Properties of Logarithms
The key properties of logarithms are:
- Logarithm of a Product
The logarithm of a product equals the sum of the logarithms of the individual factors. $$ log_b(x \cdot y) = \log_b x + \log_b y \ \ \ with \ x>0 \ , \ y>0 $$Example $$ log_2(8 \cdot 4) = \log_2 8 + \log_2 4 $$
- Logarithm of a Quotient
The logarithm of a quotient equals the difference between the logarithms of the numerator and the denominator. $$ log_b( \frac{x}{y} ) = \log_b x - \log_b y \ \ \ with \ x>0 \ , \ y>0 $$Example $$ log_2( \frac{8}{4}) = \log_2 8 - \log_2 4 $$
- Logarithm of a Power
The logarithm of a number raised to a power equals the exponent multiplied by the logarithm of the base number. $$ log_b( x^y ) = y \cdot \log_b x \ \ \ with \ x>0 $$Example $$ log_2( 8^2 ) = 2 \cdot \log_2 8 $$
- Logarithm of a Radical
The logarithm of a root is equal to the logarithm of the radicand divided by the index of the root. $$ log_b \ \sqrt[n]{x} = \frac{1}{n} \cdot \log_b x \ \ \ with \ x>0 $$Example $$ log_2 \ \sqrt[3]{8} = \frac{1}{3} \cdot \log_2 8 $$
A Practical Example
Example 1
Consider the following calculation:
$$ log_2(8 \cdot 4) = \log_2 8 + \log_2 4 $$
The logarithm on the left side simplifies to the logarithm of 32 base 2:
$$ log_2(32) = \log_2 8 + \log_2 4 $$
The logarithm of 32 base 2 is 5, since 25 = 32.
$$ 5 = \log_2 8 + \log_2 4 $$
On the right side, the logarithm of 8 base 2 equals 3, because 23 = 8.
$$ 5 = 3 + \log_2 4 $$
And the logarithm of 4 base 2 equals 2, since 22 = 4.
$$ 5 = 3 + 2 $$
$$ 5 = 5 $$
This confirms the identity.
Example 2
Consider this example:
$$ log_2( \frac{8}{4} ) = \log_2 8 - \log_2 4 $$
The left side simplifies to the logarithm of 2 base 2:
$$ log_2(2) = \log_2 8 - \log_2 4 $$
The logarithm of 2 base 2 equals 1, since 21 = 2.
$$ 1 = \log_2 8 - \log_2 4 $$
On the right side, the logarithm of 8 base 2 is 3, because 23 = 8.
$$ 1 = 3 - \log_2 4 $$
And the logarithm of 4 base 2 is 2, since 22 = 4.
$$ 1 = 3 - 2 $$
$$ 1 = 1 $$
So the identity checks out.
Example 3
Consider this example:
$$ log_2( 8^2 ) = 2 \cdot \log_2 8 $$
The left side simplifies to the logarithm of 64 base 2:
$$ log_2(64) = 2 \cdot \log_2 8 $$
The logarithm of 64 base 2 equals 6, since 26 = 64.
$$ 6 = 2 \cdot \log_2 8 $$
On the right side, the logarithm of 8 base 2 equals 3, because 23 = 8.
$$ 6 = 2 \cdot 3 $$
$$ 6 = 6 $$
This verifies the identity.
Example 4
Consider this example:
$$ log_2 \ \sqrt[3]{8} = \frac{1}{3} \cdot \log_2 8 $$
The cube root of 8 equals 2, since 23 = 8.
$$ log_2 \ 2 = \frac{1}{3} \cdot \log_2 8 $$
The logarithm of 2 base 2 equals 1, because 21 = 2.
$$ 1 = \frac{1}{3} \cdot \log_2 8 $$
And the logarithm of 8 base 2 equals 3, since 23 = 8.
$$ 1 = \frac{1}{3} \cdot 3 $$
$$ 1 = 1 $$
So the identity holds true.
The Proof
In this section, we’ll prove each of the logarithm properties.
1] Logarithm of a Product
The property we want to prove is:
$$ \log_b (x \cdot y) = \log_b x + \log_b y $$
Let’s define:
$$ A = log_b x $$
$$ B = log_b y $$
From these, we can express x and y as:
$$ x = b^A $$
$$ y = b^B $$
Multiplying x and y then gives:
$$ x \cdot y = b^A \cdot b^B $$
Using the exponent rule for the same base:
$$ x \cdot y = b^{A+B} $$
Taking the logarithm of both sides yields:
$$ \log_b (x \cdot y) = \log_b (b^{A+B}) $$
Applying the rule that logb bA+B = A + B, we get:
$$ \log_b (x \cdot y) = A + B $$
Substituting back for A and B, we have:
$$ \log_b (x \cdot y) = \log_b x + \log_b y $$
Thus, the property is proven.
2] Logarithm of a Quotient
The property we want to prove is:
$$ \log_b ( \frac{x}{y} ) = \log_b x - \log_b y $$
Let’s define:
$$ A = log_b x $$
$$ B = log_b y $$
So:
$$ x = b^A $$
$$ y = b^B $$
Therefore:
$$ \frac{x}{y} = \frac{b^A}{b^B} $$
Applying the exponent rule gives:
$$ \frac{x}{y} = b^{A-B} $$
Taking the logarithm of both sides:
$$ \log_b \frac{x}{y} = \log_b b^{A-B} $$
Which simplifies to:
$$ \log_b \frac{x}{y} = A - B $$
Substituting back for A and B:
$$ \log_b \frac{x}{y} = \log_b x - \log_b y $$
Thus, the property is proven.
3] Logarithm of a Power
The property we want to prove is:
$$ log_b( x^y ) = y \cdot \log_b x $$
Let’s define:
$$ A = \log_b x $$
So:
$$ x = b^A $$
Raising x to the power y gives:
$$ x^y = ( b^A )^y $$
Using exponent rules:
$$ x^y = b^{A \cdot y} $$
Taking the logarithm of both sides:
$$ \log_b (x^y) = \log_b b^{A \cdot y} $$
Which simplifies to:
$$ \log_b (x^y) = A \cdot y $$
Since A = logb x, we get:
$$ \log_b (x^y) = (\log_b x) \cdot y $$
$$ \log_b (x^y) = y \cdot \log_b x $$
Thus, the property is proven.
4] Logarithm of a Radical
The property we want to prove is:
$$ log_b \ \sqrt[n]{x} = \frac{1}{n} \cdot \log_b x $$
We can write the nth root as a power with an exponent of 1/n:
$$ log_b \ x^{\frac{1}{n}} = \frac{1}{n} \cdot \log_b x $$
By letting y = 1/n, this becomes the same as the logarithm of a power, which we’ve already shown.
$$ log_b \ x^y = y \cdot \log_b x $$
And so on.
