Problem 49 - Algebra 2 and Trigonometry

Question

In \(48-55,\) if \(\log a=c,\) express each of the following in terms of \(c\) $$ \log 10 a $$

Step-by-Step Solution

Answer

\( \log (10a) = 1 + c \)

Step 1: Identify Given Information

We are given that \( \log a = c \). Our task is to express \( \log (10a) \) in terms of \( c \).

Step 2: Use Logarithmic Product Rule

Recall that the logarithmic product rule states: \( \log(xy) = \log x + \log y \). Apply it here: \( \log (10a) = \log 10 + \log a \).

Step 3: Simplify the Expression

We know \( \log a = c \). We also know \( \log 10 = 1 \) because 10 is the base of the common logarithm. Substitute these values to simplify: \( \log (10a) = 1 + c \).

Key Concepts

Understanding the Logarithmic Product Rule

Logarithms are mathematical tools used to deal with exponential relationships. One important property of logarithms is the logarithmic product rule. This rule helps simplify expressions involving the product of two numbers. The product rule states:

• \( \log(xy) = \log x + \log y \)

This essentially means that the logarithm of a product is equal to the sum of the logarithms of its factors. It's handy when you want to break down complex logarithmic expressions into simpler parts. For instance, in the expression \( \log(10a) \), we can apply the rule and split it into \( \log 10 + \log a \). Each part can often be simplified further, especially if one of the terms is a common logarithm or an already known value, like \( \log a = c \) in our problem.

Skills in Simplifying Logarithmic Expressions

Simplifying logarithmic expressions might look complicated at first, but it's just a matter of applying rules and properties correctly. With the expression \( \log(10a) \), here's a breakdown of how simplification works: - Begin by applying the logarithmic product rule, as discussed, to separate the terms: \( \log(10a) = \log 10 + \log a \).- Next, use known values to replace parts of the expression. Since \( \log 10 = 1 \), substitute this value in: \( \log(10a) = 1 + \log a \).- Remember that we were given \( \log a = c \), so substitute \( c \) into the equation: \( \log(10a) = 1 + c \).By following these steps, you simplify the expression from \( \log(10a) \) to a much more manageable form, \( 1 + c \). Each step ensures the expression is reduced to its simplest terms by using known values and properties effectively.

Exploring Common Logarithms

Common logarithms are logarithms with a base of 10. They're used frequently in real-world applications because they're relatively easy to work with. This is because the common logarithm of ten is exactly 1. For instance, in the expression \( \log(10) \), since its base is already 10, the value equals 1. This is a crucial property that makes expressions involving multiples of ten simpler to manage.

• This property stems from the fact that \( 10^1 = 10 \), which fits perfectly into the base-exponent relationship that defines logarithms.

When simplifying logarithmic expressions, it's beneficial to identify parts where common logarithms occur. They can often help condense complex expressions quickly, as seen when simplifying \( \log(10a) \) to \( 1 + c \). Recognizing common logarithms is a powerful tool in efficiently solving math problems involving logs.