3.1.7 Log Rules

Logarithms are usually tested on the third and fourth columns of the test, however, if logarithm rules are fully understood these can be some of the simplest problems on the test. The following is a collection of log rules which are actively tested:

• Definition: $\log_a b = x \Leftrightarrow a^x = b$
• Power Rule: $\log_a b^n = n \log_a b$
• Addition of Logs: $\log_a b + \log_a c = \log_a (bc)$
• Subtraction of Logs: $\log_a b - \log_a c = \log_a (\frac{b}{c})$
• Change of Bases: $\log_a b = \frac{\log b}{\log a}$

In the above table $\log_{10} a$ is represented as $\log a$. The following are some sample problems illustrating how each one of the rules might be tested:

Example: Find $\log_4 .0625$. Solution: Applying the definition we know that $4^x = .0625 = \frac{1}{16}$. Therefore, our answer is $x = -2$.

Example: Find $\log_8 16$. Solution: Again, applying the definition, $8^x = 16$, which can be changed to $2^{3x} = 2^4 \Rightarrow x = \frac{4}{3}$.

Example: Find $\log_{12} 16 + \log_{12} 36 - \log_{12} 4$. Solution: We know from the addition/subtraction of logs that the above expression can be written as $\log_{12} \frac{16 \cdot 36}{4} = \log_{12} 16 \cdot 9 = \log_{12} 144 \Rightarrow 12^x = 144 \Rightarrow x = 2$.

Example: Find $\log_5 8 \div \log_{25} 16$ Solution: These are probably the most challenging logarithm problems you will see on the exam. They involved changing bases and performing the power rule. Let’s look at what happens when we change bases: $\log_5 8 \div \log_{25} 16 = \frac{\log 8}{\log 5} \div \frac{\log 16}{\log 25} = \frac{\log 2^3}{\log 5} \times \frac{\log 5^2}{\log 2^4} = \frac{3 \cdot \log 2}{\log 5} \times \frac{2 \cdot \log 5}{4 \cdot \log 2} = 3 \times \frac{1}{2} = \frac{3}{2}$.

In addition to the above problems, there are some approximations of logarithms which pop up. For those, there are some quantities which would be nice to have memorized to compute a more accurate approximations. Those are:

• $\log_{10} 2 \approx .3$
• $\log_{10} 5 \approx .7$
• $\ln 2 \approx .7$
• $\ln 10 \approx 2.3$

Where $\ln x = \log_e x$.

The following is example of how approximations of logs can be calculated: $200 \log 200 = 200 \log(2 \cdot 100) = 200 \cdot (\log 2 + \log 100) \approx 200 \cdot (.3 + 2) = 460$

The following are some more practice problems:

Problem Set 3.1.7