/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 Use the properties of logarithms... [FREE SOLUTION] | 91Ó°ÊÓ

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Use the properties of logarithms to expand the expression as a sum, difference, and/or multiple of logarithms. (Assume all variables are positive.)\(\ln \sqrt{z}\)

Short Answer

Expert verified
\(\ln \sqrt{z} = \frac{1}{2} * \ln{z}\)

Step by step solution

01

Identify the properties of logarithms to use

For the current problem, the property of logarithms to be used is: \(\ln \sqrt{x} = \frac{1}{2}* \ln{x}\). This property is derived from a more general rule: \(\ln{a^b} = b * \ln{a}\).
02

Apply the property to the expression

Apply this rule to the term \(\sqrt{z}\) and write \( \sqrt{z} \) as \( z^{1/2}\). Substituting it in the expression, we get: \( \ln(\sqrt{z}) = \ln(z^{1/2}) \)
03

Expand the logarithm

Using the property \(\ln{a^b} = b * \ln{a}\), the expression becomes: \( \frac{1}{2}* \ln{z}\).

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Key Concepts

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

Logarithmic Expansion
Logarithmic expansion involves using the properties of logarithms to rewrite complex logarithmic expressions in a simpler form. This is often done to make calculations easier or to better understand the structure of the expression.

In the given exercise, we're asked to expand the expression \(\ln \sqrt{z}\). First, we recognize that a square root can be written as an exponent: \(\sqrt{z} = z^{1/2}\). Using this property helps transform the original problem into something we can further work with.
  • Expression: \(\ln \sqrt{z}\)
  • Re-written as: \(\ln(z^{1/2})\)
Understanding this rewrite is essential in manipulating the expression to reveal its expanded form, as the square root becomes a power, leading us to utilize further logarithmic properties.
Power Rule of Logarithms
The power rule of logarithms is a key tool for expanding logarithmic expressions. It is based on the principle that an exponent inside a logarithm can be moved in front as a multiplier.

Mathematically, this is expressed as \(\ln{a^b} = b \cdot \ln{a}\). This rule is powerful because it allows us to take a complex expression and simplify it by separating the power from the logarithm itself.
For our exercise, applying the power rule to \(\ln(z^{1/2})\), we shift the \(1/2\) in front of the logarithm, transforming the expression into:
  • \(\ln(z^{1/2}) = \frac{1}{2} \cdot \ln z\)
By doing this, the logarithm is expanded into a simpler and more manageable form. The multiplier \(1/2\) provides a clear indication of how the original term relates to \(\ln z\). This makes further manipulation and interpretation straightforward.
Variable Restrictions in Logarithms
When working with logarithms, it's crucial to understand the restrictions on variables. Logarithms are undefined for non-positive numbers, meaning the variables must be positive to ensure the expression is valid.

In our exercise, the assumption is that all variables are positive, which means we can safely expand and simplify without encountering undefined behavior.
Here's why it matters:
  • Logarithm of zero or a negative number is undefined.
  • Ensuring \(z > 0\) allows \(\ln z\) and by extension \(\ln(z^{1/2})\) to be valid expressions.
This restriction is not merely a formality; it ensures that our mathematical operations hold true across all steps. Misunderstanding this can lead to errors in both reasoning and calculation.

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Most popular questions from this chapter

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(2-6 \ln x=10\)

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(\frac{500}{100-e^{x / 2}}=20\)

Automobiles are designed with crumple zones that help protect their occupants in crashes. The crumple zones allow the occupants to move short distances when the automobiles come to abrupt stops. The greater the distance moved, the fewer g's the crash victims experience. (One \(\mathrm{g}\) is equal to the acceleration due to gravity. For very short periods of time, humans have withstood as much as 40 g's.) In crash tests with vehicles moving at 90 kilometers per hour, analysts measured the numbers of g's experienced during deceleration by crash dummies that were permitted to move \(x\) meters during impact. The data are shown in the table. $$ \begin{array}{|l|l|l|l|l|l|} \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\ \hline g \text { 's } & 158 & 80 & 53 & 40 & 32 \\ \hline \end{array} $$A model for these data is given by \(y=-3.00+11.88 \ln x+\frac{36.94}{x}\) where \(y\) is the number of g's. (a) Complete the table using the model.$$ \begin{array}{|l|l|l|l|l|l|} \hline x & 0.2 & 0.4 & 0.6 & 0.8 & 1.0 \\ \hline y & & & & & \\ \hline \end{array} $$(b) Use a graphing utility to graph the data points and the model in the same viewing window. How do they compare? (c) Use the model to estimate the least distance traveled during impact for which the passenger does not experience more than \(30 \mathrm{~g}\) 's. (d) Do you think it is practical to lower the number of g's experienced during impact to fewer than 23 ? Explain your reasoning.

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(\frac{3000}{2+e^{2 x}}=2\)

Solve the logarithmic equation algebraically. Approximate the result to three decimal places.\(\log _{10} x+\log _{10}(x+1)=\log _{10}(x+3)\)

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