## Logarithms

### Exponential Functions

A function in the form of y = a^{x}, where a >= 1 is called an exponential function.

**E.g.**

y = 2^{x}; y = 3^{x}

The following image shows the nature of two exponential functions:

Exponential functions show the following features, regardless of the number:

- They go through (0,1) on a grid.
- For a small increase in x, there is a steep increase in y - hence, the change becomes exponential.
- As x approaches negative infinity, y approaches 0 - creating an asymptote.
- The values of the functions remain positive, regardless of the value of x.

### The Exponential Function - e^{x}

This is a special exponential function, which is widely used in mathematics. It's special, because the **gradient** of the curve at
any point is the same as the **value** of the function at the same point. It's still an exponential function, because it shows all the features that we discussed before.

The following image shows this feature:

### The Exponential Function - e^{x} - and the Natural Logarithm Function - ln(x)

The following image shows the exponential function and the natural logarithm function on the same grid. You can see they are closely related and worth studying together.

The following features can be seen in the two graphs:

- They swap around x and y values - e
^{0}= 1; ln(1) = 0. - They are symmetrical around y = x line.
- ln(x) is not defined for x = 0 or negative x values.
- The values of the function ln(x) can be positive, negative or even 0.

### Logarithms

Consider the following indices and how they produce the corresponding logarithm values:

8^{2} = 64 => we say that log_{8}64 = 2; - this is read as *log 64 to the base 8 is 2.*

4^{3} = 64 => we say that log_{4}64 = 3; - this is read as *log 64 to the base 4 is 3.*

Note the positions of the bases and indices in each case.

You may have noticed that in both cases the function is log(64), yet the values are different, because of the different bases. It shows the
significance of the *base*, when it comes to dealing with logarithms. In short, log without a base is like a laptop without a keypad! The base is really important in logarithms.

In theory, a logarithm function can take any base value; the most used bases, however, are *10* and *e*.

log_{10}x = log(x) or lg(x); log_{e}x = ln(x);

Let's solve some problems involving logarithms now; the following approach is highly recommended to master this topic:

**E.g.1**

Find the value of log_{3}27

Let log_{3}27 = x

3^{x} = 27

Now write down 27 in index form with the same base.

3^{x} = 3^{3}

So, x = 3.

log_{3}27 = 3.

**E.g.2**

Find the value of log_{2}32

Let log_{2}32 = x

2^{x} = 32

Now write down 32 in index form with the same base.

2^{x} = 2^{5}

So, x = 5.

log_{2}32 = 5.

**E.g.3**

Find the value of log_{10}100

Let log_{10}100 = x

10^{x} = 100

Now write down 100 in index form with the same base.

10^{x} = 10^{2}

So, x = 2.

log_{10}100 = 2.

**E.g.4**

Find the value of log_{2}2

Let log_{2}2 = x

2^{x} = 2

Now write down 2 in index form with the same base.

2^{x} = 2^{1}

So, x = 1.

log_{2}2 = 1.

**E.g.5**

Find the value of log_{3}1

Let log_{3}1 = x

3^{x} = 1

Now write down 1 in index form with the same base.

3^{x} = 3^{0}

So, x = 0.

log_{3}1 = 0.

**E.g.6**

Find the value of log_{2}1/2

Let log_{2}1/2 = x

2^{x} = 1/2

Now write down 1/2 in index form with the same base.

2^{x} = 2^{-1}

So, x = -1.

log_{2}1/2 = -1.

**E.g.7**

Find the value of log_{4}8

Let log_{4}8 = x

4^{x} = 8

Now write down both 4 and 8 in a common base, such as 2, in index form.

(2^{2})^{x} = 2^{3}

2^{2x} = 2^{3}

So, 2x = 3.

x = 3/2

log_{4}8 = 3/2.

**E.g.8**

Find the value of log_{36}6

Let log_{36}6 = x

36^{x} = 6

Now write down both in index form with the base 6.

36^{x} = 6^{1}

(6^{2})^{x} = 6^{1}

So, 6^{2x} = 6^{1}

2x = 1 => x = 1/2

log_{36}6 = 1/2.

**E.g.9**

Find the value of log_{0.25}4

Let log_{0.25}4 = x

0.25^{x} = 4

Now write down both in index form with the base 4.

(1/4)^{x} = 4^{1}

(4^{-1})^{x} = 4^{1}

So, 4^{-x} = 4^{1}

-x = 1 => x = -1

log_{0.25}4 = -1.

**E.g.10**

Find the value of log_{16}8

Let log_{16}8 = x

16^{x} = 8

Now write down both in index form with the base 2.

(2^{4})^{x} = 2^{3}

So, 2^{4x} = 2^{3}

4x = 3 => x = 3/4

log_{16}8 = 3/4.

**E.g.11**

Solve log_{x}125 = 3

x^{3} = 125

Now write down 125 with the base 5.

x^{3} = 5^{3}

x = 5

**E.g.12**

Solve log_{x}9 = 0.5

x^{1/2} = 9

Now write down 9 with the base 81.

x^{1/2} = 81^{1/2}

x = 81

**Laws of Indices**

Indices and logarithms go hand in hand. So, it's really important to understand the *rules of indices*, in order to deal with logarithms. The following examples may be helpful, when it comes to revising the basic rules of indices:

**1. a ^{x} . a^{y} = a^{x + y}**

**E.g.**

2^{3} . 2^{4} = 2^{3 + 4} = 2^{7}

**2. a ^{x} / a^{y} = a^{x - y}**

**E.g.**

2^{7} / 2^{4} = 2^{7 - 4} = 2^{3}

**3. (a ^{x})^{y} = a^{xy} **

**E.g.**

(2^{3})^{2} = 2^{3X2} = 2^{6}

### Laws of Logarithms

**1. log _{a}(xy) = log_{a}x + log_{a}y**

*Proof:*

Let log_{a}(x) = k and log_{a}(y) = l

So, a^{k} = x and a^{l} = y

By multiplying together, xy = a^{k} X a^{l} = a^{k + l}

log_{a}(xy) = k + l = log_{a}(x) + log_{a}(y).

**E.g.**

Find log_{a}(4X3).

log_{a}(4X3) = log_{a}4 + log_{a}3.

**2. log _{a}(x/y) = log_{a}x - log_{a}y**

*Proof:*

Let log_{a}(x) = k and log_{a}(y) = l

So, a^{k} = x and a^{l} = y

By dividing, x/y = a^{k} :- a^{l} = a^{k - l}

log_{a}(x/y) = k - l = log_{a}(x) - log_{a}(y).

**E.g.**

Find log_{a}(5/2).

log_{a}(5/2) = log_{a}5 - log_{a}2.

**3. log _{a}(x)^{y} = ylog_{a}x**

*Proof:*

Let log_{a}(x) = k

a^{k} = x

(a^{k})^{y} = x^{y}

a^{ky} = x^{y}

log_{a}(x^{y}) = ky

log_{a}(x^{y}) = ylog_{a}(x).

**E.g.**

Find log_{a}(4)^{3}.

log_{a}(4)^{3} = 3log_{a}4.

**Additional Rule 1: log _{a}(1/x)= -log_{a}x**

*Proof:*

log_{a}(1/x) = log_{a}(1) - log_{a}(x)

= 0 - log_{a}(x)

- log_{a}(x).

**E.g.**

Express log_{a}(1/4) in an alternate form.

log_{a}(1/4) = -log_{a}4.

**Additional Rule 2: log _{b}(x)= log_{a}x/log_{a}b**

*Proof:*

Let log_{b}(x) = k

b^{k} = x

Taking log to the base *a* on both sides,

log_{a}(b^{k}) = log_{a}(x)

k log_{a}(b) = log_{a}(x)

k = log_{a}(x)/log_{a}(b)

log_{b}(x) = log_{a}(x)/log_{a}(b).

**E.g.**

Find log_{4}(8) as a logarithm of the base 2.

log_{4}8 = log_{2}8/log_{2}4

Let x = log_{2}8 => 2^{x} = 8 => x = 3

Let y = log_{2}4 => 2^{y} = 4 => y = 2

log_{4}8 = 3/2

**Additional Rule 3: log _{x}(y)= 1/log_{y}x**

*Proof:*

From the additional rule 2,

log_{x}(y) = log_{y}(y)/log_{y}(x)

= 1/log_{y}(x).

**E.g.**

Express log_{4}(8) in an alternate form.

log_{4}(8) = 1/log_{8}4.

**Logarithms Problem Solving**

**E.g.1 **

Write log_{3}8 + log_{3}7 as a single logarithm.

log_{3}8 + log_{3}7 = log_{3}8X7

log_{3}56.

**E.g.2 **

Write log_{3}5 + log_{3}4 - log_{3}2 as a single logarithm.

log_{3}5 + log_{3}4 - log_{3}2 = log_{3}(5X4/2)

log_{3}10.

**E.g.3 **

Simplify log_{3}81 - log_{3}9

log_{3}81 - log_{3}9 = log_{3}(81/9)

= log_{3}9

Let x = log_{3}9

3^{x} = 9 = 3^{2}

x = 2

log_{3}81 - log_{3}9 = 2.

**E.g.4 **

Simplify 2 log_{3}9 - log_{3}27

2 log_{3}9 - log_{3}27 = log_{3}9^{2} - log_{3}27

= log_{3}(81/27)

= log_{3}3

Let x = log_{3}3

3^{x} = 3 = 3^{1}

x = 1

2 log_{3}9 - log_{3}27 = 1.

**E.g.5 **

Simplify 3 log_{3}5 - 2 log_{3}5

3 log_{3}5 - 2 log_{3}5 = log_{3}5^{3} - log_{3}5^{2}

= log_{3}(125/25)

= log_{3}5.

**E.g.6 **

Simplify 3 log_{6}6 - 2 log_{6}6

3 log_{6}6 - 2 log_{6}6 = log_{6}6^{3} - log_{6}6^{2}

= log_{6}(216/36)

= log_{6}6

Let x = log_{6}6

6^{x} = 6 = 6^{1}

x = 1

3 log_{6}6 - 2 log_{6}6 = 1.

**E.g.7 **

Simplify 3 log_{4}4 + 1/2 log_{4}4 - 2 log_{4}4

3 log_{4}4 + 1/2 log_{4}4 - 2 log_{4}4 = log_{4}4^{3} + log_{4}4^{1/2} - log_{4}4^{2}

= log_{4}(64X2/16)

= log_{4}8

Let x = log_{4}8

4^{x} = 8

(2^{2})^{x} = 2^{3}

2x = 3 => x = 3/2

3 log_{4}4 + 1/2 log_{4}4 - 2 log_{4}4 = 3/2.

**E.g.8 **

Simplify 4 log_{10}10 - (2 log_{10}5 + 2 log_{10}2)

4 log_{10}10 - (2 log_{10}5 + 2 log_{10}2 = log_{10}10^{4} - (log_{10}5^{2} + log_{10}2^{2})

= log_{10}(10000/25X4)

= log_{10}100

Let x = log_{10}100

10^{x} = 100

10^{x} = 10^{2}

x = 2

4 log_{10}10 - (2 log_{10}5 + 2 log_{10}2) = 2.

**E.g.9 **

Write in terms of log_{x}a, log_{x}b and log_{x}c, the simplified term of log_{x}(a^{2}b / √c).

log_{x}(a^{2}b / √c) = log_{x}a^{2} + log_{x}b - log_{x}√ c

= 2 log_{x}a + log_{x}b - 1/2 log_{x}c

**E.g.10 **

Write in terms of log_{x}a and log_{x}b, the simplified term of log_{x}(a^{2}b^{2}).

log_{x}(a^{2}b^{2}) = log_{x}a^{2} + log_{x}b^{2}

= 2 log_{x}a + 2 log_{x}b

= 2(log_{x}a + log_{x}b).

### Solving equations involving logarithms

**E.g.1 **

Solve 10^{x} = 102.

Before finding the value of x, we can estimate it; since 10^{2} = 100, x must be slightly bigger than 2.

10^{x} = 102 => log_{10}102 = x

From calculator, log_{10}102 = 2.009

So, x = 2.009(3 d.p.).

**E.g.2 **

Solve 10^{2x - 1} = 400.

10^{2x - 1} = 400 => log_{10}400 = 2x - 1

From calculator, log_{10}400 = 2.6

So, 2x - 1 = 2.6

2x = 3.6

x = 1.8

**E.g.3 **

Solve 5^{x} = 130.

Since 5^{3} = 125, x must be slightly bigger than 3. Let's solve this by logarithms of different bases.

**Method 1 - using log to the base 10**

log_{10}5^{x} = log_{10}130

x log_{10}5 = _{10}130

x = _{10}130/_{10}5

= 3.02(2dp)

**Method 2 - using log to the base e**

log_{e}5^{x} = log_{e}130

x log_{e}5 = log_{e}130

x ln(5) = ln(130)

x = ln(130)/ln(5)

= 3.02(2 d.p.)

The answer is almost the same regardless of the log base.

**E.g.4 **

Solve 5^{(2x + 1)} = 3^{(x - 1)} .

log(5^{2x + 1}) = 3^{(x - 1)}

(2x + 1)log 5 = (x - 1)log 3

(2x + 1)/(x - 1) = log 3 / log 5

(2x + 1)/(x - 1) = log 3 / log 5 = 0.6826

2x + 1 = 0.6826x - 0.6826

1.3174x = -1.6826

x = -1.28

**E.g.5 **

Solve 3^{2x} + 4(3^{x}) - 12 = 0.

Let y = 3^{x} => 3^{2x} = 3^{x} . 3^{x} = y^{2}

So, the equation becomes, y^{2} + 4y - 12 = 0

(y + 6)(y - 2) = 0

y = -6 or y = 2

Since, y, 3^{x} - an exponential function - cannot be negative,

y = 2 => 3^{x} = 2

log 3^{x} = log 2

x log 3 = log 2

x = log 2/ log 3

x = 0.63(2 d.p.)

**E.g.6 **

log_{3}x + 8/log_{3}x = 6

Let y = log_{3}x

y + 8/y = 6

y^{2} + 8 = 6y

y^{2} - 6y + 8 = 0

(y - 4)(y - 2)=0

y = 4 or y = 2

log_{3}x = 4 or log_{3}x = 2

x = 3^{4} or x = 3^{2}

x = 81 or x = 9

**E.g.6 **

log_{5}(3 - 2x) = log_{25}(5x^{2} - 13x + 3)

= log_{5}(5x^{2} - 13x + 3) / log_{5}25

Since log_{5}25 = 2,

log_{5}(5x^{2} - 13x + 3) / log_{5}25 = log_{5}(5x^{2} - 13x + 3) / 2

2 log_{5}(3 - 2x) = log_{5}(5x^{2} - 13x + 3)

log_{5}(3 - 2x)^{2} = log_{5}(5x^{2} - 13x + 3)

log_{5}(3 - 2x)^{2} - log_{5}(5x^{2} - 13x + 3) = 0

log_{5}[(3 - 2x)^{2} / (5x^{2} - 13x + 3)] = 0

Since log(1) = 0 => [(3 - 2x)^{2} / (5x^{2} - 13x + 3)] = 1

9 - 12x + 4x^{2} = 5x^{2} - 13x + 3

x^{2} - x - 6 = 0

(x - 3)(x + 2) = 0

x = 3 or x = -2

Since x = 3 is not valid for log(3 - 2x),

x = -2

### Sketching Straight Lines from Curves

Thanks to logarithms, a complex mathematical relationship can be sketched as straight line, after a transformation in the format. In practice, it is tremendously useful and easy to analyse the relationship.

Suppose, the relationship involved is y = x^{n}, where n is a constant.

So, log y = log x^{n}

log y = n log x

This is in the form of *y = mx*, which is a straight line, as follows:

In the same way, we can turn the following curves into straight lines easily by converting the variables into logarithms.

y = x^{2} => log y = 2 log x

y = x^{3} => log y = 3 log x

y = x^{4} => log y = 4 log x

The gradient of each line is the index number of each function of the corresponding curve. The straight lines are as follows:

In addition, even a complex relationship can be turned into a corresponding logarithm function.

Suppose, y = px^{n}

log y = log px^{n}

log y = log p + log x^{n}

log y = log p + n log x

log y = n log x + log p

y = mx + c

The line is as follows:

The gradient, m = n; y-intercept = log p

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