Problem 32 - College Algebra with Corequisite Support

Question

For the following exercises, use properties of logarithms to evaluate without using a calculator. \(2 \log _{9}(3)-4 \log _{9}(3)+\log _{9}\left(\frac{1}{729}\right)\)

Step-by-Step Solution

Answer

The value of the expression is -4.

Step 1: Combine like terms

The expression given is: \[2 \log _{9}(3) - 4 \log _{9}(3) + \log _{9}\left(\frac{1}{729}\right)\]The first two terms can be combined because they are like terms: \[(2 \log _{9}(3) - 4 \log _{9}(3) = -2 \log _{9}(3)\]

Step 2: Apply power rule for logarithms

Using the power rule of logarithms which states \(a \log_b(x) = \log_b(x^a)\), we transform the term:\[-2 \log _{9}(3) = \log _{9}(3^{-2}) = \log _{9}\left(\frac{1}{9}\right)\]

Step 3: Simplify the expression

The expression is now:\[\log _{9}\left(\frac{1}{9}\right) + \log _{9}\left(\frac{1}{729}\right)\]We can use the product property of logarithms, \(\log_b(x) + \log_b(y) = \log_b(xy)\), to combine these logs:\[\log _{9}\left(\frac{1}{9} \times \frac{1}{729}\right)\]

Step 4: Evaluate the exponential expression

First, simplify the multiplication inside the log:\[\frac{1}{9} = 9^{-1} \quad \text{and} \quad \frac{1}{729} = 9^{-3}\]Therefore, \(\frac{1}{9} \times \frac{1}{729} = 9^{-1} \times 9^{-3} = 9^{-4}\).Thus, the expression simplifies further to:\[\log _{9}(9^{-4})\]

Step 5: Apply the inverse property

We use the inverse property of logarithms, which states \(\log_b(b^x) = x\), to simplify:\[\log _{9}(9^{-4}) = -4\]

Key Concepts

Properties of Logarithms

Logarithms have several properties that help us simplify and evaluate expressions. These include rules like the product, quotient, and power rules. These properties make calculations easier by transforming complicated expressions into simpler forms.

When dealing with logarithms, remember these helpful tips:

• The **product property** allows us to combine logarithms of multiplied values: \(\log_b(x) + \log_b(y) = \log_b(xy)\).
• The **quotient property** helps with logarithms of divided values: \(\log_b(x) - \log_b(y) = \log_b\left(\frac{x}{y}\right)\).
• The **power property** lets us simplify when we have logs with powers: \(a \cdot \log_b(x) = \log_b(x^a)\).

Understanding these basic properties allows you to manipulate logarithmic expressions effectively without needing a calculator. Always look to see which properties can be applied to solve the problem faster and more efficiently.

Power Rule for Logarithms

The power rule is a simple yet powerful tool for dealing with logarithms involving exponents. It states that if you have a logarithm with an exponent, you can move the exponent out in front of the log as a multiplication.

For instance, given an expression like \(a \log_b(x)\), it can be rewritten as \(\log_b(x^a)\). This transformation makes it easier to handle logarithms when the base number is involved with a power.

This rule came in handy in the original exercise where the expression \(-2 \log _{9}(3)\) is transformed into \(\log _{9}(3^{-2})\) or \(\log _{9}\left(\frac{1}{9}\right)\). This makes the evaluation process smoother, allowing us to combine terms with other log expressions more straightforwardly.

Product Property of Logarithms

The product property is a fundamental concept that allows us to combine logarithms of two multiplied values into a single logarithm. This property states that \(\log_b(x) + \log_b(y) = \log_b(xy)\).

Using the product property simplifies expressions significantly, helping to reduce them to fewer terms. This is because instead of evaluating multiple logs separately, you condense them into a singular calculation.

In our example, after applying the power rule, the expression \(\log _{9}\left(\frac{1}{9} \right) + \log _{9}\left(\frac{1}{729} \right)\) could be merged into \(\log _{9}\left(\frac{1}{9} \times \frac{1}{729} \right)\). This step is crucial to continue simplifying logarithmic expressions effectively.

Inverse Property of Logarithms

The inverse property of logarithms is the idea that a logarithm can cancel out an exponent with the same base. In simpler terms, for any base \(b\), \(\log_b(b^x) = x\).

This powerful property allows you to evaluate the log of a number when that number is the base raised to some power. It effectively "undoes" the exponentiation, simplifying evaluations significantly.

In the exercise, once the expression reached \(\log _{9}(9^{-4})\), applying the inverse property readily gave us \(-4\). This step effectively concludes the problem by simplifying a potentially daunting logarithmic expression to a simple integer or rational result.