Logarithm
See how to solve a logarithm
(expression/equation/inequality/function).
33 examples and their solutions.
Logarithmic Form
Definition
2m = 3
→ m = log2 3
Logarithm (log) is a way→ m = log2 3
to write the exponent of a number.
log2 3 is read as
[log base 2 of 3].
Exponent Rules
Example
24 = 16
→ Logarithmic form?
Solution → Logarithmic form?
24 = 16
4 = log2 16
4 = log2 16
Close
Example
Example
Example
2 = log3 9
→ Exponential form?
Solution → Exponential form?
2 = log3 9
32 = 9
32 = 9
Close
Example
-5 = log2 132
→ Exponential form?
Solution → Exponential form?
-5 = log2 132
2-5 = 132
2-5 = 132
Close
Example
23 = log7 3√49
→ Exponential form?
Solution → Exponential form?
23 = log7 3√49
3√49 = 723
3√49 = 723
Close
Logarithm of 1
Formula
loga 1 = 0
Logarithm of Itself
Formula
loga a = 1
loga xm
Formula
loga xm
= m loga x
= m loga x
Example
log2 8
Solution log2 8 = log2 23
= 3 log2 2
= 3⋅1
= 3
= 3 log2 2
= 3⋅1
= 3
Close
Example
log3 181
Solution log3 181 = log3 3-4
= -4 log3 3
= -4⋅1
= -4
= -4 log3 3
= -4⋅1
= -4
Close
Example
log3 2 = a
log3 32 = ?
Solution log3 32 = ?
log3 32 = log3 25
= 5 log3 2
= 5a
= 5 log3 2
= 5a
Close
loga xy
Formula
loga x⋅y
= loga x + loga y
= loga x + loga y
Example
log2 3 = a
log2 24 = ?
Solution log2 24 = ?
log2 24
= log2 23⋅3 - [1]
= log2 23 + log2 3
= 3 log2 2 + a
= 3⋅1 + a
= a + 3
= log2 23⋅3 - [1]
= log2 23 + log2 3
= 3 log2 2 + a
= 3⋅1 + a
= a + 3
Close
loga xy
Formula
loga xy
= loga x - loga y
= loga x - loga y
Example
log2 32√8
Solution log2 32√8
= log2 32 - log2 √8
= log2 25 - log2 √23
= log2 25 - log2 232 - [1]
= 5 log2 2 - 32 log2 2
= 5⋅1 - 32⋅1
= 5 - 32
= 102 - 32
= 72
= log2 32 - log2 √8
= log2 25 - log2 √23
= log2 25 - log2 232 - [1]
= 5 log2 2 - 32 log2 2
= 5⋅1 - 32⋅1
= 5 - 32
= 102 - 32
= 72
Close
Example
log6 9 - log6 15 + log6 10
Solution log6 9 - log6 15 + log6 10
= log6 9⋅1015
= log6 32⋅2⋅53⋅5
= log6 3⋅2
= log6 6
= 1
= log6 9⋅1015
= log6 32⋅2⋅53⋅5
= log6 3⋅2
= log6 6
= 1
Close
Logarithmic Equation
Formula
loga x
x > 0
0 < a < 1, a > 1
x and a should satisfy these conditions. x > 0
0 < a < 1, a > 1
Example
log3 x = 4
Solution log3 x = 4
x > 0
x = 34
= 81
x = 81
x > 0
x = 34
= 81
x = 81
[1]
Draw x > 0 on a number line.
See if x = 81 is in the colored region.
See if x = 81 is in the colored region.
Close
Example
logx 64 = 3
Solution logx 64 = 3
0 < x < 1, x > 1
x3 = 64
= 43
x = 4
x = 4
0 < x < 1, x > 1
x3 = 64
= 43
x = 4
x = 4
[1]
Draw 0 < x < 1, x > 1
on a number line.
See if x = 4 is in the colored region.
on a number line.
See if x = 4 is in the colored region.
Close
Example
log2 (log3 (log5 x)) = 0
Solution log2 (log3 (log5 x)) = 0
x > 0
log3 (log5 x) = 20
log3 (log5 x) = 1
log5 x = 31
log5 x = 3
x = 53
= 125
x = 125
x > 0
log3 (log5 x) = 20
log3 (log5 x) = 1
log5 x = 31
log5 x = 3
x = 53
= 125
x = 125
[1]
Draw x > 0 on a number line.
See if x = 125 is in the colored region.
See if x = 125 is in the colored region.
Close
Example
log2 x + log2 (x - 1) = log2 12
Solution log2 x + log2 (x - 1) = log2 12
x > 0
x - 1 > 0
x > 1
→ x > 1
log2 x(x - 1) = log2 12
x(x - 1) = 12
x2 - x = 12
x2 - x - 12 = 0 - [1]
1) x - 4 = 0
x = 4
2) x + 3 = 0
x = -3
x = 4
x > 0
x - 1 > 0
x > 1
→ x > 1
log2 x(x - 1) = log2 12
x(x - 1) = 12
x2 - x = 12
x2 - x - 12 = 0 - [1]
1) x - 4 = 0
x = 4
2) x + 3 = 0
x = -3
x = 4
[2]
Draw x > 1
on a number line.
See if x = 4, -3 are in the colored region.
on a number line.
See if x = 4, -3 are in the colored region.
Close
Example
(log5 x)2 - log5 x2 - 3 = 0
Solution (log5 x)2 - log5 x2 - 3 = 0
x > 0
x2 > 0
x ≠ 0 - [1]
→ x > 0 - [2]
(log5 x)2 - 2 log5 x - 3 = 0
(log5 x - 3)(log5 x + 1) = 0 - [3]
1) log5 x - 3 = 0
log5 x = 3
= 125
2) log5 x + 1 = 0
log5 x = -1
x = 5-1
= 15
x = 15, 125
x > 0
x2 > 0
x ≠ 0 - [1]
→ x > 0 - [2]
(log5 x)2 - 2 log5 x - 3 = 0
(log5 x - 3)(log5 x + 1) = 0 - [3]
1) log5 x - 3 = 0
log5 x = 3
= 125
2) log5 x + 1 = 0
log5 x = -1
x = 5-1
= 15
x = 15, 125
[1]
x2 > 0 is true when x ≠ 0.
[2]
Draw x > 0 and x ≠ 0
on a number line.
x > 0 covers x ≠ 0.
So draw x > 0.
The colored region is x > 0.
So x should satisfy x > 0.
on a number line.
x > 0 covers x ≠ 0.
So draw x > 0.
The colored region is x > 0.
So x should satisfy x > 0.
[4]
Draw x > 0
on a number line.
See if x = 125, 1/5 are in the colored region.
on a number line.
See if x = 125, 1/5 are in the colored region.
Close
Logarithmic Inequality
Example
log7 (x + 2) ≤ 1
Solution log7 (x + 2) ≤ 1
x + 2 > 0 - [1]
x > -2
x + 2 ≤ 71 - [2]
x + 2 ≤ 7
x ≤ 5
-2 < x ≤ 5
x + 2 > 0 - [1]
x > -2
x + 2 ≤ 71 - [2]
x + 2 ≤ 7
x ≤ 5
-2 < x ≤ 5
[2]
log7 (x + 2) ≤ 1
Change this to logarithmic form.
The base 2 is in
7 > 1.
So the order of the inequality sign
doesn't change.
≤ → ≤
Change this to logarithmic form.
The base 2 is in
7 > 1.
So the order of the inequality sign
doesn't change.
≤ → ≤
[3]
x > -2
x ≤ 5
Draw these inequalities
on a number line.
Color the intersecting region.
x ≤ 5
Draw these inequalities
on a number line.
Color the intersecting region.
Close
Example
log0.1 (x - 3) > 2
Solution log0.1 (x - 3) > 2
x - 3 > 0
x > 3
x - 3 < 0.12 - [1]
x - 3 < 0.01
x < 3.01
3 < x < 3.01
x - 3 > 0
x > 3
x - 3 < 0.12 - [1]
x - 3 < 0.01
x < 3.01
3 < x < 3.01
[1]
log0.1 (x - 3) > 2
Change this to logarithmic form.
The base 0.1 is in
0 < 0.1 < 1.
So the order of the inequality sign
does change.
< → >
Change this to logarithmic form.
The base 0.1 is in
0 < 0.1 < 1.
So the order of the inequality sign
does change.
< → >
[2]
x > 3
x < 3.01
Draw these inequalities
on a number line.
Color the intersecting region.
x < 3.01
Draw these inequalities
on a number line.
Color the intersecting region.
Close
Example
2 log3 x ≥ log3 (x + 6) + 1
Solution 2 log3 x ≥ log3 (x + 6) + 1
x > 0
x + 6 > 0
x > -6
→ x > 0 - [1]
log3 x2 ≥ log3 (x + 6) + log3 3
log3 x2 ≥ log3 (x + 6)⋅3
x2 ≥ (x + 6)⋅3 - [2] [3]
x2 ≥ 3x + 18
x2 - 3x - 18 ≥ 0
(x + 3)(x - 6) ≥ 0 - [4]
x = -3, 6
x ≤ -3 or x ≥ 6 - [5]
x ≥ 6
x > 0
x + 6 > 0
x > -6
→ x > 0 - [1]
log3 x2 ≥ log3 (x + 6) + log3 3
log3 x2 ≥ log3 (x + 6)⋅3
x2 ≥ (x + 6)⋅3 - [2] [3]
x2 ≥ 3x + 18
x2 - 3x - 18 ≥ 0
(x + 3)(x - 6) ≥ 0 - [4]
x = -3, 6
x ≤ -3 or x ≥ 6 - [5]
x ≥ 6
[1]
Draw x > 0 and x > -6
on a number line.
Color the intersecting region.
The intersecting region is x > 0.
x should satisfy this condition.
on a number line.
Color the intersecting region.
The intersecting region is x > 0.
x should satisfy this condition.
[2]
The bases are the same.
Then the numbers in the logs,
x2 and (x + 6)⋅3,
are equal.
Then the numbers in the logs,
x2 and (x + 6)⋅3,
are equal.
[3]
The bases 3 are in
3 > 1.
So the order of the inequality sign
doesn't change.
≥ → ≥
3 > 1.
So the order of the inequality sign
doesn't change.
≥ → ≥
[6]
x > 0
x ≤ -3 or x ≥ 6
Draw these inequalities
on a number line.
Color the intersecting region.
x ≤ -3 or x ≥ 6
Draw these inequalities
on a number line.
Color the intersecting region.
Close
Logarithmic Function: Graph
Graph: y = loga x (a > 1)
2. The asymptote of the graph is the y-axis.
(= The graph follows the y-axis.)
So the domain x is always (+): x > 0.
Graph: y = loga x (0 < a < 1)
2. The asymptote of the graph is the y-axis.
(= The graph follows the y-axis.)
So the domain x is always (+): x > 0.
Graph: y = ax and y = loga x
are symmetric about y = x.
So these two function are inverse functions.
Exponential Function: Graph
Example
Graph y = log3 x.
Solution Draw the asymptote y-axis.
And draw (1, 0).
And draw (1, 0).
↓
The base 3 is in 3 > 1.
So draw the graph
that goes ↗.
So draw the graph
that goes ↗.
Close
Example
Graph y = log2 (x - 3).
Solution y = log2 (x - 3)
x - 3 = 0
x = 3 - [1] [2]
x - 3 = 0
x = 3 - [1] [2]
[1]
To find the asymptote,
set (x - 3) = 0.
→ x = 3
set (x - 3) = 0.
→ x = 3
[2]
To find the domain,
set (x - 3) > 0.
→ {x|x > 3}
set (x - 3) > 0.
→ {x|x > 3}
[3]
Draw the asymptote x = 3.
And draw (4, 0).
And draw (4, 0).
↓
[4]
The base 2 is in
2 > 1.
So draw the graph
that goes ↗.
2 > 1.
So draw the graph
that goes ↗.
Close
Common Logarithm
Definition
log x = log10 x
A common logarithm (common log)is a logarithm
whose base is 10.
In high school math,
log10 = log.
Example
log 5 = ?
(Assume log 2 = 0.301.)
Solution (Assume log 2 = 0.301.)
log 5 = log 102
= log 10 - log 2
= 1 - 0.301
= 0.699 - [1]
= log 10 - log 2
= 1 - 0.301
= 0.699 - [1]
[1]
5 = 100.699
Close
Example
log 120 = ?
(Assume log 2 = 0.301, log 3 = 0.477.)
Solution (Assume log 2 = 0.301, log 3 = 0.477.)
log 120
= log 12⋅10
= log 22⋅3⋅10
= log 22 + log 3 + log 10
= 2 log 2 + log 3 + log 10
= 2⋅0.301 + 0.477 + 1
= 0.602 + 1.477
= 2.079 - [1]
= log 12⋅10
= log 22⋅3⋅10
= log 22 + log 3 + log 10
= 2 log 2 + log 3 + log 10
= 2⋅0.301 + 0.477 + 1
= 0.602 + 1.477
= 2.079 - [1]
[1]
120 = 102.079
Close
Change of Base Formula
Formula
loga x = logb xlogb a
Example
log2 70 = ?
(Assume log 2 = 0.301, log 7 = 0.845.)
Solution (Assume log 2 = 0.301, log 7 = 0.845.)
log2 70
= log 70log 2 - [1]
= log 7⋅10log 2
= log 7 + log 10log 2
= 0.845 + 10.301
= 1.8450.301
= 1845301
= log 70log 2 - [1]
= log 7⋅10log 2
= log 7 + log 10log 2
= 0.845 + 10.301
= 1.8450.301
= 1845301
[1]
Change the base to 10:
Common Logarithm.
Common Logarithm.
Close
Example
log2 3 = a
log12 18 = ?
Solution log12 18 = ?
log12 18
= log2 18log2 12
= log2 2⋅32log2 22⋅3 - [1]
= log2 2 + log2 32log2 22 + log2 3
= log2 2 + 2 log2 32 log2 2 + log2 3
= 1 + 2⋅a2⋅1 + a
= 2a + 1a + 2
= log2 18log2 12
= log2 2⋅32log2 22⋅3 - [1]
= log2 2 + log2 32log2 22 + log2 3
= log2 2 + 2 log2 32 log2 2 + log2 3
= 1 + 2⋅a2⋅1 + a
= 2a + 1a + 2
Close
Example
(log2 27)(log9 16)
Solution (log2 27)(log9 16)
= log2 27 ⋅ log2 16log2 9
= log2 33 ⋅ log2 24log2 32
= 3 log2 3 ⋅ 4 log2 22 log2 3
= 3⋅4⋅12
= 3⋅2
= 6
= log2 27 ⋅ log2 16log2 9
= log2 33 ⋅ log2 24log2 32
= 3 log2 3 ⋅ 4 log2 22 log2 3
= 3⋅4⋅12
= 3⋅2
= 6
Close
Natural Logarithm
Definition
ln x = loge x
A natural logarithm (natural log)is a logarithm
whose base is e.
(e is a constant number.
e = 2.71828...)
In high school math,
loge = ln.
Constant e
Example
ln 2ex + 1 = ?
(Assume ln 2 = 0.69.)
Solution (Assume ln 2 = 0.69.)
ln 2ex + 1
= ln 2 + ln ex + 1
= ln 2 + (x + 1) ln e
= 0.69 + (x + 1)⋅1
= 0.69 + x + 1
= x + 1.69
= ln 2 + ln ex + 1
= ln 2 + (x + 1) ln e
= 0.69 + (x + 1)⋅1
= 0.69 + x + 1
= x + 1.69
Close
Exponential Growth/Decay: Time
Formula
A = A0(1 + r)t
A: final value
A0: initial value
r: rate of change
t: time
By using logarithm,A: final value
A0: initial value
r: rate of change
t: time
you can find the time t.
Exponential Growth/Decay: Final Value
Compound Interest
Example
The population of a town is 10,000. If it increases at a rate of 8% per year, after how many years will the population be more than 24,000?
(Assume log 2.4 = 0.380, log 1.08 = 0.033.)
Solution (Assume log 2.4 = 0.380, log 1.08 = 0.033.)
A0 = 10000
A = 24000
r = 0.08 /year
10000⋅(1 + 0.08)t = 24000
(1 + 0.08)t = 2.4
1.08t = 2.4
log 1.08t = log 2.4 - [1]
t log 1.08 = log 2.4
t⋅0.033 = 0.380
t = 0.3800.033
= 38033
= 11.xx
→ 12 years - [2]
A = 24000
r = 0.08 /year
10000⋅(1 + 0.08)t = 24000
(1 + 0.08)t = 2.4
1.08t = 2.4
log 1.08t = log 2.4 - [1]
t log 1.08 = log 2.4
t⋅0.033 = 0.380
t = 0.3800.033
= 38033
= 11.xx
→ 12 years - [2]
[1]
Common log both sides.
[2]
t = 11.xx
→ After 11.xx years,
the population will be 24,000.
→ After 12 years,
the population will be more than 24,000.
→ After 11.xx years,
the population will be 24,000.
→ After 12 years,
the population will be more than 24,000.
Close
Example
A radioactive substance weighs 100g. If it decreases at a rate of 14% per week, after how many weeks will the weight be less than 10g?
(Assume log 0.86 = -0.066.)
Solution (Assume log 0.86 = -0.066.)
A0 = 100
A = 10
r = -0.14 /week
100⋅(1 - 0.14)t = 10
(1 - 0.14)t = 110
0.86t = 10-1
log 0.86t = log 10-1 - [1]
t log 0.86 = -log 10
t⋅(-0.066) = -1
0.066t = 1
t = 10.066
= 100066
= 15.xx
→ 16 weeks - [2]
A = 10
r = -0.14 /week
100⋅(1 - 0.14)t = 10
(1 - 0.14)t = 110
0.86t = 10-1
log 0.86t = log 10-1 - [1]
t log 0.86 = -log 10
t⋅(-0.066) = -1
0.066t = 1
t = 10.066
= 100066
= 15.xx
→ 16 weeks - [2]
[1]
Common log both sides.
[2]
t = 15.xx
→ After 15.xx weeks,
the weight will be 10g.
→ After 16 weeks,
the weight will be less than 10g.
→ After 15.xx weeks,
the weight will be 10g.
→ After 16 weeks,
the weight will be less than 10g.
Close
Continuous Exponential Growth/Decay: Time
Formula
A = A0ert
A: final value
A0: initial value
e: constant number (= 2.71828...)
r: rate of change
t: time
Continuous Exponential Growth/Decay: Final ValueA: final value
A0: initial value
e: constant number (= 2.71828...)
r: rate of change
t: time
Compound Interest
Constant e
Example
A substance weighs 10g. If it continuously increases at a rate of 5% per minute, after how many minutes will the weight be more than 30g?
(Assume ln 3 = 1.099.)
Solution (Assume ln 3 = 1.099.)
A0 = 10
A = 30
r = 0.05 /minute
10⋅e0.05⋅t = 30
e0.05t = 3
0.05t = ln 3 - [1]
0.05t = 1.099
t = 1.0990.05
= 109.95
= 21.xx
→ 22 minutes - [2]
A = 30
r = 0.05 /minute
10⋅e0.05⋅t = 30
e0.05t = 3
0.05t = ln 3 - [1]
0.05t = 1.099
t = 1.0990.05
= 109.95
= 21.xx
→ 22 minutes - [2]
[2]
t = 21.xx
→ After 21.xx minutes,
the weight will be 30g.
→ After 22 minutes,
the weight will be more than 30g.
→ After 21.xx minutes,
the weight will be 30g.
→ After 22 minutes,
the weight will be more than 30g.
Close
Example
A radioactive substance weighs 50g. If it continuously decreases at a rate of 3% per second, after how many seconds will the weight be less than 20g?
(Assume ln 0.4 = -0.916.)
Solution (Assume ln 0.4 = -0.916.)
A0 = 50
A = 20
r = -0.03 /second
50⋅e-0.03⋅t = 20
e-0.03t = 0.4
-0.03t = ln 0.4
-0.03t = -0.916
0.03t = 0.916
t = 0.9160.03
= 91.63
= 30.xx
→ 31 seconds - [1]
A = 20
r = -0.03 /second
50⋅e-0.03⋅t = 20
e-0.03t = 0.4
-0.03t = ln 0.4
-0.03t = -0.916
0.03t = 0.916
t = 0.9160.03
= 91.63
= 30.xx
→ 31 seconds - [1]
[1]
t = 30.xx
→ After 30.xx seconds,
the weight will be 20g.
→ After 31 seconds,
the weight will be less than 20g.
→ After 30.xx seconds,
the weight will be 20g.
→ After 31 seconds,
the weight will be less than 20g.
Close
Half-Life
Formula
-rt = ln 2
r: Rate of changet: Half-life
The half-life is the amount of time
when a value
continuously & exponentially decreases
to one half.
A0 → A0/2
The formula came from
A0ert = A = A0/2.
Continuous Exponential Growth/Decay: Time
Example
A weight of a radioactive substance is continuously decreasing at a rate of 4% per second. Find the half-life of the substance.
(Assume ln 2 = 0.69.)
Solution (Assume ln 2 = 0.69.)
r = -0.04 /second
-(-0.04)⋅t = ln 2
0.04t = 0.69
t = 0.690.04
= 694 seconds - [1]
-(-0.04)⋅t = ln 2
0.04t = 0.69
t = 0.690.04
= 694 seconds - [1]
[1]
A half-life does not have to be an integer.
Close