Question

Rewrite the equation in exponential form. (See Example 1.) \(\log _{1 / 2} 16=-4\)

Short answer

The exponential form of the equation \(\log _{1 / 2} 16=-4\) is \((1 / 2)^{-4} = 16\).

Step-by-step solution

Analyze the Equation

The given equation is \(\log _{1 / 2} 16=-4\), where the logarithm base is \(1 / 2\), the number is \(16\), and the logarithm value, which is also the exponent in the exponential form, is \(-4\).

Convert to Exponential Form

Using the relationship between logarithmic and exponential forms, the given equation can be rewritten as \((1 / 2)^{-4} = 16\)

Check the Result

To confirm that this is correct, calculate \((1 / 2)^{-4}\), which equals \(16\), confirming that the original logarithmic equation and rewritten exponential equation are equivalent.

Key concepts

Logarithm Basics

Understanding logarithms is essential for interpreting the relationship between exponentiation and its reverse process. A logarithm answers the question: To what power must we raise a given base to obtain a specific number? In the format \( \log_b a = c \) it translates to: what power \( c \) do we raise the base \( b \) to get \( a \) as the result? For instance, in the exercise \( \log _{1 / 2} 16=-4 \) the base is \( \frac{1}{2} \) and we are asked to find the power that when \( \frac{1}{2} \) is raised to it, we obtain 16. This relationship is a cornerstone of logarithmic to exponential form conversion.

When working with logarithms, remember these basic properties: the logarithm of one (\( \log_b 1 \) is always 0 regardless of the base because any number raised to the power of 0 is 1. Similarly, the logarithm of the base itself (\( \log_b b \) is always 1 because any number raised to the power of 1 is the number itself. These properties help in simplifying logarithmic expressions and in understanding their behaviors in equations.

Exponents and Powers

Exponents and powers are a fundamental concept in algebra and are particularly important when dealing with logarithmic expressions. An exponent represents the number of times a base is multiplied by itself. For instance, \( 2^3 \) means that 2 is multiplied by itself 3 times (2*2*2), equaling 8. Now, when we have negative exponents, such as \( (1/2)^{-4} \) from our exercise, it signifies an inverse operation; that is, the base is now on the denominator raised to the positive power: \( (1/2)^{-4} = (1^4)/(2^4) = 1/16 \).

Understanding how to manipulate exponents is key when converting logarithmic expressions to exponential form. The 'power' of a logarithm, indicated in the exercise as -4, is actually the exponent to which the base \( 1/2 \) must be raised to obtain the number 16. Hence, the expression \( \log _{1 / 2} 16 \) neatly transforms into the exponential equation \( (1/2)^{-4} = 16 \) when we apply what we know about exponents.

Algebraic Expressions

Algebraic expressions are the backbone of algebra and encompass numbers, variables, and arithmetic operations. They can take various forms, from simple terms like \( 5x \) to more complex equations. When dealing with logarithmic equations, we handle algebraic expressions that involve logarithms. For instance, in the step-by-step solution of our exercise, the exponential form \( (1 / 2)^{-4} = 16 \) is itself an algebraic expression.

It's important to be comfortable in rewriting and manipulating these expressions as different forms might be required for simplification or solving equations. In our example, the logarithmic expression was modified to an equivalent exponential form to facilitate the understanding and solving of the equation. Mastery of algebraic expressions allows for flexibility in mathematical operations and is fundamental for success in various realms of mathematics and applied sciences.