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Chapter 4: Problem 114

a. Evaluate \(\log _{7} 7^{6}\) b. Evaluate \(6 \cdot \log _{7} 7\) c. How do the values of the expressions in parts (a) and (b) compare?

Short Answer

Expert verified
\(\b \log _{7} 7^{6} = 6\) and \(\b_6 \cdot \log _{7} 7 = 6\). The values are equal.

Step by step solution

01

Evaluate \(\log _{7} 7^{6}\)

Use the logarithm property \(\log_b (b^x) = x\). Here, \(\b = 7\) and \(\b^x = 7^{6}\). Therefore, \(\/text{ \(\log _{7} 7^{6} = 6\)}\)
02

Evaluate \(\b_6 \cdot \log _{7} 7\)

Use the fact that \(\b \log_b (b) = 1\). Here, \(\b = 7\), so \(\b \log _{7} 7 = 1\). Therefore, \(\/6 \cdot \log _{7} 7 = 6 \cdot 1 = 6 \)
03

Compare the Values

Both expressions \(\b \log _{7} 7^{6}\) and \(\b_6 \cdot \log _{7} 7\) evaluate to 6. Therefore, the values are equal.

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Key Concepts

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

Logarithm Properties
Logarithm properties simplify complex expressions involving logarithms. These properties are essential in algebra, making it easier to solve equations and evaluate expressions. Key properties include the **power rule** and the **basic logarithm rule**. For instance, when we have \(\b \log_b (b^x) = x\), it reveals that any logarithm of a base raised to a power equals that power. Let's see these properties in action with the exercise.

The first part of the exercise, \(\b \log_7 7^6\), utilizes the power rule. Because of the property \(\b \log_b (b^x) = x\), \(\b \log_7 7^6\) equals \(\b 6\). Understanding and applying logarithm properties make these evaluations straightforward and intuitive.
Exponents
Exponents represent the number of times a base is multiplied by itself. For example, in \(\b 7^6\), 7 is the base, and 6 is the exponent, meaning 7 is multiplied by itself six times. To intertwine this with logarithms, consider the first part of the exercise again.

In \(\b \log_7 7^6\), we see the base 7 raised to the exponent 6. By the power rule of logarithms \(\b \log_b (b^x) = x\), this simplifies to 6, translating the exponent directly into the output of the logarithmic expression. This direct relationship between logarithms and exponents is crucial for simplifying and solving logarithmic equations. Another application is seen in \(\b 6 \cdot \log_7 7\). By recognizing \(\b \log_7 (7) \) evaluates to 1 thanks to the property \(\b \log_b(b) = 1\), multiplying by 6 leads straightforwardly to 6.
Comparative Evaluation
Comparative evaluation of expressions checks their equality or inequality. It is a useful skill for verifying calculations and understanding relationships between mathematical expressions.

In the exercise, we compared \(\b \log_7 7^6\) and \(\b 6 \cdot \log_7 7\). Both yield the result 6 though they are approached differently. The first uses the power rule to simplify directly to the exponent. The second multiplies a simple logarithmic solution by 6. This comparison showcases that different methods can yield the same result, reinforcing the robustness of logarithmic properties in handling various forms of expressions.

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Most popular questions from this chapter

Use the model \(A=P e^{r t} .\) The variable \(A\) represents the future value of \(P\) dollars invested at an interest rate \(r\) compounded continuously for \(t\) years. If \(\$ 10,000\) is invested in an account earning \(5.5 \%\) interest compounded continuously, determine how long it will take the money to triple. Round to the nearest year.

The isotope of plutonium \({ }^{238} \mathrm{Pu}\) is used to make thermoelectric power sources for spacecraft. Suppose that a space probe is launched in 2012 with \(2.0 \mathrm{~kg}\) of \({ }^{238} \mathrm{Pu}\) a. If the half-life of \({ }^{238} \mathrm{Pu}\) is \(87.7 \mathrm{yr}\), write a function of the form \(Q(t)=Q_{0} e^{-k t}\) to model the quantity \(Q(t)\) of \({ }^{238} \mathrm{Pu}\) left after \(t\) years. b. If \(1.6 \mathrm{~kg}\) of \({ }^{238} \mathrm{Pu}\) is required to power the spacecraft's data transmitter, for how long will scientists be able to receive data? Round to the nearest year.

a. The populations of two countries are given for January 1,2000 , and for January 1,2010 . Write a function of the form \(P(t)=P_{0} e^{k t}\) to model each population \(P(t)\) (in millions) \(t\) years after January 1, 2000.$$ \begin{array}{|l|c|c|c|} \hline & \begin{array}{c} \text { Population } \\ \text { in 2000 } \\ \text { (millions) } \end{array} & \begin{array}{c} \text { Population } \\ \text { in 2010 } \\ \text { (millions) } \end{array} & \boldsymbol{P}(t)=\boldsymbol{P}_{0} e^{k t} \\ \hline \text { Switzerland } & 7.3 & 7.8 & \\ \hline \text { Israel } & 6.7 & 7.7 & \\ \hline \end{array}$$ b. Use the models from part (a) to predict the population on January \(1,2020,\) for each country. Round to the nearest hundred thousand. c. Israel had fewer people than Switzerland in the year 2000 , yet from the result of part (b), Israel will have more people in the year \(2020 ?\) Why? d. Use the models from part (a) to predict the year during which each population will reach 10 million if this trend continues.

Determine if the statement is true or false. For each false statement, provide a counterexample. For example, \(\log (x+y) \neq \log x+\log y\) because \(\log (2+8) \neq \log 2+\log 8\) (the left side is 1 and the right side is approximately 1.204 ). $$ \log (x y)=(\log x)(\log y) $$

Given the functions defined by \(f(x)=2 x-1\) and \(g(x)=\frac{x+1}{2}\), a. Graph \(y=f(x), y=g(x),\) and the line \(y=x .\) Does the graph suggest that \(f\) and \(g\) are inverses? Why? b. Enter the following functions into the graphing editor. ( $$\mathrm{Y}_{1}=2 x-1$$ \(\mathrm{Y}_{2}=(x+1) / 2\) \(\mathrm{Y}_{3}=\mathrm{Y}_{1}\left(\mathrm{Y}_{2}\right)\) \(\mathrm{Y}_{4}=\mathrm{Y}_{2}\left(\mathrm{Y}_{1}\right)\) c. Create a table of points showing \(Y_{3}\) and \(Y_{4}\) for several values of \(x\). (Hint: Use the right and left arrows to scroll through the table editor to show functions \(Y_{3}\) and \(Y_{4}\).) Does the table suggest that \(f\) and \(g\) are inverses? Why?
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