Problem 40 - Calculus with Concepts in Calculus

Question

Let \(a>0 .\) Using the Change of Base Formula, show that \(\log _{1 / a} x=-\log _{a} x\) for \(x>0\)

Step-by-Step Solution

Answer

The change of base formula simplifies to \(\log_{1/a} x = -\log_a x\), confirming the equality for \(x > 0\).

Step 1: Understanding the Task

We need to show that the logarithm of a number to the reciprocal base is equal to the negative of the logarithm of the number to the original base. The task involves applying the change of base formula to demonstrate this property.

Step 2: Applying the Change of Base Formula

By the change of base formula, we can express \(\log_{1/a} x = \frac{\log_b x}{\log_b (1/a)}\) where \(b\) is any positive number. We will use \(b = a\), resulting in:\(\log_{1/a} x = \frac{\log_a x}{\log_a (1/a)}.\)

Step 3: Simplifying the Denominator

Recognizing \(1/a\) as \(a^{-1}\), we apply the power rule for logarithms, which states that \(\log_a (a^n) = n\log_a a\). This gives:\(\log_a (1/a) = \log_a (a^{-1}) = -1\cdot\log_a a = -1.\)

Step 4: Completing the Proof

Substitute \(\log_a (1/a) = -1\) back into the equation from the second step. This gives:\(\log_{1/a} x = \frac{\log_a x}{-1} = -\log_a x.\)Therefore, we've shown that \(\log_{1/a} x = -\log_a x.\)

Key Concepts

Understanding Logarithms

Logarithms form the foundational concept needed to tackle problems like the one above. Essentially, logarithms answer the question: "To what exponent must we raise a specific base to obtain a given number?" For example, in the expression \(\log_b x\), this denotes the power to which the base \(b\) must be raised to yield \(x\). When the base \(b\) equals 10, we refer to it as the common logarithm, and when \(b\) is \(e\) (approximately 2.718), it's the natural logarithm.

• This operation is the inverse of exponentiation.
• It allows us to express exponential relationships using simpler terms.
• Logarithms can help solve equations involving exponential terms, making them invaluable in many fields like mathematics, physics, and engineering.

Understanding these core principles of logarithms allows you to manipulate and transform equations that appear complex at first glance.

Reciprocal Base and Change of Base Formula

The concept of a reciprocal base is vital when understanding how to evaluate logarithms like \(\log_{1/a} x\). A reciprocal base means inverting the original base \(a\), giving us \(1/a\). The change of base formula is a powerful tool to rewrite logarithms with a new base. This rule is represented as:\[\log_{c} x = \frac{\log_b x}{\log_b c}\]where \(b\) is the new base you are transforming to.
By transforming the base, we can simplify calculations or convert them into a more familiar base, thereby easing the computational process.
In our exercise, applying the change of base formula lets us calculate \(\log_{1/a} x\) and demonstrates that it's equivalent to \(-\log_{a} x\) after simplification, especially when you employ properties like the power rule for logarithms.

Power Rule for Logarithms

The power rule is a key property of logarithms allowing transformations of expressions into simpler forms. It states: \(\log_b (a^n) = n\log_b a\). Essentially, this indicates that an exponent within the logarithmic expression can be taken out and multiplied. This concept becomes critically important when dealing with complex exponents or when using logarithms to solve exponential equations. Applying this rule in the exercise:

• Translates the denominator \(\log_a (1/a) = \log_a (a^{-1})\).
• This leads to \(\log_a (a^{-1}) = -1 \cdot \log_a a = -1\).

Using the power rule simplifies the expression, allowing us to easily demonstrate that \(\log_{1/a} x = -\log_a x\). This is a perfect example of how foundational principles like the power rule can unlock solutions to seemingly complicated logarithmic problems.