# Logarithm

Logarithm, often called â€˜logs,â€™ is the power to which a number must be raised to get the result. It is thus the inverse of the exponent and is written as:

ba = x â‡” logbx = a

Here,

• â€˜bâ€™ is the base
• â€˜aâ€™ is the exponent
• â€˜xâ€™ is the argument

are the 3 parts of a logarithm.

Thus, the logarithm represents the exponent to which a base is raised to yield a given number.

## How to Undo Logarithms

For example, we know 43 = 64

Here, using the logarithm, we can answer how many 4s multiply to get 64.

Since 4 Ã— 4 Ã— 4 = 64, we multiply three 4s to get 64, which is written in the logarithmic form as log4(64) = 3, read as â€˜log base 4 of 64 is 3.â€™

Thus, 43 = 64 â‡” log4(64) = 3, where the base is 4, and the exponent or power is 3

Here are some examples of conversions from exponential to logarithmic form and vice-versa.

Find the value of log7(343).

Solution:

As we know, 7 Ã— 7 Ã— 7 = 73 = 343
Thus, log7(343) = 3

Convert 35 = 243 in its logarithmic form.

Solution:

As we know, ba = x â‡’ logbx = a
Here, 35 = 243
â‡’ log3(243) = 5, the required logarithmic form.

However, the expression logbx has some restrictions, which are as follows:

## Restriction on the Variable

### For Base

The base â€˜bâ€™ of a logarithm is always a positive real number (b > 0) and does not equal 1 (b â‰  1).

For negative bases, logarithm leads to complex results. Now, let us assume the base is 1, and the equation is:

log17 = x â‡’ 1x = 7

Since 1 raised to any power yields 1, 1x = 7 is false. Thus, the base does not equal 1.

### For Argument

The argument â€˜xâ€™ is always a positive real number (a > 0).

Since a positive number (b > 0) is raised to any power, it yields a positive number (bx > 0).

Thus, bx = a follows that a > 0.

## Common Logarithms (log) vs Natural Logarithm (ln)

The base of a logarithm can have many positive values except 1. However, two of them are frequently used.

### Common Logarithm

The logarithm whose base is 10 is known as the common logarithm or base-10-logarithm. It is often denoted as log(x) without a subscript.

For example, log10(10000) = log(10000) = 4 â‡” 104 = 10000

### Natural Logarithm

The logarithm with the base â€˜eâ€™ (â‰ˆ 2.718â€¦, Eulerâ€™s number) is the natural logarithm or base-e-logarithm, denoted by ln(x) or loge(x).

For example, ln(e2) = 2 â‡” e2 = e Ã— e, ln(9) = c â‡” ec = 9

## Rules

Certain rules (also properties or identities) of logarithms are used to simplify, expand, or condense them.

### Simplifying

Let us simplify a logarithmic function log35 using the appropriate rule(s).

Since logb(xy) = logbx + logby and 35 = 5 Ã— 7 (Using Product rule)

Thus, log35 = log(5 Ã— 7) = log 5 + log 7 â‰ˆ 0.699 + 0.845 (rounded to 3 decimal places)

â‡’ log35 â‰ˆ 1.544 (rounded to 3 decimal places)

We observe that logarithms also have decimal values like 1.544, which means 101.544â€¦ = 35, and is graphically represented as:

However, negative logarithms are formed when the argument is between 0 and 1. In logbx < 0, for 0 < x < 1, â€˜bâ€™ is the base, and â€˜xâ€™ is the argument.

For example, log(0.0001) = -4 gives a negative value, and its exponential form, 10-4 = 0.0001, gives a decimal.

### Expanding

Let us expand the logarithm expression log(5x4y5).

Here, log(5x4y5)

Using the product rule, we get

log(5) + log(x4) + log(y5)

Using the power rule, we get

log(5) + 4 log(x) + 5 log(y)

Thus, log(5x4y5) = log(5) + 4 log(x) + 5 log(y)

### Condensing

Considering the above sum of logarithms log(5) + 4 log(x) + 5 log(y) and condensing it into a single logarithm, we get:

log(5) + 4 log(x) + 5 log(y)

Using the power rule, we get

log(5) + log(x4) + log(y5)

Using the product rule, we get

log(5x4y5)

Thus, log(5) + 4 log(x) + 5 log(y) = log(5x4y5)

Evaluate ${\log \left( \dfrac{100x^{4}}{10y^{2}}\right) +2\log \left( y\right)}$

Solution:

Here, ${\log \left( \dfrac{100x^{4}}{10y^{2}}\right) +2\log \left( y\right)}$
= ${\log \left( \dfrac{10x^{4}}{y^{2}}\right) +2\log \left( y\right)}$
Using the quotient rule, we get
${\log \left( 10x^{4}\right) -\log \left( y^{2}\right) +2\log \left( y\right)}$
Using the product rule, we get
${\log \left( 10\right) +\log \left( x^{4}\right) -\log \left( y^{2}\right) +2\log \left( y\right)}$
Using the power rule, we get
${\log \left( 10\right) +4\log \left( x\right) -2\log \left( y\right) +2\log \left( y\right)}$
Since logbb = 1, log(10) = 1
Thus, ${\log \left( \dfrac{100x^{4}}{10y^{2}}\right) +2\log \left( y\right)}$ = 1 + 4log(x)

Expand the expression log2(16x8)

Solution:

Here, log2(16x8)
Using the product law of logarithms, we get
log2(16) + log2(x8)
= 4 + log2(x8)
Using the power law of logarithms, we get
4 + 8 log2(x)
Thus, log2(16x8) = 4 + 8 log2(x)

Condense the expression log5(9x2) – log5(x) + log5(8x4)

Solution:

Here, log5(9x2) – log5(x) + log5(8x4)
First, we use the quotient rule by subtracting the logarithms. The expression is:Â
${\log _{5}\left( \dfrac{9x^{2}}{x}\right) +\log _{5}\left( 8x^{4}\right)}$
= ${\log _{5}\left( 9x\right) +\log _{5}\left( 8x^{4}\right)}$
Now, we use the product rule by adding the logarithms. The expression becomes
${\log _{5}\left( 9x\cdot 8x^{4}\right)}$
= ${\log _{5}\left( 72x^{5}\right)}$
Thus, log5(9x2) – log5(x) + log5(8x4) = ${\log _{5}\left( 72x^{5}\right)}$

Solve the equation log(x) + log(3) = log(51) and find the value of x.

Solution:

Here, log(x) + log(3) = log(51)
â‡’ log(3x) = log(51) (by the product rule)
Now, canceling the log from both sides, we get
3x = 51
â‡’ x = 17
Thus, x = 17

## Solving Nested Logarithm

A nested logarithm, also called iterated logarithm or repeated logarithm, represents a logarithm within another logarithm and is denoted as log(log(logâ€¦(logx))) or logn(x), where â€˜nâ€™ is the level of nesting.

Let us solve the nested logarithm log2(log8(64))

First, we solve the inner logarithm log8(64). Since 82 = 64, log8(64) = 2

Now, solving the outer logarithm, we get log2(2) = 1

Thus, log2(log8(64)) is simplified to 1.

Simplify log5(log3(log7(343)))

Solution:

Here, log5(log3(log7(343)))
Since 73 = 343, log7(343) = 3
Solving the inner logarithm, we get
log5(log3(3))
Since logb(b) = 1, log3(3) = 1
Now, solving further logarithms, we get
log5(1)
= 0