Problem 30 - Calculus and its applications

Question

Given \(\log _{b} 3=1.099\) and \(\log _{b} 5=1.609\), find each value. $$ \log _{b}(5 b) $$

Step-by-Step Solution

Answer

The value is 2.609.

Step 1: Use Logarithmic Properties

Recall the property of logarithms that states the logarithm of a product is the sum of the logarithms. This can be written as: \[ \log_b (mn) = \log_b m + \log_b n \] Apply this property to \( \log_b (5b) \), so we have: \[ \log_b (5b) = \log_b 5 + \log_b b \]

Step 2: Evaluate Logarithm of the Base

Remember that the logarithm of a base to itself is 1. In other words, \[ \log_b b = 1 \] Substitute this into the expression from Step 1: \[ \log_b (5b) = \log_b 5 + 1 \]

Step 3: Plug in Given Values

From the problem statement, we have \( \log_b 5 = 1.609 \). Substitute this value into our expression: \[ \log_b (5b) = 1.609 + 1 \]

Step 4: Calculate the Final Expression

Perform the addition to find the final answer: \[ \log_b (5b) = 2.609 \]

Key Concepts

Product Property of Logarithms

Logarithms have a neat way of turning multiplication inside an argument into addition, making them extremely useful. This is called the "product property of logarithms." When you encounter a logarithmic expression that involves the product of two numbers, you can split it into the sum of two separate logarithms.

The key rule is as follows:

• \( \log_b (mn) = \log_b m + \log_b n \)

This tells you that no matter the base of the logarithm, if you multiply two numbers inside the logarithm function, you can separate them by adding their individual logarithms.

For example, using our exercise

• \( \log_b (5b) \) can be split into \( \log_b 5 + \log_b b \)

This transformation converts a more complicated problem into a simpler and often more solvable form.

Evaluating Logarithms

To evaluate a logarithm, you typically need to either know basic values or use properties and given values. In our exercise, this involved simplifying the expression using known values.

Once we apply the product property of logarithms, the task comes down to simply evaluating known values:

• We know \( \log_b 5 = 1.609 \)
• For \( \log_b b \), we recognize it is a special value of 1

Using these values, evaluating the logarithm becomes a straightforward calculation of adding them together.

Thus, \( \log_b (5b) = 1.609 + 1 \), simplifying our efforts down to basic arithmetic, like adding 1.609 and 1 to arrive at the answer, which is 2.609.

Base Properties in Logarithms

A unique property of logarithms is that

• \( \log_b b = 1 \)

This means when the argument of a logarithm is the same as the base, it results in an output of 1. It's like asking how many times do you need to multiply the base by itself to get the base – the answer is once.

This property becomes very handy when you are simplifying and evaluating expressions. It simplifies logarithmic expressions involving the base itself and was crucial during our exercise. In other words, any time you see something like \( \log_b b \) in your problem, you can confidently replace it with 1, simplifying your work immediately.

Remembering this property can help in effortlessly solving portions of exercises involving logarithms, as shown when we computed \( \log_b (5b) \) by recognizing \( \log_b b = 1 \). This allows problems to be re-framed in terms of known numeric calculations.