Problem 18 - Calculus with Concepts in Calculus

Question

Calculate the logarithm by using \((11)\). $$ \log _{1 / 2} \frac{1}{3} $$

Step-by-Step Solution

Answer

The logarithm is approximately 1.585.

Step 1: Understand the Problem

We need to calculate the logarithm \( \log_{1/2} \frac{1}{3} \). This means we want to find the power \( x \) such that \( (1/2)^x = 1/3 \).

Step 2: Use the Change of Base Formula

Using the change of base formula, we can express the logarithm as \( \log_{1/2} \frac{1}{3} = \frac{\log_{10} (1/3)}{\log_{10} (1/2)} \). This formula helps us compute the logarithm using a base that we can easily evaluate with a calculator, like base 10.

Step 3: Calculate \( \log_{10} \frac{1}{3} \)

Using a calculator, find \( \log_{10} \frac{1}{3} \). It evaluates to approximately \( -0.4771 \).

Step 4: Calculate \( \log_{10} \frac{1}{2} \)

Next, find \( \log_{10} \frac{1}{2} \) using a calculator. It evaluates to approximately \( -0.3010 \).

Step 5: Divide the Logarithms

Substitute the values back into the formula from Step 2: \( \log_{1/2} \frac{1}{3} = \frac{-0.4771}{-0.3010} \).

Step 6: Simplify the Division

Perform the division: \( \frac{-0.4771}{-0.3010} \approx 1.585 \).

Step 7: Verify the Result

Verify by calculating \( (1/2)^{1.585} \approx \frac{1}{3} \). This confirms the calculation is correct.

Key Concepts

Change of Base Formula

In the context of logarithms, the change of base formula is an important tool used to simplify calculations, especially when the base of the logarithm isn't equal to 10 or isn't a natural number. This formula allows us to transform logarithms from one base to another using bases that are more convenient for computation.

The formula is expressed as:

• \(\log_b a = \frac{\log_k a}{\log_k b}\)

Where \(b\) and \(a\) are the original base and the argument of the logarithm, respectively, and \(k\) is the new base we decide to use. Often, \(k\) is chosen to be 10 or \(e\) (approximately 2.718), because these are standard bases available on most scientific calculators.

This approach is particularly useful in cases such as \( \log_{1/2} \frac{1}{3} \). By changing the base to 10, we can utilize our calculator to compute:

• \( \log_{1/2} \frac{1}{3} = \frac{\log_{10} (1/3)}{\log_{10} (1/2)} \)

This simple yet powerful formula significantly simplifies the process of solving logarithms, making it more accessible to students.

Powers and Exponents

Understanding powers and exponents is crucial when working with logarithms. Logarithms themselves are the inverse operations of exponents. If you remember basic exponent rules, such as \((b^x = a)\), logarithms take the form \(\log_b a = x\).

In our example of \(\log_{1/2} \frac{1}{3}\), it is necessary to find the exponent \(x\) such that multiplying \((1/2)\) by itself \(x\) times results in \(\frac{1}{3}\). This is again a conceptual flip from when we use powers to build numbers:

• Exponential form example: if \(2^3 = 8\), then \(\log_2 8 = 3\)

This reciprocal relationship between logarithms and exponents helps us understand the base and the power to which that base must be raised. It's crucial in interpreting and solving logarithmic equations.

Calculator Use in Mathematics

Scientific calculators play an immense role in computational mathematics, especially with tasks involving logarithms, powers, and exponents. These calculators are equipped to handle standard bases like 10 (common logarithms) and \(e\) (natural logarithms), allowing us to use the change of base formula effectively.

When calculating \(\log_{1/2} \frac{1}{3}\), for example, a calculator can quickly provide values such as:

• Finding \(\log_{10} \frac{1}{3}\), yielding approximately \(-0.4771\)
• Finding \(\log_{10} \frac{1}{2}\), yielding approximately \(-0.3010\)

These outputs from a calculator facilitate completion of the division step efficiently. Afterwards, ensuring the results make sense can involve using the calculator again to raise the base to the found exponent, confirming approximate equality.

Thus, calculators not only streamline calculations, but also aid in reinforcing understanding and verifying the correctness of solutions in mathematical exercises.