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Last modified on May 24th, 2024

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.

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

**log**_{b}**(xy) = log**_{b}**(x) + log**_{b}**(y)**

For example, log_{8}(63) = log_{8}(7 × 9) = log_{8}(7) + log_{8}(9)

Now, let us derive this rule.

Let us assume log_{b}(x) = p and log_{b}(y) = q.

Converting the logarithms into exponential forms, we get

b^{p} = x …..(i)

b^{q} = y …..(ii)

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

b^{p} × b^{q} = xy

⇒ b^{p + q} = xy …..(iii)

Converting (iii) into the logarithmic form, we get

log_{b}(xy) = p + q

By putting the values of p and q, we get

**log**_{b}**(xy) = log**_{b}**(x) + log**_{b}**(y)**

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

Let us assume log_{b}(x) = p and log_{b}(y) = q.

Converting the logarithms into exponential forms, we get

b^{p} = x …..(i) and b^{q} = 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

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

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

**log**_{b}**(x**^{n}**) = n log**_{b}**x**

For example, log_{7}(3^{8}) = 8 log_{7}(3)

Let us assume log_{b}(x) = p

Converting the logarithm into its exponential form, we get

b^{p} = x

Raising the power by n, we get

(b^{p})^{n} = x^{n}

⇒ b^{np} = x^{n}

Now, converting into the logarithmic form, we get

log_{b}(x^{n}) = np

By putting the value of p, we get

**log**_{b}**(x**^{n}**) = n log**_{b}**(x)**

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.’

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

**⇒ log**_{b}**x ⋅ log**_{c}**b = log**_{c}**x**

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

Let us assume log_{b}(x) = p, log_{c}(x) = r, and log_{c}(b) = s

Converting the logarithms into their exponential forms, we get

b^{p} = x …..(i)

c^{r} = x …..(ii)

c^{s} = b …..(iii)

From (i) and (ii), b^{p} = c^{r} …..(iv)

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

(c^{s})^{p} = c^{r}

⇒ c^{sp} = c^{r}

⇒ sp = r

⇒ ${p=\dfrac{r}{s}}$

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

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

**⇒ log**_{b}**x ⋅ log**_{c}**b = log**_{c}**x, **another form of the 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 ${\log _{3}\left( 3^{7}\right) =7}$

Let us assume log_{b}(x) = p …..(i)

Converting the logarithm into its exponential form, we get

b^{p} = 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

log_{b}(b^{p}) = p

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

The logarithm of 1 to any base is zero.

**log**_{b}**(1) = 0**

For example, log_{15}(1) = 0, log_{2}(1) = 0

Let us consider the logarithmic equation log_{b}(1) = x

Converting into its exponential form, we get

b^{x} = 1

⇒ b^{x} = b^{0}

⇒ x = 0

Thus, **log**_{b}**(1) = 0**

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

**log**_{b}**(b) = 1**

For example, log_{9}(9) = 1, log_{4}(4) = 1

Let us consider the logarithmic equation log_{b}(b) = x

Converting into its exponential form, we get

b^{x} = b

⇒ b^{x} = b^{1}

⇒ x = 1

Thus, **log**_{b}**(b) = 1**

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

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

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

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

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

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

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

Here are some more special properties of natural logarithm:

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

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

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:

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

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

**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, ${\log _{5}\left[ \dfrac{15\left( x-3\right) ^{2}\left( x+4\right) ^{3}}{7\left( x+1\right) }\right]}$ = ${1+\log _{5}3-\log _{5}7+2\log _{5}\left( x-3\right) +3\log _{5}\left( x+4\right) -\log _{5}\left( x+1\right)}$

Last modified on May 24th, 2024