Problem 15 - Algebra and Trigonometry

Question

Express the equation in logarithmic form. $$ \begin{array}{ll}{\text { (a) } 8^{-1}=\frac{1}{8}} & {\text { (b) } 2^{-3}=\frac{1}{8}}\end{array} $$

Step-by-Step Solution

Answer

(a) \(\log_8 \left(\frac{1}{8}\right) = -1\); (b) \(\log_2 \left(\frac{1}{8}\right) = -3\).

Step 1: Understand the problem statement

Determine what it means to express an equation in logarithmic form. It involves rewriting the exponential equation in the form of a logarithm. This helps identify a relationship where the base of the logarithm raises to a power to produce a number.

Step 2: Convert equation (a) into logarithmic form

Given the equation (a): \(8^{-1} = \frac{1}{8}\). In the exponential form, 8 raised to the power of -1 equals \(\frac{1}{8}\). In logarithmic form, it is written as \(\log_8 \left(\frac{1}{8}\right) = -1\).

Step 3: Convert equation (b) into logarithmic form

Given the equation (b): \(2^{-3} = \frac{1}{8}\). In exponential form, 2 raised to the power of -3 equals \(\frac{1}{8}\). In logarithmic form, it is expressed as \(\log_2 \left(\frac{1}{8}\right) = -3\).

Key Concepts

Exponential Equations

An exponential equation is an equation in which a variable appears in the exponent. Such equations can often describe growth and decay processes. For example, the equations provided in the exercise \(8^{-1} = \frac{1}{8}\) and \(2^{-3} = \frac{1}{8}\) are exponential equations.

Exponential equations involve expressions where a base is raised to a power. Here are some key points:

• The base is the number that is being raised to the power, such as 8 or 2 in our examples.
• The exponent (or power) indicates how many times the base is used in a multiplication, like -1 or -3 here.
• The result is the value you get after applying the exponent to the base.

When working with such equations, converting them to logarithmic form can simplify solving them by allowing comparison of exponents directly.

Logarithmic Form

Converting an equation to logarithmic form involves shifting from the format of exponential equations. This conversion hinges on the definition of logarithms: if \(b^x = y\), then \(\log_b(y) = x\).

In the exercise, to convert \(8^{-1} = \frac{1}{8}\) into logarithmic form, we recognize or rearrange the components as: the base \(8\), the exponent \(-1\), and the resultant value \(\frac{1}{8}\). In simpler terms, we ask: "To what power must 8 be raised to get \(\frac{1}{8}\)?" This is encapsulated in \(\log_8 \left(\frac{1}{8}\right) = -1\).

Similarly, take equation \(2^{-3} = \frac{1}{8}\) which would translate to the logarithmic form \(\log_2 \left(\frac{1}{8}\right) = -3\).

Understanding this transformation from exponential to logarithmic is crucial because it simplifies solving for unknown exponents and helps us analyze problems involving large scale growth or decay.

Base of a Logarithm

The base of a logarithm is fundamental because it determines the scale and outcome of the logarithmic relationship. Simply put, the base is the number that is repeatedly multiplied in the exponential equation.

For instance, in \(\log_8 \left(\frac{1}{8}\right) = -1\), 8 is the base indicating that 8 needs to be raised to the power of -1 to get \(\frac{1}{8}\). Understanding the base is crucial because:

• It conveys which base (number) we're predicting the exponential result from.
• It differs from exponentiation where it is the resultant of a base raised to a power.
• Using the correct base in equations like \(\log_2 \left(\frac{1}{8}\right) = -3\) helps us understand how the scaling factor influences the calculation.

Choosing the proper base, in essence, guides us in interpreting and managing exponential growth or decay models across various fields of study, including science, finance, and technology.