Problem 87 - College Algebra

Question

In Exercises \(83-88,\) let \(\log _{b} 2=A\) and \(\log _{b} 3=\) C. Write each expression in terms of \(A\) and \(C\). $$ \log _{b} \sqrt{\frac{2}{27}} $$

Step-by-Step Solution

Answer

The simplified form of \(\log _{b} \sqrt{\frac{2}{27}}\) in terms of \(A\) and \(C\) is \(0.5 \cdot A - 1.5 \cdot C\)

Step 1: Recognize the given terms

The values given are already defined as \(\log _{b} 2=A\) and \(\log _{b} 3=C\). Thus, we have to express \(\log _{b} \sqrt{\frac{2}{27}}\) in terms of \(A\) and \(C\).

Step 2: Rewrite the square root as an exponent

Remember that a square root can be rewritten as an exponent of \(1/2\). So, \(\log _{b} \sqrt{\frac{2}{27}} = \log _{b} ( (2/27) ^ {\frac{1}{2}})\).

Step 3: Use the properties of Logarithms

The property \(\log _{b}(m/n) = \log _{b} m - \log _{b} n\) splits the fraction into a difference of two logs: \(\log _{b} (2) ^ {\frac{1}{2}} - \log _{b} (27) ^ {\frac{1}{2}}\). Further the property \(\log _{b} m^{n} = n \cdot \log _{b} m\), moves the exponent in front of the log: \( 0.5 \cdot \log _{b} 2 - 0.5 \cdot \log _{b} 27\)

Step 4: Simplify terms

As per the given values \(\log _{b} 2 = A\) and also write \(27\) as \(3^{3}\), we get: \( 0.5 \cdot A - 0.5 \cdot \log_{b}(3^{3})\), which simplifies to \( 0.5 \cdot A - 0.5 \cdot 3 \cdot \log_{b}3\), which can be further simplified to \( 0.5 \cdot A - 1.5 \cdot C\), since \(\log _{b}3 = C\) by given values.

Key Concepts

Logarithmic Identities

Logarithmic identities are essential in helping you manipulate and simplify expressions involving logarithms. These identities are a set of algebraic rules that allow transformations and simplifications:

• Product Rule: This rule states that \( \log_{b}(mn) = \log_{b}m + \log_{b}n \). It shows that the logarithm of a product is the sum of the logarithms.
• Quotient Rule: This identity is \( \log_{b}(m/n) = \log_{b}m - \log_{b}n \). It expresses that the logarithm of a quotient is the difference of the logarithms.
• Power Rule: Given by \( \log_{b}(m^n) = n \cdot \log_{b}m \). It indicates that the logarithm of a power is the exponent times the logarithm of the base.
• Change of Base Formula: Useful for converting logarithms from one base to another: \( \log_{b}m = \frac{\log_{k}m}{\log_{k}b} \).

Applying these identities makes solving logarithmic equations or evaluating logarithmic expressions much more manageable. In the given exercise, recognizing and applying the quotient and power rules were fundamental to simplifying the expression \( \log_{b} \sqrt{\frac{2}{27}} \) in terms of \( A \) and \( C \).

Exponent Rules

Exponent rules are closely linked with logarithms because logarithms are essentially exponents. These rules help in understanding how to manipulate expressions involving powers:

• Product of Powers: If bases are the same, add the exponents: \( a^m \cdot a^n = a^{m+n} \).
• Quotient of Powers: Subtract the exponents when dividing similar bases: \( \frac{a^m}{a^n} = a^{m-n} \).
• Power of a Power: Multiply the exponents: \( (a^m)^n = a^{m \cdot n} \).
• Power of a Product: Distribute the exponent: \( (ab)^n = a^n b^n \).
• Negative Exponent Rule: Represents the reciprocal: \( a^{-n} = \frac{1}{a^n} \).
• Zero Exponent Rule: Any non-zero base raised to the power of zero is one: \( a^0 = 1 \).

In the original exercise, understanding that a square root can be written as an exponent (\( \sqrt{2} = 2^{1/2} \)) was crucial to transforming the log expression \( \log_{b}(\sqrt{\frac{2}{27}}) \) into a more workable form.

Logarithmic Functions

Logarithmic functions help describe the inverse relationship of exponential functions. The function takes the form \( y = \log_{b}(x) \) and can be understood as finding the power to which \( b \) must be raised to yield \( x \):

• These functions have graphs that pass through the point \( (1,0) \) because \( \log_{b}(1) = 0 \).
• They have a vertical asymptote at \( x = 0 \), meaning the graph approaches but never touches this line.
• The domain of a logarithmic function is \( x > 0 \), as logarithms are undefined for non-positive values.
• The range extends over all real numbers, \( y \in \mathbb{R} \).

Understanding these properties is crucial for graphing and solving equations that involve logarithmic functions. In the exercise context, transforming the expression using log properties ultimately derived a function representation using terms \( A \) and \( C \), showcasing how logarithmic functions can elegantly express complex combinations.