Logarithmic Properties

There are various properties of logarithmic functions that are useful in manipulating them. They enable us to simplify expressions and solve equations involving logarithms.

Product Rule

The product rule for logarithms states that the logarithm of a product is the sum of the logarithms of the factors. This means that $\log_b(MN) = \log_b(M) + \log_b(N)$ for $b > 0$.

Given the logarithm of a product, we can write an equivalent sum of logarithms through the following steps...

1. Factor the argument of the logarithm completely, expressing each whole number factor as a product of primes.
2. Write the equivalent expression by summing the logarithms of each factor.

$\underline{Example}$

Expand $\log_b(8k)$.

The argument of the logarithm is $8k$ which can be factored into $8$ and $k$. There is a whole number $8$ which we can factorize till we only have prime numbers. This gives us $8 = 2 * 2 * 2$.

All the factors include $2$, $2$, $2$, and $k$. The product rule states that logarithm of the product is the sum of the logarithms of the factors which means that $\log_b(8k) = \log_b(2) + \log_b(2) + \log_b(2) + \log_b(k) = 3\log_b(2) + \log_b(k)$.

Quotient Rule

The quotient rule for logarithms states that the logarithm of a quotient is the difference of individual logarithms. This means that $\log_b(\frac{M}{N}) = \log_b(M) - \log_b(N)$.

Given the logarithm of a quotient, we can write an equivalent difference of logarithms through the following steps...

1. Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
2. Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
3. Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.

$\underline{Example}$

Expand $\log_3(\frac{7x^2 + 21x}{7x(x - 1)(x - 2)})$.

We can factor the numerator further to get $7x^2 + 21x = 7x(x + 3)$. However, the denominator is already factored. This gives us the fraction $\frac{7x(x + 3)}{7x(x - 1)(x - 2)}$. In order to simplify the fraction, we can cancel out the common terms in the numerator and denominator to get $\frac{x + 3}{(x - 1)(x - 2)}$. This is the argument in lowest terms.

The quotient rule states that the logarithm of a quotient is the difference of the logarithms of the numerator and denominator. This means that $\log_3(\frac{x + 3}{(x - 1)(x - 2)}) = \log_3(x + 3) - \log_3((x - 1)(x - 2))$.

We can expand the logarithm even further by applying the product rule for $\log_3((x - 1)(x - 2))$ to get $\log_3(x + 3) - (\log_3(x - 1) + \log_3(x - 2)) = \log_3(x + 3) - \log_3(x - 1) - \log_3(x - 2)$.

When fully expanded, the logarithm of the given expression is $\log_3(x + 3) - \log_3(x - 1) - \log_3(x - 2)$.

Power Rule

The power rule for logarithms states that the logarithm of a power is the product of the exponent and the logarithm of the base. This means that $\log_b(M^n) = n\log_b(M)$.

Given the logarithm of a power, we can write an equivalent product of a factor and a logarithm through the following steps...

1. Express the argument as a power, if needed.
2. Write the equivalent expression by multiplying the exponent times the logarithm of the base.

$\underline{Example}$

Expand $\log_3(25)$ using the power rule for logs.

We can rewrite $25$ as $5^2$. The power rule states that the logarithm of a power is the product of the exponent and the logarithm of the base. This means that $\log_3(25) = \log_3(5^2) = 2\log_3(5)$.

Law of Logs

Taken together, the product, quotient, and power rules are often called the law of logs and there are many cases where we use them together to simplify expressions.

$\underline{Example}$

Expand $\ln(\frac{x^4y}{7})$

Applying the quotient rule, we get $\ln(x^4y) - \ln(7)$.

We can further expand $\ln(x^4y)$ using the product rule to get $\ln(x^4) + \ln(y)$.

Finally, we can expand $\ln(x^4)$ using the power rule to get $4\ln(x)$.

Putting it all together, we get $4\ln(x) + \ln(y) - \ln(7)$.

We can also work backwards to combine logarithms using the properties of logarithms. Given a sum, difference, or product of logarithms with the same base, we can write an equivalent expression as a single logarithm using the following steps...

1. Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.
2. Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.
3. Finally, apply the quotient property. Rewrite differences of logarithms as the logarithm of a quotient. Note that addition and subtraction of logarithms are condensed from left to right.

$\underline{Example}$

Condense $\log(5) + 0.5\log(x) - \log(7x - 1) + 3\log(x - 1)$ as a single logarithm.

We can begin by applying the power property to rewrite $0.5\log(x)$ as $\log(x^{0.5}) = \log(\sqrt{x})$. We can also rewrite $3\log(x - 1)$ as $\log((x - 1)^3)$ which together gives us $\log(5) + \log(\sqrt{x}) - \log(7x - 1) + \log((x - 1)^3)$.

Now that we only have sums and differences of logarithms, we can start simplifying the expression left to right. We can rewrite $\log(5) + \log(\sqrt{x})$ as $\log(5\sqrt{x})$ which gives us $\log(5\sqrt{x}) - \log(7x - 1) + \log((x - 1)^3)$.

Next, we can rewrite $\log(5\sqrt{x}) - \log(7x - 1)$ as $\log(\frac{5\sqrt{x}}{7x - 1})$ which gives us $\log(\frac{5\sqrt{x}}{7x - 1}) + \log((x - 1)^3)$.

Finally, we can rewrite $\log(\frac{5\sqrt{x}}{7x - 1}) + \log((x - 1)^3)$ as $\log(\frac{5(x - 1)^3\sqrt{x}}{7x - 1})$.

Our final condensed expression is $\log(\frac{5(x - 1)^3\sqrt{x}}{7x - 1})$.

Change of Base Formula

Most calculators and computer software can only compute logarithms with base $10$ or base $e$. However, we can use the change-of-base formula to rewrite any logarithm as the quotient of logarithms of any other base. Given any positive real numbers $M$, $b$, and $n$, where $n \neq 1$ and $b \neq 1$, $\log_b(M) = \frac{\log_n(M)}{\log_n(b)}$.

We can derive the change-of-base formula by using the definition and properties of logarithms. Let $y = \log_b(M)$, we can rewrite this as $b^y = M$. Using this we get...

$$\begin{array}{llllllllll} b^y &=& M \\ \log_n(b^y) &=& \log_n(M) \\ y\log_n(b) &=& \log_n(M) \\ y &=& \frac{\log_n(M)}{\log_n(b)} \\ \log_b(M) &=& \frac{\log_n(M)}{\log_n(b)} \\ \end{array}$$

$\underline{Example}$

Change $\log_{0.5}(8)$ to a quotient of natural logarithms.

Using the change-of-base formula, we can rewrite $\log_{0.5}(8) = \frac{\ln(8)}{\ln(0.5)}$. We can further simplify this to get $\frac{\ln(8)}{\ln(0.5)} = \frac{\ln(2^3)}{\ln(2^{-1})} = \frac{3\ln(2)}{-1\ln(2)} = -3$.