Problem 2 - Algebra and Trigonometry

Question

Evaluate the expression. $$ \log _{2} 160-\log _{2} 5 $$

Step-by-Step Solution

Answer

The answer is 5.

Step 1: Recall the Logarithm Property

First, let's recall the logarithmic property which states that for any real numbers \(a\), \(b\), and base \(c > 0\), then \(\log_c a - \log_c b = \log_c \left( \frac{a}{b} \right)\). This property will be useful to simplify the given expression.

Step 2: Apply the Logarithm Property

Apply the logarithmic property to the given expression \(\log_{2} 160 - \log_{2} 5\). Using the property: \(\log_{2} 160 - \log_{2} 5 = \log_{2} \left( \frac{160}{5} \right)\).

Step 3: Simplify the Inside of the Logarithm

Now, compute the division inside the logarithm: \(\frac{160}{5}\). Performing the calculation, \(\frac{160}{5} = 32\). So we have \(\log_{2} 32\).

Step 4: Evaluate the Logarithm

We need to find the power to which 2 must be raised to yield 32. Since \(2^5 = 32\), \(\log_{2} 32 = 5\).

Key Concepts

Logarithmic Subtraction

Logarithmic subtraction is a powerful tool that simplifies expressions by implementing the property of logarithms. When you see a subtraction between two logarithms with the same base, you can combine them into a single logarithm. This is achieved by dividing the inside terms of the logarithms. For instance, given two logarithms, \( \log_c a \) and \( \log_c b \), the property goes like this:

• \( \log_c a - \log_c b = \log_c \left( \frac{a}{b} \right) \)

This transformation is immensely useful because it can often result in a simpler expression that is much easier to evaluate further. In our example with \( \log_{2} 160 - \log_{2} 5 \), we used this property to rewrite the expression into \( \log_{2} \left( \frac{160}{5} \right) \).
This application of logarithmic subtraction lays the groundwork for assessing more complex logarithmic problems with ease.

Simplification of Logarithms

Once you've applied logarithmic subtraction, the next essential step is simplification. This involves ensuring that the expression is as straightforward as possible. Simplification usually makes further calculations and evaluations easier to perform.
When simplifying \( \log_{2} \left( \frac{160}{5} \right) \), calculate the division inside the logarithm:

• \( \frac{160}{5} = 32 \)

Now, the new expression is \( \log_{2} 32 \). Simplification not only makes the expression easier to understand but also helps you recognize identifiable patterns or straightforward results. It’s always advisable to break down the expressions into their simplest elements before diving into more complex evaluations.

Logarithmic Expression Evaluation

The final step in dealing with logarithmic expressions, after subtraction and simplification, is evaluation. Depending on your base and the simplified form of the expression, you find the result of the logarithm.
From the step where we had \( \log_{2} 32 \), evaluate it by finding the power to which the base must be raised to reach the simplified result. Recall the property of exponents:

• For \( \log_{2} 32 \), think: What power must 2 be raised to yield 32?
• Since we know \( 2^5 = 32 \), it follows that \( \log_{2} 32 = 5 \).

This evaluation provides a definitive numerical answer and completes the process of solving the original logarithmic expression.