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Write the given logarithm in terms of logarithms of \(x, y,\) and \(z\). $$\log _{2} \frac{\sqrt{x}}{y^{3}}$$

Short Answer

Expert verified
\[ 0.5\log_{2}x - 3\log_{2}y \]

Step by step solution

01

Apply the Quotient Property

The logarithm of a quotient is the difference of the logarithms of the numerator and denominator. Therefore, apply this property to the given logarithm. This leads to: \[ \log_{2}\sqrt{x} - \log_{2}y^{3} \]
02

Apply the Root Property

The logarithm of a root is the quotient of the logarithm of the number and the root index. It can therefore be applied to the first part of the equation. This transforms the equation to: \[ 0.5\log_{2}x - \log_{2}y^{3} \]
03

Distribute the Logarithm to the Power

In the next step, distribute the logarithm to the power in the second part of the equation. For this, use the Power Property, which states that the logarithm of a number raised to a power equals the power times the logarithm of the number itself. This gives: \[ 0.5\log_{2}x - 3\log_{2}y \]

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Key Concepts

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

Logarithm Properties
Understanding logarithm properties is essential for manipulating and simplifying logarithmic expressions. A key property to remember is the *Quotient Property*. This property states that the logarithm of a quotient of two numbers is the difference between the logarithm of the numerator and the logarithm of the denominator. For instance, in the expression \( \log_2 \frac{\sqrt{x}}{y^3} \), the Quotient Property lets us separate it into two parts: \( \log_2 \sqrt{x} - \log_2 y^3 \).
Additionally, remember the *Product Property* and the *Power Property*, which are useful for different situations. The Product Property says the logarithm of a product is the sum of logarithms, while the Power Property, which we will discuss in more detail, allows conversion of powers in logarithms.
Change of Base Formula
The Change of Base Formula is quite useful, especially when working with logarithms that aren't in base 10 or base \(e\). This formula lets us convert a logarithm to a different base. It's stated as: \( \log_b a = \frac{\log_k a}{\log_k b} \), where \( k \) can be any positive number.
This formula proves handy when using calculators, which commonly have buttons for \( \log_{10} \) or \( \ln \) (natural logarithm, base \( e \)). For instance, the log base 2 of a number can be converted into base 10 or \( e \) for easier computation.
In our problem, converting wasn't necessary because base 2 directly worked with the expressions we needed to simplify. But knowing how to change bases keeps a lot of options open when solving real-world problems.
Power Property
The Power Property is a cornerstone in simplifying expressions involving logarithms. It tells us that given a logarithm with a number raised to a power, we can "pull" that power out front. Mathematically, it means \( \log_b (a^c) = c \cdot \log_b a \).
In the solution, the term \( \log_2 y^3 \) uses this property. By transforming it into \( 3 \cdot \log_2 y \), calculations become more manageable.
This trick also applies to numbers under a root, as roots can be seen as raising numbers to a fractional power. Thus, understanding the Power Property is not just about reducing powers but includes handling roots efficiently, as we'll see in the next section.
Root Property
Roots in logarithms can be handled gracefully with the Root Property, which is closely related to the Power Property. When dealing with roots, remember that the root can be expressed as a fractional exponent.
For example, the square root of \( x \) is written as \( x^{1/2} \). Thus, \( \log_b \sqrt{x} \) becomes \( \frac{1}{2} \cdot \log_b x \) by applying the Power Property. This manipulation makes it easier to work with roots within logarithmic expressions.
In our exercise, \( \sqrt{x} \) in \( \log_2 \sqrt{x} \) is converted to \( 0.5 \cdot \log_2 x \). This simplifies the expression significantly and is critical for ensuring all parts are in their simplest form.

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Most popular questions from this chapter

The following argument seems to indicate that \(0.125>0.25\) Find the first incorrect statement in the argument. $$\begin{aligned} 3 &>2 \\ 3(\log 0.5) &>2(\log 0.5) \\ \log 0.5^{3} &>\log 0.5^{2} \\ 0.5^{3} &>0.5^{2} \\ 0.125 &>0.25 \end{aligned}$$

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