# Logarithm Rules (Properties)

Logarithm rules are the properties or the identities of the logarithm that are used to simplify complex logarithmic expressions and solve logarithmic equations involving variables. They are derived from the exponent rules, as they are just the opposite of writing an exponent.

Here is the list of all the logarithmic identities.

These rules are used to simplify complex logarithmic expressions and solve logarithmic equations involving variables.

## Product Rule

It states that the logarithm of a product is the sum of the logarithms of the numbers being multiplied.

logb(xy) = logb(x) + logb(y)

For example, log8(63) = log8(7 × 9) = log8(7) + log8(9)

Now, let us derive this rule.

### Derivation

Let us assume logb(x) = p and logb(y) = q.

Converting the logarithms into exponential forms, we get

bp = x …..(i)

bq = y …..(ii)

Now, by multiplying (i) and (ii), we get

bp × bq = xy

⇒ bp + q = xy …..(iii)

Converting (iii) into the logarithmic form, we get

logb(xy) = p + q

By putting the values of p and q, we get

logb(xy) = logb(x) + logb(y)

## Quotient Rule

It states that the logarithm of the quotient is equal to the difference between the logarithm of the dividend minus the logarithm of the divisor.

${\log _{b}\left( \dfrac{x}{y}\right) =\log _{b}x-\log _{b}y}$

For example, ${\log _{6}\left( \dfrac{28}{25}\right) =\log _{6}\left( 28\right) -\log _{6}\left( 25\right)}$

### Derivation

Let us assume logb(x) = p and logb(y) = q.

Converting the logarithms into exponential forms, we get

bp = x …..(i) and bq = y …..(ii)

On dividing (i) by (ii), we get

${\dfrac{b^{p}}{b^{q}}=\dfrac{x}{y}}$

⇒ ${b^{p-q}=\dfrac{x}{y}}$ …..(iii)

Converting (iii) into the logarithmic form, we get

${\log _{b}\left( \dfrac{x}{y}\right) =p-q}$

By putting the values of p and q, we get

${\log _{b}\left( \dfrac{x}{y}\right) =\log _{b}x-\log _{b}y}$

## Power Rule

It states that the logarithm of a number raised to a power is the power times the logarithm of the number.

logb(xn) = n logbx

For example, log7(38) = 8 log7(3)

### Derivation

Let us assume logb(x) = p

Converting the logarithm into its exponential form, we get

bp = x

Raising the power by n, we get

(bp)n = xn

⇒ bnp = xn

Now, converting into the logarithmic form, we get

logb(xn) = np

By putting the value of p, we get

logb(xn) = n logb(x)

## Change of Base Rule

It states that the logarithm of a given number ‘x’ to a base ‘b’ can be expressed as the ratio of the logarithm of ‘x’ to any other base ‘c’ divided by the logarithm of ‘b’ to the base ‘c.’

${\log _{b}x=\dfrac{\log _{c}x}{\log _{c}b}}$

⇒ logbx ⋅ logcb = logcx

For example, log4(15) can be written as ${\dfrac{\log _{c}15}{\log _{c}4}}$, where ‘c’ is any other base.

### Derivation

Let us assume logb(x) = p, logc(x) = r, and logc(b) = s

Converting the logarithms into their exponential forms, we get

bp = x …..(i)

cr = x …..(ii)

cs = b …..(iii)

From (i) and (ii), bp = cr …..(iv)

Substituting the value of (iii) in (iv), we get

(cs)p = cr

⇒ csp = cr

⇒ sp = r

⇒ ${p=\dfrac{r}{s}}$

Now, substituting the values of p, r, and s, we get

${\log _{b}x=\dfrac{\log _{c}x}{\log _{c}b}}$

⇒ logbx ⋅ logcb = logcx, another form of the rule.

## Inverse Rule

It states that:

1. The logarithm of a base raised to a number is that number.

${\log _{b}\left( b^{x}\right) =x}$

2. A base raised to the logarithm of a number is equal to that number.

${b^{\log _{b}x}=x}$

For example, ${8^{\log _{8}5}=5}$ and ${\log _{3}\left( 3^{7}\right) =7}$

### Derivation

Let us assume logb(x) = p …..(i)

Converting the logarithm into its exponential form, we get

bp = x …..(ii)

Now, by putting the value of p in (ii), we get

${b^{\log _{b}x}=x}$

Again, by putting the value of (ii) in (i), we get

logb(bp) = p

Thus, mathematically, it is ${\log _{b}\left( b^{x}\right) =x}$

## Zero Rule

The logarithm of 1 to any base is zero.

logb(1) = 0

For example, log15(1) = 0, log2(1) = 0

### Derivation

Let us consider the logarithmic equation logb(1) = x

Converting into its exponential form, we get

bx = 1

⇒ bx = b0

⇒ x = 0

Thus, logb(1) = 0

## Identity Rule

It states that the logarithm of a base to itself is 1.

logb(b) = 1

For example, log9(9) = 1, log4(4) = 1

### Derivation

Let us consider the logarithmic equation logb(b) = x

Converting into its exponential form, we get

bx = b

⇒ bx = b1

⇒ x = 1

Thus, logb(b) = 1

## Reciprocal Rule

It states that the logarithm of a number’s reciprocal is equal to the negative of the logarithm of that number. ${\log _{b}\left( \dfrac{1}{x}\right) =-\log _{b}\left( x\right)}$

For example, ${\log _{3}\left( \dfrac{1}{11}\right) =-\log _{3}\left( 11\right)}$

### Derivation

Let ${x}$ and ${\dfrac{1}{x}}$ be the number and its reciprocal.

Since ${x\cdot \dfrac{1}{x}=1}$, then

${\log _{b}\left( x\cdot \dfrac{1}{x}\right) =\log _{b}\left( 1\right)}$

⇒ ${\log _{b}\left( x\right) +\log _{b}\left( \dfrac{1}{x}\right) =0}$

${\log _{b}\left( \dfrac{1}{x}\right) =-\log _{b}\left( x\right)}$

## Properties of Natural Logarithm

It states that the natural logarithm (denoted by loge(x) or ln(x)) follows all the above properties of base logarithms.

Here are some more special properties of natural logarithm:

### Derivative Rule

It states that the derivative of a natural logarithm of a number with respect to it is equal to ${\dfrac{1}{x}}$.

${\dfrac{d}{dx}\left( \ln \left( x\right) \right) =\dfrac{1}{x}}$

### Integration Rule

It states that the integration of ${\dfrac{1}{x}}$ with respect to ‘x’ is equal to the natural logarithm of the absolute value of x, plus a constant of integration ‘C.’

${\int \dfrac{1}{x}dx=\ln \left| x\right| +C}$, here ‘C’ is a constant.

Also, while integrating ln(x) with respect to x, the formula is as follows:

${\int \ln \left( x\right) dx=x\cdot \ln \left( x\right) -x+C}$

⇒ ${\int \ln \left( x\right) dx=x\left( \ln \left( x\right) -1\right) +C}$

## Solved Example

Use the rules of logarithms to simplify ${\log _{5}\left[ \dfrac{15\left( x-3\right) ^{2}\left( x+4\right) ^{3}}{7\left( x+1\right) }\right]}$

Solution:

Here, ${\log _{5}\left[ \dfrac{15\left( x-3\right) ^{2}\left( x+4\right) ^{3}}{7\left( x+1\right) }\right]}$
= ${\log _{5}\left[ 15\left( x-3\right) ^{2}\left( x+4\right) ^{3}\right] -\log _{5}\left[ 7\left( x+1\right) \right]}$ (by the quotient rule)
= ${\log _{5}15+\log _{5}\left[ \left( x-3\right) ^{2}\right] +\log _{5}\left[ \left( x+4\right) ^{3}\right] -\log _{5}7-\log _{5}\left( x+1\right)}$ (by the product rule)
= ${\log _{5}\left( 3\times 5\right) +2\log _{5}\left( x-3\right) +3\log _{5}\left( x+4\right) -\log _{5}7-\log _{5}\left( x+1\right)}$ (by the power rule)
= ${\log _{5}5+\log _{5}3+2\log _{5}\left( x-3\right) +3\log _{5}\left( x+4\right) -\log _{5}7-\log _{5}\left( x+1\right)}$ (by the product rule)
= ${1+\log _{5}3-\log _{5}7+2\log _{5}\left( x-3\right) +3\log _{5}\left( x+4\right) -\log _{5}\left( x+1\right)}$ (by the identity rule)
Thus, ${\log _{5}\left[ \dfrac{15\left( x-3\right) ^{2}\left( x+4\right) ^{3}}{7\left( x+1\right) }\right]}$ = ${1+\log _{5}3-\log _{5}7+2\log _{5}\left( x-3\right) +3\log _{5}\left( x+4\right) -\log _{5}\left( x+1\right)}$