Problem 30 - College Algebra

Question

For the following exercises, use properties of logarithms to evaluate without using a calculator. $$ \log _{3}\left(\frac{1}{9}\right)-3 \log _{3}(3) $$

Step-by-Step Solution

Answer

-5

Step 1: Simplify the First Term Using the Change of Base Formula

For the term \( \log _{3}\left(\frac{1}{9}\right) \), we recognize that \( \frac{1}{9} \) can be expressed as \( 3^{-2} \). Thus,\[ \log_{3} \left( \frac{1}{9} \right) = \log_{3} (3^{-2}) \]Using the logarithmic power rule \( \log_{b}(a^n) = n \cdot \log_{b}(a) \), this becomes:\[ \log_{3} (3^{-2}) = -2 \cdot \log_{3}(3) \]Since \( \log_{3}(3) = 1 \), it simplifies to \(-2 \cdot 1 = -2\).

Step 2: Simplify the Second Term Using the Logarithmic Power Rule

The term \( 3 \log_{3}(3) \) can also be simplified using the property \( n \log_{b}(a) = \log_{b}(a^n) \). Therefore,\[ 3 \log_{3}(3) = \log_{3}(3^3) \]Since \( \log_{3}(3) = 1 \), it ultimately evaluates to \( 3 \cdot 1 = 3 \).

Step 3: Combine the Simplified Terms

Now combine the results from Step 1 and Step 2:\[ \log_{3}\left(\frac{1}{9}\right) - 3 \log_{3}(3) = -2 - 3 \]Simplifying this gives:\[ -2 - 3 = -5 \].

Key Concepts

Change of Base Formula

The change of base formula is very useful when dealing with logarithms. It allows us to rewrite a logarithm in terms of logs with a different base. This is especially helpful when we need to work with logarithms that aren't conveniently expressed in our chosen base.
The general change of base formula is:

• \( \log_{b}(a) = \frac{\log_{k}(a)}{\log_{k}(b)} \)

This formula implies that any logarithm \( \log_{b}(a) \) can be written as a fraction of two logarithms with common base \( k \).
In the original exercise, however, we used a direct simplification because the expression involved \( \log_{3} \) which is already in terms of a single base. Nevertheless, knowing how to use change of base can rescue you when a calculator is not an option!

Logarithmic Power Rule

The logarithmic power rule is a handy property used in simplifying expressions that involve exponents inside the logarithms.
It states that if you have a logarithmic expression of something raised to a power, you can bring that power in front as a multiplier. The formula is straightforward:

• \( \log_{b}(a^n) = n \cdot \log_{b}(a) \)

In simpler terms, you can "move" the power to the front, turning it into a coefficient.

In our exercise, we used this rule to simplify both terms:

• For \( \log_{3}(3^{-2}) \), the exponent \(-2\) comes forward, resulting in \(-2 \cdot \log_{3}(3) = -2\).
• For \( 3 \log_{3}(3) \), we think of it as \( \log_{3}(3^3) \), hence simplifying to \( 3 \).

This property is extremely powerful when handling large exponents or fractions within logarithms.

Evaluating Logarithms

Evaluating logarithms often require reducing complex logarithmic expressions to simpler terms by leveraging various logarithm properties.
These properties include the product, quotient, and power rules, as previously described.
In terms of evaluating simple logarithms, it's important to recognize some key expressions:

• For example, \( \log_{b}(b) = 1 \) since \( b^1 = b \).
• Another important one is \( \log_{b}(1) = 0 \) because any base raised to the power of 0 gives us 1.

Knowing these can greatly speed up evaluating logarithms. In our exercise, once we simplified both of our terms separately (using the power rule and the concept \( \log_{3}(3) = 1 \)), all that was left was straightforward arithmetic: subtracting the results from each other.
Understanding these foundational rules is key to mastering logarithms and solving exercises like the one tackled.