Problem 78 - Precalculus

Question

Show that \(\log _{b} 1=0\) and \(\log _{b} b=1\) for every \(b>0, b \neq 1\).

Step-by-Step Solution

Answer

For any base \(b > 0\) and \(b \neq 1\), \(\log_b 1 = 0\) and \(\log_b b = 1\).

Step 1: Understanding the Logarithm of 1

We need to show that the logarithm of 1 with any base greater than 0 and not equal to 1 is 0. Recall the definition: \[\log_b a = c \Rightarrow b^c = a\]For \(\log_b 1\), we have:\[b^c = 1\]The only exponent \(c\) that satisfies this equation for any base \(b\), provided \(b eq 1\), is \(c = 0\). Hence, \(\log_b 1 = 0\).

Step 2: Confirming the Logarithm of the Base

We next need to prove that \(\log_b b = 1\). With the same logarithmic definition, \[\log_b b = c \Rightarrow b^c = b\]This equation is satisfied only when \(c = 1\), since any number raised to the power of 1 is the number itself. Therefore, \(\log_b b = 1\) for any base \(b\) greater than zero and not equal to 1.

Key Concepts

Logarithmic Properties

Logarithms possess a set of properties that make them extremely useful in both mathematics and real-world applications. These properties facilitate the simplification and solving of various mathematical problems.

• The power property of logarithms states that \(\log_b (a^c) = c \cdot \log_b a\). This means you can "bring down" the exponent in front of the log, simplifying complex expressions.
• Another key property is the product property: \(\log_b (mn) = \log_b m + \log_b n\). This allows the decomposition of a log expression into a sum of individual logs.
• Finally, the quotient property notes that \(\log_b (m/n) = \log_b m - \log_b n\), helping break down logs of fractions.

Each property holds simple, identifiable patterns in mathematical problems and aids in understanding how logarithms function.

Exponentiation

Exponentiation is the mathematical operation where a base is raised to a power or exponent. It is denoted as \(b^c\), where \(b\) is the base and \(c\) is the exponent.
This operation is key to understanding logarithms because they are the inverse of exponentiation. For example, when you have \(b^c = a\), taking the logarithm base \(b\) of both sides gives \(\log_b a = c\).
The rules of exponentiation include:

• Multiplying exponents when raising a power to a power: \((b^a)^c = b^{a \cdot c}\)
• Adding exponents when multiplying: \(b^a \times b^c = b^{a+c}\)
• Subtracting exponents when dividing: \(b^a / b^c = b^{a-c}\)
• Understanding zero exponents: \(b^0 = 1\), critical for grasping that \(\log_b 1 = 0\) as explained in the initial exercise and solution.

Together, these insights into exponentiation make it easier to comprehend the fundamental nature of logarithms.

Base-Exponent Relationship

The base-exponent relationship is a fundamental concept that ties together the ideas of exponentiation and logarithms. When you see an expression like \(b^c = a\), you're witnessing this relationship in action.
Here, the base \(b\) is the number being repeatedly multiplied by itself \(c\) times, resulting in \(a\). Understanding this relationship is crucial to reversing the process through logarithms, where \(\log_b a = c\).

This relationship shows why logarithms and exponentiation are inverse operations. The base remains consistent, while the logarithmic and exponent values swap roles. For \(\log_b b = 1\), it shows that \(b\) to the power of 1 always returns the base itself. Similarly for \(\log_b 1 = 0\), it shows any base raised to the power of 0 yields 1.

• The balance of the base-exponent relationship ensures calculations involving logarithms and exponentials flow seamlessly back and forth.
• Recognizing this interplay helps solve mathematical problems more intuitively.

Such harmony is why these concepts are widely applied not just mathematically, but also in scientific formulas and calculations.