Problem 84 - Precalculus with Limits

Question

Comparing Logarithmic Quantities In Exercises 83 and \(84,\) compare the logarithmic quantities. If two are equal, then explain why. $$ \log _{7} \sqrt{70}, \quad \log _{7} 35, \quad \frac{1}{2}+\log _{7} \sqrt{10} $$

Step-by-Step Solution

Answer

The logarithmic quantities \( \log _{7} \sqrt{70} \) and \( \frac{1}{2}+\log _{7} \sqrt{10} \) are equal.

Step 1: Simplify First Logarithm Log_7 sqrt(70)

Let's begin by simplifying the first quantity. Using the property of logarithms, the square root can be rewritten as a power of 1/2, yielding \( \log _{7} 70^{1/2} \). We can further simplify this as \( \frac{1}{2} \log _{7} 70 \).

Step 2: Simplify Second Logarithm Log_7 35

The second quantity is already in its simplest form, which is \( \log _{7} 35 \). There's no need to modify it further.

Step 3: Simplify Third Logarithm 1/2 + Log_7 sqrt(10)

The third logarithmic quantity can be simplified in a similar manner as the first. Here we have \( \frac{1}{2}+\log _{7} \sqrt{10} \), which simplifies to \( \frac{1}{2}+\log _{7} 10^{1/2} = \frac{1}{2}+\frac{1}{2} \log _{7} 10 \).

Step 4: Compare the Logarithmic Quantities

Now we have three simplified logarithmic expressions, \( \frac{1}{2} \log _{7} 70, \log _{7} 35, \frac{1}{2}+\frac{1}{2} \log _{7} 10 \). We can see that the first and the third expressions are equal because log base 7 of 70 is the same as log base 7 of 10 times 7, resulting in the same logarithmic value. Hence, \( \frac{1}{2} \log _{7} 70 = \frac{1}{2}+\frac{1}{2} \log _{7} 10 \).

Key Concepts

Logarithm Properties

Logarithms are powerful mathematical tools that allow us to simplify and solve equations involving exponential growth and decay. A key property that makes logarithms extremely useful is the ability to convert multiplication into addition. This is expressed as \( \log_b(xy) = \log_b(x) + \log_b(y) \). It means that the logarithm of a product is equal to the sum of the logarithms.

Another important property is the power rule: \( \log_b(x^n) = n \cdot \log_b(x) \). This property allows you to take exponents outside of the logarithm, which can simplify expressions and make calculations easier. For example, a square root can be viewed as a power of \( \frac{1}{2} \). Hence, \( \log_b(\sqrt{x}) = \frac{1}{2} \cdot \log_b(x) \).

• Product Property: \( \log_b(xy) = \log_b(x) + \log_b(y) \)
• Power Rule: \( \log_b(x^n) = n \cdot \log_b(x) \)
• Change of Base Formula: \( \log_b(x) = \frac{\log_c(x)}{\log_c(b)} \)

Logarithmic Expressions

Logarithmic expressions can initially seem complex but understanding their structure can make them much easier to handle. A logarithmic expression typically includes a base, an argument (the value inside the log), and the logarithm itself. The base tells you what kind of logarithm it is (e.g., base 10 or base 7).

In the expression \( \log_7 35 \), 35 is the argument, and 7 is the base. Simplifying these expressions involves using properties of logarithms. For instance, in the original exercise when we see \( \log_7 \sqrt{70} \), we recognize that \( \sqrt{70} \) can be rewritten using the power property: it turns into \( \log_7(70^{1/2}) \), which can be further simplified.

Often, you might also encounter terms like \( \frac{1}{2} \), reflecting another part outside the log function that needs to be dealt with, usually involving addition or multiplication within the simplified form. With practice, interpreting and converting logarithmic expressions becomes more intuitive, allowing for more straightforward solutions to complex problems.

Simplifying Logarithms

Simplifying logarithms is a crucial skill in mathematics that makes equations more manageable. The process involves using properties like the power and product properties to reduce complex expressions to simpler forms.

For example, let's take the expression \( \log_7 \sqrt{70} \). By employing the power rule, we rewrite the square root as \( 70^{1/2} \) and then simplify it as \( \frac{1}{2} \log_7 70 \).

• Convert roots to powers: \( \sqrt{a} = a^{1/2} \)
• Apply power property: \( \log_b(a^{n}) = n \cdot \log_b(a) \)
• Combine like terms where possible to reflect simplified expression

These simplifications reveal relationships between different logarithmic quantities, such as in the exercise where two expressions ended up being equal, demonstrating the beauty of symmetry in mathematics.