Problem 62 - College Algebra

Question

Is the following true: \(\frac{\log _{3}(27)}{\log _{4}\left(\frac{1}{64}\right)}=-1 ?\) Verify the result.

Step-by-Step Solution

Answer

Yes, the statement is true.

Step 1: Evaluate the numerator

To evaluate the numerator \( \log_{3}(27) \), we recognize that 27 is a power of 3 since \( 27 = 3^3 \). Thus, \( \log_{3}(27) = 3 \).

Step 2: Evaluate the denominator

Now, evaluate the denominator \( \log_{4}\left(\frac{1}{64}\right) \). We can rewrite \( \frac{1}{64} \) as \( 4^{-3} \) because \( 64 = 4^3 \), so \( \frac{1}{64} = 4^{-3} \). Therefore, \( \log_{4}\left(\frac{1}{64}\right) = \log_{4}(4^{-3}) = -3 \).

Step 3: Perform the division

Now divide the evaluated numerator by the evaluated denominator: \[ \frac{\log_{3}(27)}{\log_{4}\left(\frac{1}{64}\right)} = \frac{3}{-3} = -1 \].

Step 4: Conclusion

Since the division yields \(-1\), the given statement \( \frac{\log _{3}(27)}{\log _{4}\left(\frac{1}{64}\right)}=-1 \) is true.

Key Concepts

Numerator Evaluation

The numerator in this exercise is the expression \( \log_{3}(27) \). To solve it, we need to find the logarithm of 27 with the base 3. Let's remember that a logarithm answers the question: "To what power must the base be raised to produce that number?"
In this situation, we ask: to what power does 3 need to be raised to get 27? Knowing some basic powers can help, and in this case, we realize that 27 is actually \( 3^3 \). This simple observation allows us to say \( \log_{3}(27) = 3 \). This is because if you take 3 to the power of 3, you indeed get 27.
If you know your cube numbers, this step is often quick to compute! So, calculating the numerator was a matter of identifying that 27 is a power of 3.

Denominator Evaluation

Next, we tackle the denominator which is \( \log_{4}\left(\frac{1}{64}\right) \). Evaluating this requires understanding how negative exponents and the properties of logarithms work together. We begin by expressing \( \frac{1}{64} \) in terms of a power of 4.
First, recognize that 64 is a power of 4, because \( 64 = 4^3 \). This is helpful because fractions are often expressions of negative exponents. Therefore, \( \frac{1}{64} \) can be written as \( 4^{-3} \).
Using the property of logarithms, which states that \( \log_{b}(b^x) = x \), we see that \( \log_{4}(4^{-3}) = -3 \). So, by converting the fraction into a negative exponent, we make evaluating the denominator straightforward.

• Step 1: Express \( \frac{1}{64} \) as \( 4^{-3} \).
• Step 2: Use the logarithmic identity to find that \( \log_{4}(4^{-3}) = -3 \).

Logarithmic Properties

Logarithmic properties are like a toolkit that simplifies otherwise tricky problems. For this problem, we used a couple of key properties. First, the property that helped with the numerator: \( \log_{b}(b^x) = x \). This property states that if you have a logarithm where the base and the exponential base match, the result is simply the exponent.
Another property used was that related to fractions in exponents. For the denominator, expressing a number as a fractional power (negative exponent) opened the door to an easy evaluation. By recognizing that \( a^{-b} \) translates to \( 1/a^b \), it becomes easier to find log expressions where negative parts can correspond directly with their positive counterparts.
Bringing it all together, once both parts are evaluated using these properties, dividing the results became straightforward: \( \frac{3}{-3} = -1 \). This confirmed that the original problem statement was correct—that the complex expressions and their log evaluations indeed simplified to \(-1 \). Understanding and applying these properties effectively can turn intimidating-looking expressions into manageable calculations.

• Matching the base and the exponent leads directly to the power.
• Negative exponents transform naturally into fractional logarithmic expressions.