# Logarithm

See how to solve a logarithm

(expression/equation/inequality/function).

33 examples and their solutions.

## Logarithmic Form

### Definition

2

→ m = log

Logarithm (log) is a way^{m}= 3→ m = log

_{2}3to write the exponent of a number.

log

_{2}3 is read as

[log base 2 of 3].

Exponent Rules

### Example

2

→ Logarithmic form?

Solution ^{4}= 16→ Logarithmic form?

2

4 = log

^{4}= 164 = log

_{2}16Close

### Example

### Example

### Example

2 = log

→ Exponential form?

Solution _{3}9→ Exponential form?

2 = log

3

_{3}93

^{2}= 9Close

### Example

-5 = log

→ Exponential form?

Solution _{2}132→ Exponential form?

-5 = log

2

_{2}1322

^{-5}= 132Close

### Example

23 = log

→ Exponential form?

Solution _{7}^{3}√49→ Exponential form?

23 = log

_{7}^{3}√49^{3}√49 = 7^{23}Close

## Logarithm of 1

### Formula

log

_{a}1 = 0## Logarithm of Itself

### Formula

log

_{a}a = 1## log_{a} x^{m}

### Formula

log

= m log

_{a}x^{m}= m log

_{a}x### Example

log

Solution _{2}8 log

= 3 log

= 3⋅1

= 3

_{2}8 = log_{2}2^{3}= 3 log

_{2}2= 3⋅1

= 3

Close

### Example

log

Solution _{3}181 log

= -4 log

= -4⋅1

= -4

_{3}181 = log_{3}3^{-4}= -4 log

_{3}3= -4⋅1

= -4

Close

### Example

log

log

Solution _{3}2 = alog

_{3}32 = ? log

= 5 log

= 5a

_{3}32 = log_{3}2^{5}= 5 log

_{3}2= 5a

Close

## log_{a} xy

### Formula

log

= log

_{a}x⋅y= log

_{a}x + log_{a}y### Example

log

log

Solution _{2}3 = alog

_{2}24 = ? log

= log

= log

= 3 log

= 3⋅1 + a

= a + 3

_{2}24= log

_{2}2^{3}⋅3 - [1]= log

_{2}2^{3}+ log_{2}3= 3 log

_{2}2 + a= 3⋅1 + a

= a + 3

Close

## log_{a} xy

### Formula

log

= log

_{a}xy= log

_{a}x - log_{a}y### Example

log

Solution _{2}32√8 log

= log

= log

= log

= 5 log

= 5⋅1 - 32⋅1

= 5 - 32

= 102 - 32

= 72

_{2}32√8= log

_{2}32 - log_{2}√8= log

_{2}2^{5}- log_{2}√2^{3}= log

_{2}2^{5}- log_{2}2^{32}- [1]= 5 log

_{2}2 - 32 log_{2}2= 5⋅1 - 32⋅1

= 5 - 32

= 102 - 32

= 72

Close

### Example

log

Solution _{6}9 - log_{6}15 + log_{6}10 log

= log

= log

= log

= log

= 1

_{6}9 - log_{6}15 + log_{6}10= log

_{6}9⋅1015= log

_{6}3^{2}⋅2⋅53⋅5= log

_{6}3⋅2= log

_{6}6= 1

Close

## Logarithmic Equation

### Formula

log

x > 0

0 < a < 1, a > 1

x and a should satisfy these conditions. _{a}xx > 0

0 < a < 1, a > 1

### Example

log

Solution _{3}x = 4 log

x > 0

x = 3

= 81

- [1]

x = 81

_{3}x = 4x > 0

x = 3

^{4}= 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

log

Solution _{x}64 = 3 log

0 < x < 1, x > 1

x

= 4

x = 4

- [1]

x = 4

_{x}64 = 30 < x < 1, x > 1

x

^{3}= 64= 4

^{3}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

log

Solution _{2}(log_{3}(log_{5}x)) = 0 log

x > 0

log

log

log

log

x = 5

= 125

- [1]

x = 125

_{2}(log_{3}(log_{5}x)) = 0x > 0

log

_{3}(log_{5}x) = 2^{0}log

_{3}(log_{5}x) = 1log

_{5}x = 3^{1}log

_{5}x = 3x = 5

^{3}= 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

log

Solution _{2}x + log_{2}(x - 1) = log_{2}12 log

x > 0

x - 1 > 0

x > 1

→ x > 1

log

x(x - 1) = 12

x

x

1) x - 4 = 0

x = 4

2) x + 3 = 0

x = -3

- [2]

x = 4

_{2}x + log_{2}(x - 1) = log_{2}12x > 0

x - 1 > 0

x > 1

→ x > 1

log

_{2}x(x - 1) = log_{2}12x(x - 1) = 12

x

^{2}- x = 12x

^{2}- 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

(log

Solution _{5}x)^{2}- log_{5}x^{2}- 3 = 0 (log

x > 0

x

x ≠ 0 - [1]

→ x > 0 - [2]

(log

(log

1) log

log

= 125

2) log

log

x = 5

= 15

- [4]

x = 15, 125

_{5}x)^{2}- log_{5}x^{2}- 3 = 0x > 0

x

^{2}> 0x ≠ 0 - [1]

→ x > 0 - [2]

(log

_{5}x)^{2}- 2 log_{5}x - 3 = 0(log

_{5}x - 3)(log_{5}x + 1) = 0 - [3]1) log

_{5}x - 3 = 0log

_{5}x = 3= 125

2) log

_{5}x + 1 = 0log

_{5}x = -1x = 5

^{-1}= 15

x = 15, 125

[1]

x

^{2}> 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

log

Solution _{7}(x + 2) ≤ 1 log

x + 2 > 0 - [1]

x > -2

x + 2 ≤ 7

x + 2 ≤ 7

x ≤ 5

- [3]

-2 < x ≤ 5

_{7}(x + 2) ≤ 1x + 2 > 0 - [1]

x > -2

x + 2 ≤ 7

^{1}- [2]x + 2 ≤ 7

x ≤ 5

-2 < x ≤ 5

[2]

log

Change this to logarithmic form.

The base 2 is in

7 > 1.

So the order of the inequality sign

doesn't change.

≤ → ≤

_{7}(x + 2) ≤ 1Change 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

log

Solution _{0.1}(x - 3) > 2 log

x - 3 > 0

x > 3

x - 3 < 0.1

x - 3 < 0.01

x < 3.01

- [2]

3 < x < 3.01

_{0.1}(x - 3) > 2x - 3 > 0

x > 3

x - 3 < 0.1

^{2}- [1]x - 3 < 0.01

x < 3.01

3 < x < 3.01

[1]

log

Change this to logarithmic form.

The base 0.1 is in

0 < 0.1 < 1.

So the order of the inequality sign

does change.

< → >

_{0.1}(x - 3) > 2Change 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 log

Solution _{3}x ≥ log_{3}(x + 6) + 1 2 log

x > 0

x + 6 > 0

x > -6

→ x > 0 - [1]

log

log

x

x

x

(x + 3)(x - 6) ≥ 0 - [4]

x = -3, 6

x ≤ -3 or x ≥ 6 - [5]

- [6]

x ≥ 6

_{3}x ≥ log_{3}(x + 6) + 1x > 0

x + 6 > 0

x > -6

→ x > 0 - [1]

log

_{3}x^{2}≥ log_{3}(x + 6) + log_{3}3log

_{3}x^{2}≥ log_{3}(x + 6)⋅3x

^{2}≥ (x + 6)⋅3 - [2] [3]x

^{2}≥ 3x + 18x

^{2}- 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,

x

are equal.

Then the numbers in the logs,

x

^{2}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 = log_{a} 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 = log_{a} 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 = a^{x} and y = log_{a} x

^{x}and y = log

_{a}x

are symmetric about y = x.

So these two function are inverse functions.

Exponential Function: Graph

### Example

Graph y = log

Solution _{3}x.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 = log

Solution _{2}(x - 3). y = log

x - 3 = 0

x = 3 - [1] [2]

[3]

_{2}(x - 3)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 = log

A common logarithm (common log)_{10}xis a logarithm

whose base is 10.

In high school math,

log

_{10}= 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 = 10

^{0.699}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 2

= log 2

= 2 log 2 + log 3 + log 10

= 2⋅0.301 + 0.477 + 1

= 0.602 + 1.477

= 2.079 - [1]

= log 12⋅10

= log 2

^{2}⋅3⋅10= log 2

^{2}+ 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 = 10

^{2.079}Close

## Change of Base Formula

### Formula

log

_{a}x = log_{b}xlog_{b}a### Example

log

(Assume log 2 = 0.301, log 7 = 0.845.)

Solution _{2}70 = ?(Assume log 2 = 0.301, log 7 = 0.845.)

log

= log

= log 7⋅10log 2

= log 7 + log 10log 2

= 0.845 + 10.301

= 1.8450.301

= 1845301

_{2}70= 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

log

log

Solution _{2}3 = alog

_{12}18 = ? log

= log

= log

= log

= log

= 1 + 2⋅a2⋅1 + a

= 2a + 1a + 2

_{12}18= log

_{2}18log_{2}12= log

_{2}2⋅3^{2}log_{2}2^{2}⋅3 - [1]= log

_{2}2 + log_{2}3^{2}log_{2}2^{2}+ log_{2}3= log

_{2}2 + 2 log_{2}32 log_{2}2 + log_{2}3= 1 + 2⋅a2⋅1 + a

= 2a + 1a + 2

Close

### Example

(log

Solution _{2}27)(log_{9}16) (log

= log

= log

= 3 log

= 3⋅4⋅12

= 3⋅2

= 6

_{2}27)(log_{9}16)= log

_{2}27 ⋅ log_{2}16log_{2}9= log

_{2}3^{3}⋅ log_{2}2^{4}log_{2}3^{2}= 3 log

_{2}3 ⋅ 4 log_{2}22 log_{2}3= 3⋅4⋅12

= 3⋅2

= 6

Close

## Natural Logarithm

### Definition

ln x = log

A natural logarithm (natural log)_{e}xis a logarithm

whose base is e.

(e is a constant number.

e = 2.71828...)

In high school math,

log

_{e}= ln.

Constant e

### Example

ln 2e

(Assume ln 2 = 0.69.)

Solution ^{x + 1}= ?(Assume ln 2 = 0.69.)

ln 2e

= ln 2 + ln e

= ln 2 + (x + 1) ln e

= 0.69 + (x + 1)⋅1

= 0.69 + x + 1

= x + 1.69

^{x + 1}= ln 2 + ln e

^{x + 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 = A

A: final value

A

r: rate of change

t: time

By using logarithm,_{0}(1 + r)^{t}A: final value

A

_{0}: initial valuer: 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.)

A

A = 24000

r = 0.08 /year

10000⋅(1 + 0.08)

(1 + 0.08)

1.08

log 1.08

t log 1.08 = log 2.4

t⋅0.033 = 0.380

t = 0.3800.033

= 38033

= 11.xx

→ 12 years - [2]

_{0}= 10000A = 24000

r = 0.08 /year

10000⋅(1 + 0.08)

^{t}= 24000(1 + 0.08)

^{t}= 2.41.08

^{t}= 2.4log 1.08

^{t}= 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.)

A

A = 10

r = -0.14 /week

100⋅(1 - 0.14)

(1 - 0.14)

0.86

log 0.86

t log 0.86 = -log 10

t⋅(-0.066) = -1

0.066t = 1

t = 10.066

= 100066

= 15.xx

→ 16 weeks - [2]

_{0}= 100A = 10

r = -0.14 /week

100⋅(1 - 0.14)

^{t}= 10(1 - 0.14)

^{t}= 1100.86

^{t}= 10^{-1}log 0.86

^{t}= 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 = A

A: final value

A

e: constant number (= 2.71828...)

r: rate of change

t: time

Continuous Exponential Growth/Decay: Final Value_{0}e^{rt}A: final value

A

_{0}: initial valuee: 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.)

A

A = 30

r = 0.05 /minute

10⋅e

e

0.05t = ln 3 - [1]

0.05t = 1.099

t = 1.0990.05

= 109.95

= 21.xx

→ 22 minutes - [2]

_{0}= 10A = 30

r = 0.05 /minute

10⋅e

^{0.05⋅t}= 30e

^{0.05t}= 30.05t = 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.)

A

A = 20

r = -0.03 /second

50⋅e

e

-0.03t = ln 0.4

-0.03t = -0.916

0.03t = 0.916

t = 0.9160.03

= 91.63

= 30.xx

→ 31 seconds - [1]

_{0}= 50A = 20

r = -0.03 /second

50⋅e

^{-0.03⋅t}= 20e

^{-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.

A

_{0}→ A

_{0}/2

The formula came from

A

_{0}e

^{rt}= A = A

_{0}/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