The document discusses properties of logarithms. It begins by recalling rules of exponents and their corresponding rules of logarithms. Four basic logarithm rules are presented: 1) logb(1) = 0, 2) logb(xy) = logb(x) + logb(y), 3) logb(x/y) = logb(x) - logb(y), 4) logb(xt) = tlogb(x). It then works through an example problem to demonstrate using these rules to write the logarithm of a expression in terms of logarithms of its variables. It concludes by noting that logarithms and exponentials are inverse functions, so logb(bx) =
1. b0
= 1
Propertiesof Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
4.
1. b0
= 1
2.br
· bt
= br+t
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
5.
1. b0
= 1
2.br
· bt
= br+t
3. = br-t
bt
br
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
6.
1. b0
= 1
2.br
· bt
= br+t
3. = br-t
4. (br
)t
= brt
bt
br
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
7.
1. logb(1) =01. b0
= 1
2. br
· bt
= br+t
3. = br-t
4. (br
)t
= brt
bt
br
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
8.
1. logb(1) =0
2. logb(x·y) = logb(x)+logb(y)
1. b0
= 1
2. br
· bt
= br+t
3. = br-t
4. (br
)t
= brt
bt
br
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
9.
1. logb(1) =0
2. logb(x·y) = logb(x)+logb(y)
3. logb( ) = logb(x) – logb(y)
x
y
1. b0
= 1
2. br
· bt
= br+t
3. = br-t
4. (br
)t
= brt
bt
br
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
10.
1. logb(1) =0
2. logb(x·y) = logb(x)+logb(y)
3. logb( ) = logb(x) – logb(y)
4. logb(xt
) = t·logb(x)
x
y
1. b0
= 1
2. br
· bt
= br+t
3. = br-t
4. (br
)t
= brt
bt
br
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
11.
1. logb(1) =0
2. logb(x·y) = logb(x)+logb(y)
3. logb( ) = logb(x) – logb(y)
4. logb(xt
) = t·logb(x)
x
y
1. b0
= 1
2. br
· bt
= br+t
3. = br-t
4. (br
)t
= brt
bt
br
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
We veryify part 2: logb(xy) = logb(x) + logb(y), x, y > 0.
Proof:
12.
1. logb(1) =0
2. logb(x·y) = logb(x)+logb(y)
3. logb( ) = logb(x) – logb(y)
4. logb(xt
) = t·logb(x)
x
y
1. b0
= 1
2. br
· bt
= br+t
3. = br-t
4. (br
)t
= brt
bt
br
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
We veryify part 2: logb(xy) = logb(x) + logb(y), x, y > 0.
Proof:
Let x and y be two positive numbers.
13.
1. logb(1) =0
2. logb(x·y) = logb(x)+logb(y)
3. logb( ) = logb(x) – logb(y)
4. logb(xt
) = t·logb(x)
x
y
1. b0
= 1
2. br
· bt
= br+t
3. = br-t
4. (br
)t
= brt
bt
br
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
We veryify part 2: logb(xy) = logb(x) + logb(y), x, y > 0.
Proof:
Let x and y be two positive numbers. Let logb(x) = r
and logb(y) = t, which in exp-form are x = br
and y = bt
.
14.
1. logb(1) =0
2. logb(x·y) = logb(x)+logb(y)
3. logb( ) = logb(x) – logb(y)
4. logb(xt
) = t·logb(x)
x
y
1. b0
= 1
2. br
· bt
= br+t
3. = br-t
4. (br
)t
= brt
bt
br
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
We veryify part 2: logb(xy) = logb(x) + logb(y), x, y > 0.
Proof:
Let x and y be two positive numbers. Let logb(x) = r
and logb(y) = t, which in exp-form are x = br
and y = bt
.
Therefore x·y = br+t
,
15.
1. logb(1) =0
2. logb(x·y) = logb(x)+logb(y)
3. logb( ) = logb(x) – logb(y)
4. logb(xt
) = t·logb(x)
x
y
1. b0
= 1
2. br
· bt
= br+t
3. = br-t
4. (br
)t
= brt
bt
br
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
We veryify part 2: logb(xy) = logb(x) + logb(y), x, y > 0.
Proof:
Let x and y be two positive numbers. Let logb(x) = r
and logb(y) = t, which in exp-form are x = br
and y = bt
.
Therefore x·y = br+t
, which in log-form is
logb(x·y) = r + t = logb(x)+logb(y).
16.
1. logb(1) =0
2. logb(x·y) = logb(x)+logb(y)
3. logb( ) = logb(x) – logb(y)
4. logb(xt
) = t·logb(x)
x
y
1. b0
= 1
2. br
· bt
= br+t
3. = br-t
4. (br
)t
= brt
bt
br
Properties of Logarithm
Recall the following
Rules of Exponents:
The corresponding
Rules of Logs are:
We veryify part 2: logb(xy) = logb(x) + logb(y), x, y > 0.
Proof:
Let x and y be two positive numbers. Let logb(x) = r
and logb(y) = t, which in exp-form are x = br
and y = bt
.
Therefore x·y = br+t
, which in log-form is
logb(x·y) = r + t = logb(x)+logb(y).The other rules may be verified similarly.
3x2
√y
log( ) =log( ),3x2
√y
3x2
y1/2
Properties of Logarithm
a. Write log( ) in terms of log(x) and log(y).
Example A:
19.
3x2
√y
log( ) =log( ), by the quotient rule
= log (3x2
) – log(y1/2
)
3x2
√y
3x2
y1/2
Properties of Logarithm
a. Write log( ) in terms of log(x) and log(y).
Example A:
20.
3x2
√y
log( ) =log( ), by the quotient rule
= log (3x2
) – log(y1/2
)
product rule
= log(3) + log(x2
)
3x2
√y
3x2
y1/2
Properties of Logarithm
a. Write log( ) in terms of log(x) and log(y).
Example A:
21.
3x2
√y
log( ) =log( ), by the quotient rule
= log (3x2
) – log(y1/2
)
product rule power rule
= log(3) + log(x2
) – ½ log(y)
3x2
√y
3x2
y1/2
Properties of Logarithm
a. Write log( ) in terms of log(x) and log(y).
Example A:
22.
3x2
√y
log( ) =log( ), by the quotient rule
= log (3x2
) – log(y1/2
)
product rule power rule
= log(3) + log(x2
) – ½ log(y)
= log(3) + 2log(x) – ½ log(y)
3x2
√y
3x2
y1/2
Properties of Logarithm
a. Write log( ) in terms of log(x) and log(y).
Example A:
23.
3x2
√y
log( ) =log( ), by the quotient rule
= log (3x2
) – log(y1/2
)
product rule power rule
= log(3) + log(x2
) – ½ log(y)
= log(3) + 2log(x) – ½ log(y)
3x2
√y
3x2
y1/2
Properties of Logarithm
a. Write log( ) in terms of log(x) and log(y).
b. Combine log(3) + 2log(x) – ½ log(y) into one log.
Example A:
24.
3x2
√y
log( ) =log( ), by the quotient rule
= log (3x2
) – log(y1/2
)
product rule power rule
= log(3) + log(x2
) – ½ log(y)
= log(3) + 2log(x) – ½ log(y)
3x2
√y
3x2
y1/2
Properties of Logarithm
a. Write log( ) in terms of log(x) and log(y).
log(3) + 2log(x) – ½ log(y) power rule
= log(3) + log(x2
) – log(y1/2
)
b. Combine log(3) + 2log(x) – ½ log(y) into one log.
Example A:
25.
3x2
√y
log( ) =log( ), by the quotient rule
= log (3x2
) – log(y1/2
)
product rule power rule
= log(3) + log(x2
) – ½ log(y)
= log(3) + 2log(x) – ½ log(y)
3x2
√y
3x2
y1/2
Properties of Logarithm
a. Write log( ) in terms of log(x) and log(y).
log(3) + 2log(x) – ½ log(y) power rule
= log(3) + log(x2
) – log(y1/2
) product rule
= log (3x2
) – log(y1/2
)
b. Combine log(3) + 2log(x) – ½ log(y) into one log.
Example A:
26.
3x2
√y
log( ) =log( ), by the quotient rule
= log (3x2
) – log(y1/2
)
product rule power rule
= log(3) + log(x2
) – ½ log(y)
= log(3) + 2log(x) – ½ log(y)
3x2
√y
3x2
y1/2
Properties of Logarithm
a. Write log( ) in terms of log(x) and log(y).
log(3) + 2log(x) – ½ log(y) power rule
= log(3) + log(x2
) – log(y1/2
) product rule
= log (3x2
) – log(y1/2
)= log( )3x2
y1/2
b. Combine log(3) + 2log(x) – ½ log(y) into one log.
Example A:
27.
Recall that givena pair of inverse functions, f and f -1
,
then f(f -1
(x)) = x and f -1
(f(x)) = x.
Properties of Logarithm
28.
Recall that givena pair of inverse functions, f and f -1
,
then f(f -1
(x)) = x and f -1
(f(x)) = x.
Since expb(x) and logb(x) is a pair of inverse functions,
Properties of Logarithm
29.
Recall that givena pair of inverse functions, f and f -1
,
then f(f -1
(x)) = x and f -1
(f(x)) = x.
Since expb(x) and logb(x) is a pair of inverse functions,
we have that:
a. logb(expb(x)) = x or logb(bx
) = x
Properties of Logarithm
30.
Recall that givena pair of inverse functions, f and f -1
,
then f(f -1
(x)) = x and f -1
(f(x)) = x.
Since expb(x) and logb(x) is a pair of inverse functions,
we have that:
a. logb(expb(x)) = x or logb(bx
) = x
b. expb(logb(x)) = x or blog (x)
= x
Properties of Logarithm
b
31.
Recall that givena pair of inverse functions, f and f -1
,
then f(f -1
(x)) = x and f -1
(f(x)) = x.
Since expb(x) and logb(x) is a pair of inverse functions,
we have that:
a. logb(expb(x)) = x or logb(bx
) = x
b. expb(logb(x)) = x or blog (x)
= x
Properties of Logarithm
b
Example B: Simplify
a. log2(2-5
) =
b. 8log (xy)
=
c. e2+ln(7)
=
8
32.
Recall that givena pair of inverse functions, f and f -1
,
then f(f -1
(x)) = x and f -1
(f(x)) = x.
Since expb(x) and logb(x) is a pair of inverse functions,
we have that:
a. logb(expb(x)) = x or logb(bx
) = x
b. expb(logb(x)) = x or blog (x)
= x
Properties of Logarithm
b
Example B: Simplify
a. log2(2-5
) = -5
b. 8log (xy)
=
c. e2+ln(7)
=
8
33.
Recall that givena pair of inverse functions, f and f -1
,
then f(f -1
(x)) = x and f -1
(f(x)) = x.
Since expb(x) and logb(x) is a pair of inverse functions,
we have that:
a. logb(expb(x)) = x or logb(bx
) = x
b. expb(logb(x)) = x or blog (x)
= x
Properties of Logarithm
b
Example B: Simplify
a. log2(2-5
) = -5
b. 8log (xy)
= xy
c. e2+ln(7)
=
8
34.
Recall that givena pair of inverse functions, f and f -1
,
then f(f -1
(x)) = x and f -1
(f(x)) = x.
Since expb(x) and logb(x) is a pair of inverse functions,
we have that:
a. logb(expb(x)) = x or logb(bx
) = x
b. expb(logb(x)) = x or blog (x)
= x
Properties of Logarithm
b
Example B: Simplify
a. log2(2-5
) = -5
b. 8log (xy)
= xy
c. e2+ln(7)
= e2
·eln(7)
= 7e2
8