Problem 32 - Algebra 2

Question

Write each equation in logarithmic form. \(\left(\frac{1}{3}\right)^{-2}=9\)

Step-by-Step Solution

Answer

\(\log_{\frac{1}{3}}(9) = -2\)

Step 1: Identify the components for conversion

The given equation is \( \left(\frac{1}{3}\right)^{-2} = 9 \). This is in exponential form, where the base is \( \frac{1}{3} \), the exponent is \(-2\), and the result is \(9\).

Step 2: Recall the logarithmic form

The logarithmic form of an equation is typically written as \( \log_{b}(y) = x \), where \(b\) is the base, \(y\) is the result, and \(x\) is the exponent.

Step 3: Write the equation in logarithmic form

Using the identified components from the exponential form, convert the equation to logarithmic form. This means we write: \( \log_{\frac{1}{3}}(9) = -2 \).

Key Concepts

Exponential Form

Exponential form is a way of expressing numbers and equations where a base number is raised to the power of an exponent. This is typically written as \( b^x = y \), where \( b \) is the base, \( x \) is the exponent, and \( y \) is the result. In the context of our exercise, we have the equation \( \left(\frac{1}{3}\right)^{-2} = 9 \). Here, the base is \( \frac{1}{3} \), the exponent is \(-2\), and the result is \( 9 \).
Exponents indicate how many times the base is multiplied by itself. In this specific example, the negative exponent \(-2\) tells us to take the reciprocal of the base and then square it. This means \( \left(\frac{1}{3}\right)^{-2} = \left(\frac{3}{1}\right)^2 = 9 \).
Exponential expressions are powerful tools that simplify large or complex calculations.

Conversion Between Forms

Converting between exponential and logarithmic forms is a key skill in mathematics. It involves rewriting an equation from one form to another. This process helps in solving equations and in understanding the relationship between exponents and logarithms.
In our example, we start with the exponential form \( \left(\frac{1}{3}\right)^{-2} = 9 \). To convert this into logarithmic form, we use the general formula \( b^x = y \) to \( \log_b(y) = x \). It's like translating from one language to another by following specific rules.
Here, the conversion results in the logarithmic equation \( \log_{\frac{1}{3}}(9) = -2 \). This states that the exponent \(-2\) is what you need to raise the base \( \frac{1}{3} \) to get the result \( 9 \).
Conversions between forms provide flexibility in mathematical problem-solving, allowing you to choose the most effective approach for each scenario.

Logarithms

Logarithms are the opposite operation of exponentiation, often used to simplify complex calculations. A logarithm answers the question: "To what exponent must the base be raised, to get a certain number?" This makes it a powerful tool when dealing with exponential equations.
The general form of a logarithmic equation is \( \log_b(y) = x \), where \( b \) is the base, \( y \) is the result, and \( x \) is the exponent. In the exercise example, we rewrote \( \left(\frac{1}{3}\right)^{-2} = 9 \) as \( \log_{\frac{1}{3}}(9) = -2 \).
Logarithms simplify multiplication and division to addition and subtraction. This property is incredibly useful in fields like engineering, computer science, and finance, where exponential growth and decay occur frequently. Understanding logarithms opens a world of efficient calculation methods, making them essential in higher-level math.