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Chapter 1: Problem 10

Simplify. \(\log \frac{1}{\sqrt{10}}\)

Short Answer

Expert verified
The simplified form is \(-\frac{1}{2}\).

Step by step solution

01

Apply Logarithm Division Property

We start with the expression \( \log \frac{1}{\sqrt{10}} \). By using the property \( \log \frac{a}{b} = \log a - \log b \), we can rewrite this as \( \log 1 - \log \sqrt{10} \).
02

Evaluate Logarithm of One

We know that \( \log 1 = 0 \). So, we replace \( \log 1 \) with 0 in the expression: \( 0 - \log \sqrt{10} \).
03

Simplify Using Square Root

Recall the property that \( \sqrt{10} = 10^{1/2} \). Thus, we can write \( \log \sqrt{10} \) as \( \log 10^{1/2} \).
04

Apply Logarithm Power Rule

Use the logarithm power rule: \( \log b^{a} = a \cdot \log b \). Therefore, \( \log 10^{1/2} \) simplifies to \( \frac{1}{2} \cdot \log 10 \).
05

Substitute and Simplify the Expression

Since \( \log 10 = 1 \), we substitute this into our expression from Step 4: \( \frac{1}{2} \cdot 1 = \frac{1}{2} \). Therefore, the expression \( 0 - \frac{1}{2} \) simplifies to \(-\frac{1}{2} \).

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Key Concepts

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

Logarithm Division Property
When working with logarithms, one very useful property is the division property. This rule helps when you need to deal with a logarithm of a fraction. If you see an expression like \( \log \frac{a}{b} \), you can change it to \( \log a - \log b \). This simply means that the logarithm of a fraction is the same as the subtraction of the logarithms of the numerator and the denominator.

In the example \( \log \frac{1}{\sqrt{10}} \), applying the division property means you separate it into two parts: \( \log 1 \) and \( \log \sqrt{10} \).
  • Calculate each part separately.
  • Combine them by subtracting.
Using this property is like splitting the problem into smaller, easier chunks.
Logarithm Power Rule
The power rule of logarithms is another handy tool, especially when the logs involve exponents. The rule states that \( \log b^{a} = a \cdot \log b \). This tells us that the exponent comes out as a multiplier in front of the log.

For example, if you have \( \log 10^{1/2} \), you can apply the power rule to bring down the exponent:
  • Identify the exponent (here it's \( \frac{1}{2} \)).
  • Move it in front of the log as a multiplier.
This changes it into \( \frac{1}{2} \cdot \log 10 \). By simplifying like this, complex expressions become much more manageable.
Simplifying Logarithmic Expressions
Simplifying logarithmic expressions is all about making them as straightforward as possible. Once we have used properties like division and power, our aim is to reduce the expression to its simplest form.

Consider our example: we transformed \( \log \frac{1}{\sqrt{10}} \) initially by dealing with \( \log 1 - \log \sqrt{10} \), where \( \log 1 \) is 0. Then, by using the square root information and simplifying \( \log 10^{1/2} \) with the power rule, we ended up with \( -\frac{1}{2} \).
  • Apply properties and rules to break down complex parts.
  • Transform every piece into its simplest form.
The goal is to get to an answer like \( -\frac{1}{2} \) that is easy to understand and doesn't require further simplification.

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Most popular questions from this chapter

Inflation, as measured by Japan's consumer price index, \(^{65}\) decreased (thus the word deflation) by \(0.7 \%\) in the year \(2001 .\) If this rate were to continue for the next 10 years, use your computer or graphing calculator to determine how long before the value of a typical item would be reduced to \(95 \%\) of its value in 2001 .

Measurement Conversion Let \(x\) be the length of an object in furlongs, let \(r\) be the length in rods, and let \(y\) be the length in yards. Given that there are 40 rods to a furlong, find the function \(g\) such that \(r=g(x)\). Given that there are 5.5 yards to a rod, find the function \(f\) such that \(y=f(r)\). Now determine \(y\) as a function of \(x,\) and relate this to the composition of two functions. Explain your formula in words.

Growth Rates of Lake Trout Ruzycki and coworkers \(^{74}\) studied the growth habits of lake trout in Bear Lake, UtahIdaho. The mathematical model they created was given by the equation \(L(t)=960\left(1-e^{-0.12[t+0.45]}\right),\) where \(t\) is age in years and \(L\) is length in millimeters. Graph this equation. What seems to be happening to the length as the trout become older? Ruzycki and coworkers also created a mathematical model that connected length with weight and was given by the equation \(W(L)=1.30 \times 10^{-6} \cdot L^{3.31},\) where \(L\) is length in millimeters and \(W\) is weight in grams. Find the length of a 10 -year old lake trout. Find the weight of a 10-year old lake trout. Find \(W\) as a function of \(t\).

The population of Latvia was decreasing at the rate of \(0.52 \%\) a year \(^{68}\) from 1995 to 2000 . Assuming that the population continued to decline at this annual rate, use your computer or graphing calculator to determine how long until the population of Latvia will become \(95 \%\) of what is was in the year 2000 .

An account earns an annual rate of \(r\), expressed as a decimal, and is compounded quarterly. The account initially has \(\$ 1000\) and five years later has \(\$ 1500\). What is \(r ?\)
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