since 2−1 = 12
It’s Halloween all year round for a logarithm, and it’s costume holds clues to help you identify
which exponent it is!
y = log b x says . . .
“I am the exponent y that when placed on the base b, the result is x!”
Consider the following examples:
a) log 3 27
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b) log 16 4
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Celebrity inverses
Note that exponential and logarithmic functions are either always increasing or always decreasing over
their domains. See the graphs below for values of b corresponding to each case.
1. 0 < b < 1 →
decreasing
2. b > 1 →
increasing
Use the Cancellation Property (C.P.) to verify that f (x) = bx and g(x) = log b x are inverses. These
inverses are truly celebrities in the world of calculus!
Use the definitions, graphs above and your knowledge of inverses to determine the domain and range of
f (x) = bx and g(x) = log b x
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Because these functions are always increasing or always decreasing, they will always pass the
Horizontal Line Test! Therefore, exponential and logarithmic functions are 1-1 (one-to-one)
and consequently have inverses! Recall that for a 1-1 function: for each y-value, there is a unique x.
Algebraically, we are saying that for all x1 , x2 in the domain of a function h,
if h(x1 ) = h(x2 ), then x1 = x2 .
Applying this definition to exponential and logarithmic functions (base b), we have:
For f (x) = bx :
f (x1 ) = f (x2 ) ⇒ x1 = x2
For g(x) = log b x :
bx 1 = b x 2 ⇒ x 1 = x 2
equivalently:
equivalently:
g(x1 ) = g(x2 ) ⇒ x1 = x2
log b (x1 ) = log b (x2 ) ⇒ x1 = x2
In other words, whenever we have an equation
with a log with the same base on both sides, the
input values must be the same!
In other words, whenever we have an equation
with the same base on both sides, the exponents
must be the same!
Exercise 1: Use the definition of a logarithm and the 1-1 property of exponential
functions to evaluate the expressions:
!1"
(a) log 2 16
√
(b) log 7 ( 3 49)
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Celebrity inverses
The exponential (base b) and the logarithm (base b) form a very
popular pair of inverses and the bases with “V.I.P. status” are
b = 10 and b = e = 2.71828 . . .. We write:
1. b = 10:
2. b = e:
f (x) = 10x
f (x) = ex
and
and
g(x) = log x
g(x) = ln x
For f (x) = b x and g(x) = log b x, the Cancellation Property of Inverses gives:
(i) f (g(x)) = b (g(x)) = b (log b x) = x
(ii) g(f (x)) = log b (f (x)) = log b (b x ) = x
1 Recall that the Cancellation Property of Inverses captures the fact that when a function is composed of (i.e. has “eaten”) its inverse, they undo or cancel each other out
and return the number they are acting on. In particular:
b (log b x) = x shows us that when f (x) = b x “eats” g(x) = log b x, they
cancel each other out and leave behind just x!
log b (b x ) = x shows us that when g(x) = log b x “eats” f (x) = b x , they
also cancel each other out and leave behind just x!
2 The Cancellation Property of Inverses (C.P.) allows us to express any number in
exponential or logarithmic form with any base!
b log b □ = □
and log b (b □ ) = □
For example: suppose we want to express the number 3 in exponential and
logarithmic form, base 5. Simply set □ = 3 and b = 5!
5 log 5 3 = 3
and log 5 (5 3 ) = 3
Exercise 2: (a) Express 32 in exponential form, base 2.
√
(b) Express 7 32 as an exponential form, base 2.
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Celebrity Inverses
Exercise 2: (Continued)
(c) Express 2 in exponential form, base e.
(d) Express
2
e
in exponential form, base e.
Exercise 3: For parts (a)–(d), identify the following:
(i) the exponential function f and logarithmic function g.
(ii) the input x and whether (f ◦ g)(x) or (g ◦ f )(x). I.e. who “ate” who?
(iii) Are f and g are inverses? If so, use C.P. to evaluate. If not, explain why.
a) log 4 (47 )
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b) ln(π 6 )
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c) 7log 5 (3)
d) eln(8)
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Properties of Logarithms
First recall the relevant Properties of Exponents (P.O.E.):
1. bm · bn = bm+n
2.
bm
bn
= bm−n
3. (bm )n = bmn
Properties of Logarithms (P.O.L.) and Change of Base Formula (C.O.B.):
Let b, c ∈ R+ , such that b, c ∕= 1. Then, for r, s ∈ R+ , we have:
1. log b (r) + log b (s) = log b (rs)
proof: Let x = logb (r), y = logb (s) and z = logb (rs). Then, by def’n, we have:
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x = logb (r)
⇐⇒
r = bx
(1)
y = logb (s)
⇐⇒
s = by
(2)
z = logb (rs)
⇐⇒
rs = bz
(3)
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Properties of Logarithms
Thus, we are required to prove that x + y = z.
Substituting the results of (1) and (2) into (3), we have:
rs = bz
⇐⇒
b x · by = b z
P.O.E.
⇐⇒
bx+y = bz
1−1
x+y =z
⇐⇒
Therefore, logb (r) + logb (s) = logb (rs), as required.
2. log b (r) − log b (s) = log b
3. log b (r s ) = slog b (r)
□
!r"
s
The proofs of the remaining properties are left as an exercise for you!
Can you articulate these properties in words?
4. Change of Base Formula (C.O.B.): log b (r) =
Note: logc (b) ∕= 0, since b ∕= 1.
log c (r)
log c (b) .
Exercise 4: TRUE or FALSE: Determine if the equation is true for all possible
values of the variables.
# $
x
(a) log xy = log
log y
! "
(b) − ln A1 = ln A
! "
(c) log7 ba2 = log7 a − 2 log7 b
(d) (log x)3 = 3 log x
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Evaluating Expressions
Exercise 5: Evaluate each of the following expressions.
a) log 3
!
1
√
4
3
"
d) log(6log 6 (
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e
b) ln(ln(ln(e )))
√
5
10)
)
e)
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! "
1
c) ln
+ ln (5e)
e
log 2 (log 4 4) − log 4 (log 2 4)
eln 2 + ln 6 − 2 ln 2
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Evaluating Expressions
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Evaluating Expressions
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Solving Equations
Exercise 6: Solve for x.
a) x = log ( 1 ) 3
3
e) 54x = 3
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b) log x
!
f ) log (ln x) = 5
1
64
"
= −3
c) log 16 x =
g) 3x + 9x = 5(9 + 3x )
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1
4
d) 32x−1 =
1
9
h) ex − 12e−x − 1 = 0
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Solving Equations
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Solving Equations
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Solving Equations
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