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Simplify the expression by using the definition and properties of logarithms. $$\log _{7} \sqrt{7}$$

Short Answer

Expert verified
The expression simplifies to \( \frac{1}{2} \).

Step by step solution

01

Express the Square Root as an Exponent

Recognize that \( \sqrt{7} \) can be written as \( 7^{1/2} \). This conversion helps in utilizing the power rule of logarithms. Thus, the expression becomes \( \log_{7}(7^{1/2}) \).
02

Apply the Power Rule of Logarithms

The power rule states that \( \log_{b}(a^c) = c \cdot \log_{b} a \). Applying this rule, we have \( \log_{7}(7^{1/2}) = \frac{1}{2} \cdot \log_{7}(7) \).
03

Use the Logarithmic Identity

Recall the logarithmic identity \( \log_{b}(b) = 1 \). Since \( \log_{7}(7) = 1 \), the expression simplifies to \( \frac{1}{2} \cdot 1 \).
04

Simplify the Final Expression

Finally, simplify the expression to obtain \( \frac{1}{2} \). Thus, the simplified form of the original expression is \( \frac{1}{2} \).

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Key Concepts

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

Properties of Logarithms
Logarithms have several key properties that simplify complex expressions involving large numbers. Understanding these properties allows you to work with logarithms more efficiently. One of the main properties is the product rule, which is useful in many multiplication scenarios. The product rule states: \( \log_{b}(xy) = \log_{b}(x) + \log_{b}(y) \). This means you can split the log of a product into the sum of logs.

Another essential property is the quotient rule: \( \log_{b}\left(\frac{x}{y}\right) = \log_{b}(x) - \log_{b}(y) \). This lets you divide and represent as the difference of logs. Additionally, there's the power rule which we'll delve into next. It's particularly important when dealing with exponents, as in our current problem.
Power Rule of Logarithms
The power rule of logarithms is pivotal when simplifying expressions where exponents appear within the logarithm. It states that \( \log_{b}(a^c) = c \cdot \log_{b}(a) \).

This property allows us to bring the exponent "c" in front of the logarithm, transforming the power inside the log into a multiplication outside it. For example, in our problem, the expression \( \log_{7}(7^{1/2}) \) was simplified by applying the power rule to become \( \frac{1}{2} \cdot \log_{7}(7) \), thus reducing the complexity of evaluating the logarithm.

To use this rule effectively, always express your numbers in the form of a base to an exponent. This not only helps in simplification but also in connecting the logarithm with other mathematical operations.
Logarithmic Identity
The logarithmic identity \( \log_{b}(b) = 1 \) is a fundamental principle that makes simplifying expressions very straightforward. This identity tells us that the log of a base to itself is always equal to one. For any base \( b \), if you take \( \log_{b}(b) \), the result is 1.

In the context of our example, using \( \log_{7}(7) = 1 \) transforms the expression \( \frac{1}{2} \cdot \log_{7}(7) \) efficiently into \( \frac{1}{2} \cdot 1 \). This simplification step is crucial because it lets us find the final, most reduced form of the expression without further calculations.

Recognizing and applying this identity saves time and ensures accuracy, especially when handling expressions with similar bases.
Simplifying Expressions
Simplifying expressions is the process of reducing them to their simplest form, which often helps make calculations easier or more intelligible. This not only involves applying the properties of logarithms but also requires recognizing patterns and opportunities to use algebraic rules effectively.

In our example, simplifying the expression \( \log _{7} \sqrt{7} \) involved several steps: rewriting the square root as \( 7^{1/2} \), applying the power rule to bring down the exponent, and using the logarithmic identity to complete the simplification. Each step reduces complexity and increases clarity.

Remember, the goal of simplifying expressions is to achieve a form where unnecessary complexities are removed, allowing for quick and precise evaluations. In the end, we arrived at the simplified form \( \frac{1}{2} \), which speaks to the power of these logarithmic techniques.

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Most popular questions from this chapter

Simplify each expression. (a) \(\ln e^{4}\) (b) \(\ln (1 / e)\) (c) \(\ln \sqrt{e}\)

(a) Specify the domain of the function \(y=\ln x+\ln (x-4).\) (b) Solve the inequality \(\ln x+\ln (x-4) \leq \ln 21.\)

Exercises \(55-60\) introduce a model for population growth that takes into account limitations on food and the environment. This is the logistic growth model, named and studied by the nineteenth century Belgian mathematician and sociologist Pierre Verhulst. (The word "logistic" has Latin and Greek origins meaning "calculation" and "skilled in calculation," respectively. However, that is not why Verhulst named the curve as he did. See Exercise 56 for more about this.) In the logistic model that we "I study, the initial population growth resembles exponential growth. But then, at some point owing perhaps to food or space limitations, the growth slows down and eventually levels off, and the population approaches an equilibrium level. The basic equation that we'll use for logis- tic growth is where \(\mathcal{N}\) is the population at time \(t, P\) is the equilibrium population (or the upper limit for population), and a and b are positive constants. $$\mathcal{N}=\frac{P}{1+a e^{-b t}}$$ The following figure shows the graph of the logistic function \(\mathcal{N}(t)=4 /\left(1+8 e^{-t}\right) .\) Note that in this equation the equilibrium population \(P\) is 4 and that this corresponds to the asymptote \(\mathcal{N}=4\) in the graph. (a) Use the graph and your calculator to complete the following table. For the values that you read from the graph, estimate to the nearest \(0.25 .\) For the calculator values, round to three decimal places. (b) As indicated in the graph, the line \(\mathcal{N}=4\) appears to be an asymptote for the curve. Confirm this empirically by computing \(\mathcal{N}(10), \mathcal{N}(15),\) and \(\mathcal{N}(20) .\) Round each answer to eight decimal places. (c) Use the graph to estimate, to the nearest integer, the value of \(t\) for which \(\mathcal{N}(t)=3\) (d) Find the exact value of \(t\) for which \(\mathcal{N}(t)=3 .\) Evaluate the answer using a calculator, and check that it is consistent with the result in part (c). TABLE AND GRAPH CANT COPY

Solve for \(x\) (assuming that \(a>b>0\) ): $$\begin{aligned} \left(a^{4}-2 a^{2} b^{2}+b^{4}\right)^{x-1}=(a-b)^{2 x}(a+b)^{-2} & \\ \text {Answer:} & x=\frac{\ln (a-b)}{\ln (a+b)} \end{aligned}$$

A function \(f\) with domain \((1, \infty)\) is defined by the equation \(f(x)=\log _{x} 2\) (a) Find a value for \(x\) such that \(f(x)=2\) (b) Is the number that you found in part (a) a fixed point of the function \(f ?\)

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