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Use the properties of logarithms to expand the expression as a sum, difference, and/or constant multiple of logarithms. (Assume all variables are positive.) $$\log _{10} \frac{y}{2}$$

Short Answer

Expert verified
\(\log_{10} y - C \), where \(C = \log_{10} 2\)

Step by step solution

01

Apply the Quotient Rule

The quotient rule of logarithms states that the logarithm of a quotient is the difference of the logarithms of the numerator and denominator. This rule applies here, because we have a fraction as the argument of the logarithm. Applying the rule gives us: \( \log_{10} y - \log_{10} 2 \).
02

Express the Constant in Logarithmic Form

In step 1, we separated the logarithm into the difference of the logarithm of y and the logarithm of 2. While \(\log_{10} y\) is already fully simplified, we can further simplify \(\log_{10} 2\), because the logarithm base 10 of 2 is a constant value. This can be denoted as a single term, C, where \(C = \log_{10} 2 \).

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Key Concepts

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

Logarithm Rules
Logarithms, often referred to as 'logs', are the inverse operations of exponentiation, which means they help us find the exponent that a certain base needs to be raised to obtain a given number. Understanding the rules for logarithms is essential in precalculus mathematics and helps simplify complex expressions. Three fundamental rules are the product rule, quotient rule, and the power rule.

The product rule states that the logarithm of a product is the sum of the logarithms of the factors: \(\log_b(mn) = \log_b(m) + \log_b(n)\). Similarly, the power rule tells us that the logarithm of a number raised to an exponent is equal to the exponent times the logarithm of that number: \(\log_b(m^p) = p \cdot \log_b(m)\). These rules, along with the quotient rule, allow for the simplification of complex logarithmic expressions and are thus integral tools in solving logarithmic equations.
Quotient Rule in Logarithms
The quotient rule is a cornerstone in working with logarithms and is vital for logarithmic simplification. It states that the log of a quotient is equal to the difference of the logs of the numerator and the denominator: \(\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)\).

This rule is particularly useful when dealing with expressions that involve division within a logarithm. By changing the division into subtraction, the expression becomes easier to handle. For example, in our original exercise, applying the quotient rule allows us to expand \(\log_{10} \frac{y}{2}\) into \(\log_{10} y - \log_{10} 2\). By understanding this rule, students can effectively break down complex logarithmic problems into simpler components.
Logarithmic Simplification
Logarithmic simplification involves reducing logarithmic expressions into a more easily manageable form, often as a preliminary step for solving logarithmic equations. Simplifying an expression can involve applying the quotient rule, turning constants into logarithms, or condensing multiple logarithmic terms into one.

As shown in our exercise, by applying the quotient rule, the given expression \(\log_{10} \frac{y}{2}\) becomes \(\log_{10} y - \log_{10} 2\). Moreover, since \(\log_{10} 2\) is a constant, it can be represented as such and kept aside while we perform further algebraic manipulations on the variable part, \(\log_{10} y\). This simplification makes it easier to understand the properties of the logarithmic function involved and enhances the student's ability to manipulate and solve logarithmic expressions.
Precalculus Mathematics
Precalculus mathematics is a course that prepares students for calculus studies, covering a wide range of topics including functions, complex numbers, and importantly, logarithms. A solid grasp of logarithms, their properties, and simplification techniques is crucial in precalculus. It builds the foundation for understanding continuous growth, rates of change, and many other concepts that are explored in more depth in calculus.

Understanding the properties of logarithms and being able to simplify logarithmic expressions is not only a part of precalculus curricula but also imperative for solving real-world problems. For instance, logarithmic scales are used in measuring the intensity of earthquakes (Richter scale) and sound (decibels), highlighting the importance of mastering logarithms for a well-rounded mathematical education and practical application.

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Most popular questions from this chapter

$$\$ 2500$$ is invested in an account at interest rate \(r\), compounded continuously. Find the time required for the amount to (a) double and (b) triple. $$r=0.045$$

(a) solve for \(P\) and (b) solve for \(t\). $$A=P\left(1+\frac{r}{n}\right)^{n t}$$

Use the following information for determining sound intensity. The level of sound \(\boldsymbol{\beta}\), in decibels, with an intensity of \(I\), is given by \(\boldsymbol{\beta}=10 \log \left(I / I_{0}\right),\) where \(I_{0}\) is an intensity of \(10^{-12}\) watt per square meter, corresponding roughly to the faintest sound that can be heard by the human ear. In Exercises 65 and 66 , find the level of sound \(\boldsymbol{\beta}\). (a) \(I=10^{-10}\) watt per \(\mathrm{m}^{2}\) (quiet room) (b) \(I=10^{-5}\) watt per \(\mathrm{m}^{2}\) (busy street corner) (c) \(I=10^{-8}\) watt per \(\mathrm{m}^{2}\) (quiet radio) (d) \(I=10^{0}\) watt per \(\mathrm{m}^{2}\) (threshold of pain)

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 \(g\) 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. A model for the data is given by \(y=-3.00+11.88 \ln x+(36.94 / x),\) where \(y\) is the number of g's. $$ \begin{array}{|c|c|} \hline x & \text { g's } \\ \hline 0.2 & 158 \\ 0.4 & 80 \\ 0.6 & 53 \\ 0.8 & 40 \\ 1.0 & 32 \\ \hline \end{array} $$ (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 distance traveled during impact if the passenger deceleration must not exceed \(30 \mathrm{~g}\) '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.

A cell site is a site where electronic communications equipment is placed in a cellular network for the use of mobile phones. The numbers \(y\) of cell sites from 1985 through 2008 can be modeled by \(y=\frac{237,101}{1+1950 e^{-0.355 t}}\) where \(t\) represents the year, with \(t=5\) corresponding to \(1985 .\) (Source: CTIA-The Wireless Association) (a) Use the model to find the numbers of cell sites in the years 1985,2000 , and 2006 . (b) Use a graphing utility to graph the function. (c) Use the graph to determine the year in which the number of cell sites will reach 235,000 . (d) Confirm your answer to part (c) algebraically.

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