Log Formulas

Before learning log formulas, let us recall what are logs (logarithms). A logarithm is just another way of writing exponents. When we cannot solve a problem using the exponents, then we use logarithms. There are different logarithm formulas that are derived by using the laws of exponents. Let us learn them using a few solved examples.

What are Log Formulas?

Before going to learn the log formulas, let us recall a few things. There are two types of logarithms, common logarithm (which is written as "log" and its base is 10 if not mentioned) and natural logarithm (which is written as "ln" and its base is always "e"). The below logarithm formulas are shown for common logarithms. However, they are all applicable for natural logarithms as well. Here are the most commonly used log formulas.

• log_b 1 = 0
• log_b b = 1
• log_b (xy) = log_b x + log_b y
• log_b (x / y) = log_b x - log_b y
• log_b a^x = x log_b a
• log_b a = (log_c a) / (log_c b)

Some of these rules have specific names like log_b (xy) = log_b x + log_b y is called the product formula of logs. In the same way, all the properties along with their names are mentioned in the table below.

Logarithmic Formulas Derivation

Here is the derivation of some important log formulas. We use the laws of exponents in the derivation of log formulas.

Product Formula of logarithms

The product formula of logs is, log_b (xy) = log_b x + log_b y.

Derivation:

Let us assume that log_b x = m and log_b y = n. Then by the definition of logarithm,

x = b^m and y = bⁿ.

Then xy = b^m × bⁿ = b^{m + n} (by a law of exponents, a^m × aⁿ = a^{m + n})

Converting xy = b^{m + n} into logarithmic form, we get

m + n = log_b xy

Substituting the values log_b x = m and log_b y = n here,

log_b (xy) = log_b x + log_b y

Quotient Formula of logarithms

The quotient formula of logs is, log_b (x/y) = log_b x - log_b y.

Derivation:

Let us assume that log_b x = m and log_b y = n. Then by the definition of logarithm,

x = b^m and y = bⁿ.

Then x/y = b^m / bⁿ = b^{m - n} (by a law of exponents, a^m / aⁿ = a^{m - n})

Converting x/y = b^{m - n} into logarithmic form, we get

m - n = log_b (x/y)

Substituting the values log_b x = m and log_b y = n here,

log_b (x/y) = log_b x - log_b y

Power Formula of Logarithms

The power formula of logarithms says log_b a^x = x log_b a.

Derivation:

Let log_b a = m. Then by the definition of logarithm, a = b^m.

Raising both sides by x, we get

a^x = (b^m)^x

a^x = b^mx (by a law of exponents, (a^m)ⁿ = a^mn)

Converting this back into logarithmic form,

log_b a^x = m x

Substitute m = log_b a here,

log_b a^x = x log_b a

Change of Base Formula of Logarithms

The change of base formula of logs says log_b a = (log_c a) / (log_c b).

Derivation:

Assume that log_b a = x, log_c a = y, and log_c b = z.

Converting these into exponential forms,

a = b^x ... (1)

a = c^y ... (2)

b = c^z ... (3)

From (1) and (2),

b^x = c^y

(c^z)^x = c^y (from (3))

c^zx = c^y

Since the bases are the same, the powers also should be the same.

zx = y (or) x = y / z.

Substituting the values of x, y, and z here back,

log_b a = (log_c a) / (log_c b).

Examples Using Logarithm Formulas

Example 1: Convert the following from exponential form to logarithmic form using the log formulas. a) 5³ = 125 b) 3^-3 = 1 / 27.

Solution:

Using the definition of the logarithm,

b^x = a ⇒ log_b a = x

Using this,

a) 5³ = 125 ⇒ log₅ 125 = 3

b) 3^-3 = 1 / 27 ⇒ log₃ 1/27 = -3

Answer: a) log₅ 125 = 3; b) log₃ 1/27 = -3.

Example 2: Compress the following expression as a single logarithm by using logarithmic formulas. 5 log x + log y - 8 log z.

Solution:

To find: The compressed form of the given expression as a single logarithm using logarithm formulas.

5 log x + log y - 8 log z

= (5 log x - 8 log z) + log y (Regrouped the terms)

= (log x⁵ - log z⁸) + log y (∵ a log x = log x^a)

= log (x⁵/z⁸) + log y (∵ log x - log y = log (x/y) )

= log (x⁵y/z⁸) (∵ log x + log y = log (xy) )

Answer: 5 log x + log y - 8 log z = log (x⁵y/z⁸).

Example 3: Find the integer value of log₃ (1/9) using log formulas.

Solution:

log₃ (1/9) = log₃ 1 - log₃ 9 (∵ log_b (x / y) = log_b x - log_b y)

= 0 - log₃ 3² (∵ log_b 1 = 0)

= - 2 log₃ 3 (∵ log_b a^x = x log_b a)

= -2 (1) (∵ log_b b = 1)

= -2

Answer: log₃ (1/9) = -2.