Problem 15 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 15-16: Find the number, if possible. (a) \(\log _{5} 1\) (b) \(\log _{3} 3\) (c) \(\log _{4}(-2)\) (d) \(\log _{7} 7^{2}\) (e) \(3^{\log _{3} 8}\) (f) \(\log _{5} 125\) (g) \(\log _{4} \frac{1}{16}\)

Step-by-Step Solution

Answer

(a) 0, (b) 1, (c) undefined, (d) 2, (e) 8, (f) 3, (g) -2

Step 1: Solve part (a)

To solve \( \log_{5} 1 \), we need to find the exponent \( x \) such that \( 5^x = 1 \). Since any number raised to the power 0 is 1, \( x = 0 \). Therefore, \( \log_{5} 1 = 0 \).

Step 2: Solve part (b)

To solve \( \log_{3} 3 \), we need to find the exponent \( x \) such that \( 3^x = 3 \). Since \( 3^1 = 3 \), we have \( x = 1 \). Therefore, \( \log_{3} 3 = 1 \).

Step 3: Solve part (c)

The expression \( \log_{4}(-2) \) cannot be solved because the logarithm of a negative number is undefined in the real number system. Hence, \( \log_{4}(-2) \) is not possible.

Step 4: Solve part (d)

To solve \( \log_{7} 7^2 \), we can use the logarithmic property \( \log_{b} b^a = a \). Here, \( a = 2 \), so \( \log_{7} 7^2 = 2 \).

Step 5: Solve part (e)

To solve \( 3^{\log_{3} 8} \), we use the inverse property that \( b^{\log_{b} x} = x \). Here, \( b = 3 \) and \( x = 8 \). Therefore, \( 3^{\log_{3} 8} = 8 \).

Step 6: Solve part (f)

To solve \( \log_{5} 125 \), we express 125 as a power of 5. Since \( 125 = 5^3 \), we have \( \log_{5} 125 = 3 \).

Step 7: Solve part (g)

To solve \( \log_{4} \frac{1}{16} \), we express \( \frac{1}{16} \) as a power of 4. Since \( \frac{1}{16} = 4^{-2} \), we have \( \log_{4} \frac{1}{16} = -2 \).

Key Concepts

Understanding the Properties of Logarithms

Logarithms transform multiplicative relationships into additive ones, simplifying complex calculations. Several key properties help in this process. The power rule of logarithms states that for any logarithm of the form \( \log_b(x^a) \), it equals \( a \cdot \log_b(x) \). This was used in solving \( \log_7(7^2) \), where \( \log_7(7^2) = 2 \) since it simplifies the process by pulling the exponent down as a multiplier.
Another important property is the product rule: \( \log_b(xy) = \log_b(x) + \log_b(y) \), and similarly, the quotient rule is as follows: \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \).
The change of base formula also comes handy, \( \log_b(x) = \frac{\log_k(x)}{\log_k(b)} \), allowing different bases to be more easily compared or converted.

Grasping the Concept of Exponentiation

Exponentiation is a fundamental operation in mathematics, involving raising a number (the base) to the power of an exponent. It is the reverse of taking a logarithm and involves repeated multiplication of a base.

• For example, in the case of \( 125 = 5^3 \), the number 5 is multiplied by itself 3 times.
• This is crucial in reverse-engineering log problems such as \( \log_5(125) \), which finds the number of times the base (5) must be multiplied by itself to get 125, resulting in 3.

Exponentiation also features in expressions like \( 3^{\log_3(8)} \). Here, the logarithm finds the exponent needed to achieve the base raised to a certain power, ultimately simplifying down to any input \( x \) of the form \( b^{\log_b(x)} \).

Navigating the Real Number System

The real number system includes all the rational and irrational numbers, but it does not extend to imaginary or complex numbers in typical logarithmic functions. Thus, logarithms of negative numbers, like \( \log_4(-2) \), are undefined in the real number context as there is no real number that 4 can be raised to resulting in a negative value.
Because of this, in problems involving logarithms, it's imperative to understand that the base and the argument must reside within the real number system's bounds, requiring both to be positive, where the base must also not be equal to 1.

Exploring Inverse Properties of Logarithms

Logarithms and exponentiation act as inverse operations. Understanding this allows for simplification of equations and easier computation.

• The primary inverse property states \( b^{\log_b(x)} = x \), demonstrating how exponentiation and logarithms cancel each other out, returning the initial input when the base matches.
• Similarly, if \( \log_b(b^x) = x \), the operations undo each other, reflexively demonstrating their inverse relationship.

This inverse relationship is exemplified in the problem \( 3^{\log_3(8)} = 8 \). Here, the exponentiation nullifies the logarithmic operation, leaving the base number, 8, untouched by the exponent.