Problem 23 - Precalculus

Question

Evaluate the expression. . \(\log _{\frac{1}{6}}(216)\)

Step-by-Step Solution

Answer

The value of \( \log_{\frac{1}{6}}(216) \) is \(-3\).

Step 1: Understand the Expression

The expression given is \( \log_{\frac{1}{6}} (216) \). This means we need to find what exponent we need to raise \(\frac{1}{6}\) to in order to get 216.

Step 2: Apply the Change of Base Formula

Use the change of base formula: \( \log_b(a) = \frac{\log(a)}{\log(b)} \). Here, \( a = 216 \) and \( b = \frac{1}{6} \).

Step 3: Use Property of Logarithms

Rewrite \( \log_{\frac{1}{6}}(216) \) in terms of base 10 or base \(e\) logarithms: \( \log_{\frac{1}{6}}(216) = \frac{\log(216)}{\log(\frac{1}{6})} \).

Step 4: Simplify the Denominator

Note that \( \log(\frac{1}{6}) = \log(1) - \log(6) = 0 - \log(6) = -\log(6) \). So now, the expression becomes \( \frac{\log(216)}{-\log(6)} \).

Step 5: Simplifying Further Using Base Conversion

Observe that \( 216 = 6^3 \), thus \( \log(216) = \log(6^3) = 3 \cdot \log(6) \).

Step 6: Substitute and Simplify

Substitute \( \log(216) = 3 \cdot \log(6) \) into the expression: \( \frac{3 \cdot \log(6)}{-\log(6)} \). Simplify to get \(-3\).

Step 7: Conclude

Therefore, \( \log_{\frac{1}{6}}(216) = -3 \).

Key Concepts

Change of Base Formula

The change of base formula is a useful tool in logarithms. It allows us to evaluate logarithms with bases that aren't base 10 or base \(e\), which are typically the most convenient to use with calculators.
\[\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\]
This formula states that a logarithm with base \(b\) can be calculated using any other base \(c\), such as 10 or \(e\).
To apply this in our previous exercise:

• We calculated \(\log_{1/6}(216)\) by converting it into base 10 logarithms as \(\frac{\log(216)}{\log(1/6)}\).

This conversion is helpful particularly when the base isn't a common one, making calculations more manageable.

Properties of Logarithms

Logarithms possess several key properties that simplify complex expressions. These properties are essential for solving equations like the ones in our exercise.
Here are some important properties:

• \(\log(ab) = \log(a) + \log(b)\)
• \(\log\left(\frac{a}{b}\right) = \log(a) - \log(b)\)
• \(\log(a^c) = c \cdot \log(a)\)

In the step-by-step solution of \(\log_{1/6}(216)\), we applied:
\[\log\left(\frac{1}{6}\right) = \log(1) - \log(6)\]
to simplify the denominator and
\[\log(216) = \log(6^3) = 3 \cdot \log(6)\]
to figure out the numerator.
This usage of logarithm properties makes computations easier.

Exponentiation

Exponentiation and logarithms are closely linked. An exponent indicates how many times a number, the base, is multiplied by itself.
This concept is core when solving logarithms since:

• \(\log_b(a) = x\) means that \(b^x = a\)

In our example, we looked for the exponent that raises \(\frac{1}{6}\) to get 216. That means finding \(x\) in:
\[\left(\frac{1}{6}\right)^x = 216\]
Understanding this relationship illuminates why the end result was \(-3\), clarifying that \(\frac{1}{6}\) raised to the power of \(-3\) results in 216.

Base Conversion

Base conversion refers to changing the base of numbers often seen in logarithmic contexts.
This process facilitates easier calculation by transforming a complex base into one that is more computationally friendly, such as base 10 or 2.
When handling large numbers or unusual bases, base conversion reduces errors and simplifies calculations.
In the solution, we shifted the log base from \(\frac{1}{6}\) to base 10. This step was crucial to exploit familiar and convenient computation methods, like property application and calculator usage.
Thus, base conversion is often a preparatory step for simplifying logarithmic expressions.