Problem 81 - Precalculus

Question

$$\text { Show that } \log _{b}\left(\frac{a^{2}}{b^{3}}\right)^{-3}=9-\frac{6}{\log _{a} b}$$

Step-by-Step Solution

Answer

The expression simplifies to \(9 - \frac{6}{\log_{a} (b)}\).

Step 1: Rewrite the Logarithmic Expression

Let's start by rewriting the given logarithmic expression: \[ \log_{b}\left(\frac{a^{2}}{b^{3}}\right)^{-3} \]Using the property \(\log_{b}(x^{-n}) = -n \cdot \log_{b}(x)\), this can be rewritten as:\[ -3 \cdot \log_{b} \left( \frac{a^2}{b^3} \right) \]

Step 2: Apply Logarithmic Properties

Now apply the logarithmic properties to \(\log_{b}\left(\frac{a^2}{b^3}\right)\). Using \(\log_{b}(\frac{m}{n}) = \log_{b}(m) - \log_{b}(n)\), we have:\[ \log_{b} \left( a^2 \right) - \log_{b} \left( b^3 \right) \]

Step 3: Simplify Using Exponent Properties

Evaluate \(\log_{b} \left( a^2 \right)\) and \(\log_{b} \left( b^3 \right)\) separately using the rule \(\log_{b}(x^n) = n \cdot \log_{b}(x)\):- \(\log_{b} \left( a^2 \right) = 2 \cdot \log_{b} (a)\)- \(\log_{b} \left( b^3 \right) = 3 \cdot \log_{b} (b) = 3\) (since \(b = b\))So, the expression becomes:\[ 2 \cdot \log_{b} (a) - 3 \]

Step 4: Substitute Back Into Initial Expression

Now substitute back into the initial expression:\[ -3 \cdot (2 \cdot \log_{b} (a) - 3) = -6 \cdot \log_{b} (a) + 9 \]

Step 5: Use Change of Base Formula

We need to express \(-6 \cdot \log_{b} (a)\) in terms of \(\log_{a} (b)\). Using the change of base formula \(\log_{b} (a) = \frac{1}{\log_{a} (b)}\):\[-6 \cdot \frac{1}{\log_{a} (b)} = -\frac{6}{\log_{a} (b)}\]

Step 6: Final Combine

Substitute \(-\frac{6}{\log_{a} (b)}\) into the expression from Step 4:\[ 9 - \frac{6}{\log_{a} (b)} \]Thus, the original expression simplifies to:\[ \log_{b}\left(\frac{a^{2}}{b^{3}}\right)^{-3} = 9 - \frac{6}{\log_{a} (b)} \]

Key Concepts

Logarithmic Properties

Logarithms have several properties that help simplify complex logarithmic expressions. One of these properties is when you have a power within a logarithm, such as \( \log_{b}(x^{-n}) \). This can be simplified by bringing the exponent in front of the logarithm, using the rule \( \log_{b}(x^{-n}) = -n \cdot \log_{b}(x) \). This property is particularly useful in reducing the complexity of expressions involving powers.

Another essential property is the quotient rule. When dealing with logarithms of fractions, like \( \log_{b}\left(\frac{m}{n}\right) \), it can be rewritten as \( \log_{b}(m) - \log_{b}(n) \). This turns a division inside the log into a subtraction, making it possible to handle each part separately. These properties are foundations for breaking down complicated expressions, enabling easier manipulation and solution.

These can also be combined with multiplication properties when multiple logarithms are involved, providing a robust toolkit for solving logarithmic equations.

Exponent Properties

Exponent properties play a crucial role in simplifying logarithmic expressions. These properties allow us to deal effectively with terms like \( x^n \) within logarithms. The core property here is that \( \log_{b}(x^n) = n \cdot \log_{b}(x) \). This rule lets you move the exponent in front of the log as a multiplier, which simplifies the calculation by converting exponential growth into linear scaling.

For instance, the expression \( \log_{b}(a^2) \) can be rewritten as \( 2 \cdot \log_{b}(a) \), making it easier to substitute and solve when combined with other properties. The same simplification applies to \( \log_{b}(b^3) \), which simplifies to \( 3 \cdot \log_{b}(b) \).

Exponent properties help maintain the balance and accuracy of equations involving exponents, preparing them for more straightforward calculation or further transformation using logarithmic properties.

Change of Base Formula

The change of base formula is a helpful tool when you need to convert logarithms from one base to another. This formula is especially useful when dealing with multiple logarithm bases in one expression. The formula states that \( \log_{b}(a) \) can be converted using \( \log_{b}(a) = \frac{1}{\log_{a}(b)} \).

This allows flexibility in how you set up and solve logarithmic problems, offering a means to work exclusively in a single preferred base if needed. For example, in the original exercise, we needed to express \( -6 \cdot \log_{b}(a) \) in terms of \( \log_{a}(b) \). By using the change of base formula, we were able to rewrite it as \( -\frac{6}{\log_{a}(b)} \).

The change of base formula not only simplifies computations but also helps in transforming expressions so they can be solved using known values and properties, consolidating the expression into manageable parts.