2 Unit Bridging Course - Day 11
Logarithms
Collin Zheng
1 / 22
Definition
Recall from Days 7-8 that given the expression a x :
I
x is called the index, power, or exponent of a;
I
a is called the base.
However, an index/power/exponent is also called a logarithm.
In particular, given
y = ax,
x is called the “logarithm of y to the base a”, written as:
x = loga y .
2 / 22
Definition
Recall from Days 7-8 that given the expression a x :
I
x is called the index, power, or exponent of a;
I
a is called the base.
However, an index/power/exponent is also called a logarithm.
In particular, given
y = ax,
x is called the “logarithm of y to the base a”, written as:
x = loga y .
3 / 22
Examples
Here are some examples of logarithms to the base 2:
I
log2 4 = 2, since 4 = 22 .
I
log2 8 = 3, since 8 = 23 .
I
log2
1
8
= −3, since
1
8
= 2−3 .
Here are some examples of logs to the base 10:
I
log10 100 = 2, since 100 = 102 .
I
log10 1000 = 3, since 1000 = 103 .
I
log10 0.001 = −3, since 0.001 = 10−3 .
Simply keep in mind that when you are calculating logarithms,
you are calculating powers!
4 / 22
Examples
Here are some examples of logarithms to the base 2:
I
log2 4 = 2, since 4 = 22 .
I
log2 8 = 3, since 8 = 23 .
I
log2
1
8
= −3, since
1
8
= 2−3 .
Here are some examples of logs to the base 10:
I
log10 100 = 2, since 100 = 102 .
I
log10 1000 = 3, since 1000 = 103 .
I
log10 0.001 = −3, since 0.001 = 10−3 .
Simply keep in mind that when you are calculating logarithms,
you are calculating powers!
5 / 22
Examples
Here are some examples of logarithms to the base 2:
I
log2 4 = 2, since 4 = 22 .
I
log2 8 = 3, since 8 = 23 .
I
log2
1
8
= −3, since
1
8
= 2−3 .
Here are some examples of logs to the base 10:
I
log10 100 = 2, since 100 = 102 .
I
log10 1000 = 3, since 1000 = 103 .
I
log10 0.001 = −3, since 0.001 = 10−3 .
Simply keep in mind that when you are calculating logarithms,
you are calculating powers!
6 / 22
Examples
Here are some examples of logarithms to the base 2:
I
log2 4 = 2, since 4 = 22 .
I
log2 8 = 3, since 8 = 23 .
I
log2
1
8
= −3, since
1
8
= 2−3 .
Here are some examples of logs to the base 10:
I
log10 100 = 2, since 100 = 102 .
I
log10 1000 = 3, since 1000 = 103 .
I
log10 0.001 = −3, since 0.001 = 10−3 .
Simply keep in mind that when you are calculating logarithms,
you are calculating powers!
7 / 22
log and
ln
on your Calculator
The log button on your calculator calculates logarithms to the
base 10.
E.g. log 1000 will display 3, since 1000 = 103 .
Of particular interest to us is the
logarithms to the base e.
E.g.
ln
5
ln
button, which provides
will display roughly 1.61, since 5 ≈ e1.61 .
Logarithms to the base e are called natural logarithms.
8 / 22
log and
ln
on your Calculator
The log button on your calculator calculates logarithms to the
base 10.
E.g. log 1000 will display 3, since 1000 = 103 .
Of particular interest to us is the
logarithms to the base e.
E.g.
ln
5
ln
button, which provides
will display roughly 1.61, since 5 ≈ e1.61 .
Logarithms to the base e are called natural logarithms.
9 / 22
Some Properties of Logarithms
Note that log
1
calculator. In fact,
and ln
1
both give 0 on your
loga 1 = 0
for any base a > 0, since we know from Day 7 that a 0 = 1.
Moreover, recall from Day 8 that all exponential functions of the
form y = a x (where a > 0) output strictly positive numbers for
all x. Therefore it does not make sense to take the log of zero
or any negative number.
Indeed, trying to log or ln any number ≤ 0 will result in an
‘Err’ error message being displayed on your calculator!
10 / 22
Some Properties of Logarithms
Note that log
1
calculator. In fact,
and ln
1
both give 0 on your
loga 1 = 0
for any base a > 0, since we know from Day 7 that a 0 = 1.
Moreover, recall from Day 8 that all exponential functions of the
form y = a x (where a > 0) output strictly positive numbers for
all x. Therefore it does not make sense to take the log of zero
or any negative number.
Indeed, trying to log or ln any number ≤ 0 will result in an
‘Err’ error message being displayed on your calculator!
11 / 22
Some Properties of Logarithms
Note that log
1
calculator. In fact,
and ln
1
both give 0 on your
loga 1 = 0
for any base a > 0, since we know from Day 7 that a 0 = 1.
Moreover, recall from Day 8 that all exponential functions of the
form y = a x (where a > 0) output strictly positive numbers for
all x. Therefore it does not make sense to take the log of zero
or any negative number.
Indeed, trying to log or ln any number ≤ 0 will result in an
‘Err’ error message being displayed on your calculator!
12 / 22
The Logarithm Laws
Here are some useful ‘laws’ for manipulating log expressions:
The Logarithm Laws
The following is true for the log to any base:
1) log (AB) = log A + log B, for A > 0 and B > 0.
A
2) log
= log A − log B, for A > 0 and B > 0.
B
3) log (A B ) = B log A, for A > 0.
13 / 22
The Logarithm Laws
Here are some useful ‘laws’ for manipulating log expressions:
The Logarithm Laws
The following is true for the log to any base:
1) log (AB) = log A + log B, for A > 0 and B > 0.
A
2) log
= log A − log B, for A > 0 and B > 0.
B
3) log (A B ) = B log A, for A > 0.
14 / 22
The Logarithm Laws
Here are some useful ‘laws’ for manipulating log expressions:
The Logarithm Laws
The following is true for the log to any base:
1) log (AB) = log A + log B, for A > 0 and B > 0.
A
2) log
= log A − log B, for A > 0 and B > 0.
B
3) log (A B ) = B log A, for A > 0.
15 / 22
The Logarithm Laws (cont.)
Example
Simplify 2 ln (x + 1) + ln (2x) to a single logarithm.
2 ln (x + 1) + ln (2x)
= ln (x + 1)2 + ln (2x)
←−
using log law #3
= ln [ (x + 1)2 × (2x) ]
←−
using log law #1
= ln [ 2x(x + 1)2 ]
←−
upon simplifying the inside
16 / 22
The Logarithm Laws (cont.)
Example
Simplify 2 ln (x + 1) + ln (2x) to a single logarithm.
2 ln (x + 1) + ln (2x)
= ln (x + 1)2 + ln (2x)
←−
using log law #3
= ln [ (x + 1)2 × (2x) ]
←−
using log law #1
= ln [ 2x(x + 1)2 ]
←−
upon simplifying the inside
17 / 22
The Logarithm Laws (cont.)
Example
Simplify 2 ln (x + 1) + ln (2x) to a single logarithm.
2 ln (x + 1) + ln (2x)
= ln (x + 1)2 + ln (2x)
←−
using log law #3
= ln [ (x + 1)2 × (2x) ]
←−
using log law #1
= ln [ 2x(x + 1)2 ]
←−
upon simplifying the inside
18 / 22
The Logarithm Laws (cont.)
Example
Simplify 2 ln (x + 1) + ln (2x) to a single logarithm.
2 ln (x + 1) + ln (2x)
= ln (x + 1)2 + ln (2x)
←−
using log law #3
= ln [ (x + 1)2 × (2x) ]
←−
using log law #1
= ln [ 2x(x + 1)2 ]
←−
upon simplifying the inside
19 / 22
The Logarithm Laws (cont.)
Practice Questions
I
I
1
2
ln (x 2 ) − ln(x 3 ) into a single logarithm.
√
Write ln (x 2 3x + 1) as sums and multiples of simpler
logarithms.
Simplify
20 / 22
The Logarithm Laws (cont.)
Answers
1
I ln
.
x2
I
2 ln x +
1
2
ln (3x + 1).
21 / 22
Summary
I
y = a x ⇐⇒ x = loga y, where loga y is spoken as the
“logarithm of y to the base a.”
I
The log and ln buttons on your calculator provide
logarithms to the base 10 and e, respectively.
I
log (ab) = log a + log b, for a > 0 and b > 0.
a
= log a − log b, for a > 0 and b > 0.
log
b
I
I
log (a b ) = b log a, for a > 0.
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