Logarithm Properties

Eight Rules from Exponent Laws

Each logarithm property follows from a corresponding exponent law, so every formula below can be derived from that law.

All the properties below use these conditions:

Whenever $m$ or $b$ is used as a logarithm base, that base must also be positive and not equal to one.

The first three rules come directly from $a^1=a$, $a^0=1$, and the inverse relationship between logarithms and exponents.

Where the Logarithm Rules Come From

Every logarithm rule follows from the matching exponent rule. A logarithm reports the exponent that produces a given number, so a rule about exponents becomes a rule about logarithms.

Product Rule

Let:

Multiplying $b$ and $c$ adds their exponents:

Taking the base-$a$ logarithm gives:

Quotient Rule

Using the same substitutions, division subtracts the exponents:

The calculation gives:

Power Rule

Let $y=\log_a b$, so $b=a^y$. For any real $n$, the condition $b>0$ makes $b^n$ well-defined:

This proof covers real exponents directly. Repeated multiplication would explain only positive integer exponents.

Change of Base and Reciprocal Rule

Let $x=\log_a b$, so $b=a^x$. Take the logarithm of both sides using any valid base $m$:

Because $x=\log_a b$, this proves the change-of-base formula. Choosing $m=b$ gives the reciprocal relationship:

Chain Rule

After both logarithms are written with the same valid base, $\log_m b$ appears once in the numerator and once in the denominator, so the two factors cancel:

Evaluating a Logarithm from Exponential Form

Calculate $\log_5 125$.

Because $125=5^3$, the power and inverse rules give:

The result also answers the original exponent question: $5^3=125$.

Exercises

The four problems below practice the product, quotient, power, and change-of-base rules together. Work each one before reading the worked solutions, then compare your steps with the answers.

1. Simplify three expressions that each need one rule: $\log_9 81$, $\log_2 64-\log_2 16$, and $\log_4(16^{10})$.

2. If $\log_5 4=m$ and $\log_4 3=n$, express $\log_{12}100$ in terms of $m$ and $n$.

3. A city has $300{,}000$ residents and is modeled as growing by $6\%$ at the end of each year. Under this fixed-rate model, after how many complete years will its population first reach at least $1{,}000{,}000$?

4. Dini deposits $\text{Rp}\,2{,}000{,}000$ in an account paying $12\%$ interest compounded annually. Assume no further deposits, withdrawals, fees, or rate changes. After how many complete years will the balance first reach at least $\text{Rp}\,6{,}500{,}000$?

Worked Solutions

1. Simplify each expression.

   1. Since $81=9^2$:

      $$\log_9 81=\log_9(9^2)$$

      $$\log_9(9^2)=2$$

   2. Use the quotient rule:

      $$\log_2 64-\log_2 16=\log_2\left(\frac{64}{16}\right)$$

      $$\log_2 4=2$$

   3. Since $16=4^2$:

      $$\log_4(16^{10})=\log_4((4^2)^{10})$$

      $$\log_4(4^{20})=20$$

2. Use base $4$ for the numerator and denominator:

   $$\log_{12}100=\frac{\log_4 100}{\log_4 12}$$

   $$\begin{aligned} \log_4 100 &=\log_4(4\cdot25) \\ &=1+2\log_4 5 \end{aligned}$$

   $$\log_4 12=\log_4(4\cdot3)=1+n$$

   The reciprocal rule gives $\log_4 5=\frac1{\log_5 4}=\frac1m$. Therefore:

   $$\log_{12}100=\frac{1+\frac2m}{1+n}$$

3. The fixed-rate model is:

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

   Solve the continuous equation first:

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

   $$x=\frac{\ln(10/3)}{\ln(1.06)}\approx20.662354$$

   The model updates after complete years, so check the neighboring integers:

   $$P(20)\approx962{,}141<1{,}000{,}000$$

   $$P(21)\approx1{,}019{,}869>1{,}000{,}000$$

   The population first reaches the target after $21\text{ complete years}$. This result assumes that the stated growth rate stays fixed for all $21$ years. A changing growth rate would give a different answer.

4. The annual-compounding model is:

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

   Solve the continuous equation:

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

   $$x=\frac{\ln(3.25)}{\ln(1.12)}\approx10.400324$$

   Interest is credited only after each complete year, so compare years $10$ and $11$:

   $$B(10)\approx6{,}211{,}696<6{,}500{,}000$$

   $$B(11)\approx6{,}957{,}100>6{,}500{,}000$$

   The balance first reaches the target after $11\text{ complete years}$.