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Use properties of logarithms to expand logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. \(\log \sqrt{100 x}\)

Short Answer

Expert verified
The expanded form of \(\log \sqrt{100x}\) is \(1 + \frac{1}{2} \log x\)

Step by step solution

01

Convert Square Root to Fractional Power

The square root of a number is equivalent to raising the number to the power of \(\frac{1}{2}\). So, the expression becomes: \(\log (100x)^{\frac{1}{2}}\)
02

Using Power Property of Logarithms

The power property of logarithms states that \(\log_b m^n = n \cdot \log_b m\). Applying this property to the expression, it becomes: \(\frac{1}{2} \log (100x)\)
03

Using Product Property of Logarithms

The product property of logarithms states that \(\log_b mn = \log_b m + \log_b n\). Applying this property, the expression becomes: \(\frac{1}{2}( \log 100 + \log x)\)
04

Simplify Logarithm of 100

The value of \(\log 100\) is 2 (since 10 to the power of 2 is 100). Replacing \(\log 100\) with 2, the expression simplifies to: \(\frac{1}{2} (2 + \log x)\)
05

Simplify Expression

Finally, performing the multiplication gives: \(1 + \frac{1}{2} \log x\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Properties of Logarithms
When working with logarithmic expressions, understanding the properties of logarithms is key. Logarithms have several properties that simplify expressions and allow for easier computation. Here are some important ones:
  • Product Property: Allows you to split a product inside the logarithm into separate terms, \( \log_b(mn) = \log_b m + \log_b n \).
  • Power Property: Allows you to bring down the power in front of the logarithm, \( \log_b(m^n) = n \cdot \log_b m \).
  • Quotient Property: Allows you to break down a quotient, \( \log_b\left(\frac{m}{n}\right) = \log_b m - \log_b n \).
These properties are essential tools for expanding, simplifying, or even combining logarithmic expressions, much like in arithmetic operations with ordinary numbers. Understanding and mastering these properties will help you solve logarithmic exercises efficiently.
Power Property
In the world of logarithms, the power property is incredibly useful when dealing with expressions raised to a power within a logarithm. According to this property, if you have a logarithm of a power, \( \log_b(m^n) \), it simplifies to \( n \cdot \log_b m \).
This essentially means you can "bring down" the exponent as a factor in front of the logarithm.This is particularly helpful when the expression involves a root. For example, the square root of \( 100x \), which is \((100x)^{\frac{1}{2}}\)\, can be rewritten using the power property as \( \frac{1}{2} \log(100x) \).
Remembering this will allow you to simplify such expressions with ease.
Product Property
The product property of logarithms is a straightforward yet powerful tool for breaking down logarithms of products. This property states that \( \log_b(mn) = \log_b m + \log_b n \).
This means the logarithm of a product is the sum of the logarithms of the individual factors.Why is this useful? Well, it breaks a complex multiplication into simpler additions of logs. Consider our expression \( \frac{1}{2} \log(100x) \). By using the product property, it becomes \( \frac{1}{2}(\log 100 + \log x) \).
This is often easier to calculate or further simplify, especially when some of the logs have straightforward solutions, such as \( \log 100 = 2 \).
Simplifying Logarithms
Simplifying logarithms involves using properties to make expressions easier to work with or more straightforward in appearance. Let's consider the expression \( \frac{1}{2}(2 + \log x) \) that resulted from applying the product and power properties in our exercise. To simplify further, distribute the \( \frac{1}{2} \) across the terms inside the parentheses to get \( 1 + \frac{1}{2}\log x \).
Here, every step aimed to break down the expression into something recognizable and approachable.Simplifying helps highlight the main parts of the logarithm, making it easier to interpret or solve problems involving these expressions.
Thus, simplifying is not just an exercise in manipulation but a step toward understanding and solving larger math problems efficiently.

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Most popular questions from this chapter

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