Logarithm Properties

Basic Properties of Logarithms

Like exponents, logarithms also have several important properties that need to be understood. These properties will be very helpful in solving various logarithmic problems.

Let $a > 0$ and $a \neq 1$, $b > 0$, $c > 0$, $m > 0$, $m \neq 1$, where $a, b, c, m, n$ are real numbers $(a, b, c, m, n \in \mathbb{R})$. The following are logarithm properties:

1. $^a\log a = 1$
2. $^a\log 1 = 0$
3. $^a\log a^n = n$
4. $^a\log (b \times c) = ^a\log b + ^a\log c$
5. $^a\log \left(\frac{b}{c}\right) = ^a\log b - ^a\log c$
6. $^a\log b^n = n \cdot ^a\log b$
7. $^a\log b = \frac{^m\log b}{^m\log a} = \frac{1}{^b\log a}$
8. $^a\log b \times ^b\log c = ^a\log c$

Proving Logarithm Properties

Logarithm of Multiplication

Property $4$: $^a\log (b \times c) = ^a\log b + ^a\log c$

Proof: Let $^a\log b = m$ and $^a\log c = n$

This means:

$b = a^m$ and $c = a^n$

Using the property of exponents:

Therefore:

Logarithm of Division

Property $5$: $^a\log \left(\frac{b}{c}\right) = ^a\log b - ^a\log c$

Proof: Let $^a\log b = m$ and $^a\log c = n$

Then $b = a^m$ and $c = a^n$

Recall that $\frac{a^m}{a^n} = a^{m-n}$, so:

Therefore:

Logarithm of Power

Property $6$: $^a\log b^n = n \cdot ^a\log b$

Proof: Let $^a\log b = m$

$^a\log b^n$ means the logarithm of $b$ raised to the power of $n$

Using property $4$ repeatedly:

Change of Base

Property $7$: $^a\log b = \frac{^m\log b}{^m\log a} = \frac{1}{^b\log a}$

Proof: Based on the definition of logarithm, $^a\log b = c$ if and only if $b = a^c$

Suppose we use base $m$ for the logarithm of $b$:

Using property $6$:

Since $c = ^a\log b$, then:

Therefore:

If $m = b$, then:

Chain Rule for Logarithms

Property $8$: $^a\log b \times ^b\log c = ^a\log c$

Proof: Based on the definition:

Substitute the value of $b$ into the equation for $c$:

Since $c = a^{mn}$, then:

Application Example

Suppose we want to calculate $^5\log 125$.

Using property $6$:

Exercises

1. Simplify the following expressions:

   a. $^9\log 81$

   b. $^2\log 64 - ^2\log 16$

   c. $^4\log 16^{10}$

2. If $^5\log 4 = m$, $^4\log 3 = n$, express $^{12}\log 100$ in terms of $m$ and $n$.

3. The population of city $A$ in $2010$ was $300{,}000 \text{ people}$. The average population growth rate is $6\%$ per year. If the population growth is assumed to be the same each year, in how many years will the population of city $A$ become $1 \text{ million}$?

4. How much time is needed for Dini's money, which was initially $\text{Rp}2{,}000{,}000.00$, to become $\text{Rp}6{,}500{,}000.00$ if she saves it in a bank that gives her an interest rate of $12\%$?

Answer Key

1. Determining logarithm values

   a. Answer:

   $$^9\log 81 = ^9\log 9^2$$

   $$^9\log 81 = 2 \cdot ^9\log 9$$

   $$^9\log 81 = 2 \cdot 1 = 2$$

   b. Answer:

   $$^2\log 64 - ^2\log 16 = ^2\log \frac{64}{16}$$

   $$^2\log 64 - ^2\log 16 = ^2\log 4$$

   $$^2\log 64 - ^2\log 16 = 2$$

   c. Answer:

   $$^4\log 16^{10} = ^4\log (4^2)^{10}$$

   $$^4\log 16^{10} = ^4\log 4^{20}$$

   $$^4\log 16^{10} = 20$$

2. Given that $^5\log 4 = m$, $^4\log 3 = n$

   Then:

   $$^{12}\log 100 = \frac{^4\log 100}{^4\log 12}$$

   $$= \frac{^4\log(4 \times 25)}{^4\log(4 \times 3)}$$

   $$= \frac{^4\log 4 + ^4\log 25}{^4\log 4 + ^4\log 3}$$

   $$= \frac{^4\log 4 + 2 \cdot ^4\log 5}{^4\log 4 + ^4\log 3}$$

   $$= \frac{1 + 2 \cdot \frac{1}{m}}{1 + n}$$

   $$= \frac{1 + \frac{2}{m}}{1 + n}$$

3. The initial population is $300{,}000 \text{ people}$

   The annual population growth is $6\%$.

   The appropriate function to describe population growth in $x \text{ years}$ is:

   $$f(x) = 300{,}000(1 + 0.06)^x$$

   For a population of $1{,}000{,}000 \text{ people}$:

   $$1{,}000{,}000 = 300{,}000(1 + 0.06)^x$$

   $$1{,}000{,}000 = 300{,}000(1.06)^x$$

   $$\frac{1{,}000{,}000}{300{,}000} = (1.06)^x$$

   $$3.33 = (1.06)^x$$

   $$x = ^{1.06}\log 3.33$$

   $$x = 20.645$$

   Therefore, the population will reach $1{,}000{,}000 \text{ people}$ in $20$ or $21 \text{ years}$.

4. The initial savings are $\text{Rp}2{,}000{,}000.00$

   The final savings are $\text{Rp}6{,}500{,}000.00$

   The interest rate is $12\%$.

   The appropriate function to describe Dini's savings in $x \text{ years}$ is:

   $$f(x) = 2{,}000{,}000(1 + 0.12)^x$$

   For a final saving amount of $\text{Rp}6{,}500{,}000.00$:

   $$6{,}500{,}000 = 2{,}000{,}000(1 + 0.12)^x$$

   $$6{,}500{,}000 = 2{,}000{,}000(1.12)^x$$

   $$\frac{6{,}500{,}000}{2{,}000{,}000} = (1.12)^x$$

   $$3.25 = (1.12)^x$$

   $$x = ^{1.12}\log 3.25$$

   $$x = 10.4$$

   Therefore, Dini's savings will reach $\text{Rp}6{,}500{,}000.00$ in $10 \text{ years}$.