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Chapter 6: Problem 1

Match each exponential function in parts (a)-(d) with its logarithmic form in parts (e)-(h). a. \(y=10,000(2)^{x}\) b. \(y=1000(1.4)^{x}\) c. \(y=\left(3 \cdot 10^{6}\right)(0.8)^{x}\) d. \(y=1000(0.45)^{x}\) e. \(\log y=6.477-0.097 x\) f. \(\log y=4+0.301 x\) g. \(\log y=3-0.347 x\) h. \(\log y=3+0.146 x\)

Short Answer

Expert verified
a matches with f, b matches with h, c matches with e, d matches with g.

Step by step solution

01

Identify the logarithmic form of the exponential function

Recall that the general form of an exponential function is given by \( y = A \times B^x \). The equivalent logarithmic form is \( \text{log} y = \text{log} A + x \text{log} B \). Use this to transform each given exponential function.
02

Transform function (a)

For the function \( y = 10,000(2)^x \): \( A = 10,000 \) and \( B = 2 \). Therefore, its logarithmic form is \( \text{log} y = \text{log}(10,000) + x \text{log}(2) \). Applying logarithm values, \( \text{log}(10,000) = 4 \) and \( \text{log}(2) \approx 0.301 \). Thus, \( \text{log} y = 4 + 0.301 x \). This matches with (f).
03

Transform function (b)

For the function \( y = 1000(1.4)^x \): \( A = 1000 \) and \( B = 1.4 \). Therefore, its logarithmic form is \( \text{log} y = \text{log}(1000) + x \text{log}(1.4) \). Applying logarithm values, \( \text{log}(1000) = 3 \) and \( \text{log}(1.4) \approx 0.146 \). Thus, \( \text{log} y = 3 + 0.146 x \). This matches with (h).
04

Transform function (c)

For the function \( y = (3 \times 10^6)(0.8)^x \): \( A = 3 \times 10^6 \) and \( B = 0.8 \). Therefore, its logarithmic form is \( \text{log} y = \text{log}(3 \times 10^6) + x \text{log}(0.8) \). Applying logarithm values, \( \text{log}(3 \times 10^6) \approx 6.477 \) and \( \text{log}(0.8) \approx -0.097 \). Thus, \( \text{log} y = 6.477 - 0.097 x \). This matches with (e).
05

Transform function (d)

For the function \( y = 1000(0.45)^x \): \( A = 1000 \) and \( B = 0.45 \). Therefore, its logarithmic form is \( \text{log} y = \text{log}(1000) + x \text{log}(0.45) \). Applying logarithm values, \( \text{log}(1000) = 3 \) and \( \text{log}(0.45) \approx -0.347 \). Thus, \( \text{log} y = 3 - 0.347 x \). This matches with (g).

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Key Concepts

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

logarithmic transformation
Understanding logarithmic transformations is key to convert exponential functions into their logarithmic forms. The general equation for an exponential function is given as \( y = A \times B^x \). To transform it into its logarithmic form, we use the properties of logarithms. Start by taking the common logarithm (log base 10) of both sides: \( \log y = \log (A \times B^x) \). By applying the logarithm rules, this can be broken down into: \( \log y = \log A + \log B^x \). Use another property of logarithms to simplify the right side as: \( \log y = \log A + x \log B \). This is the logarithmic transformation of an exponential function.
matching functions
Once you have the logarithmic form of the exponential functions, the next step is to match them with the given logarithmic equations. For instance, if we have an exponential function \( y = 10,000(2)^x \), following the logarithmic transformation gives us \( \log y = 4 + 0.301x \). Compare this with the given logarithmic forms to find the matching equation. This process requires careful observation of the constants in the equations. Make sure that the coefficients of \( x \) and the constant terms match to accurately pair each function.
algebraic manipulation
Algebraic manipulation involves rearranging equations and expressions to solve for variables or to put them in a desired form. In the context of converting exponential functions to their logarithmic forms, you start by isolating the constant and base. For example, in \( y = 1000(1.4)^x \), identify \( A = 1000 \) and \( B = 1.4 \). Then apply logarithms: \( \log y = \log(1000) + x \log(1.4) \). Apply known logarithmic values like \( \log(1000) = 3 \) to simplify the equation further: \( \log y = 3 + 0.146x \). Ensuring accurate algebraic manipulations is crucial for correct transformations.
logarithmic properties
Logarithmic properties help us simplify and manipulate logarithmic expressions. Key properties include: the product rule \( \log(AB) = \log A + \log B \), the power rule \( \log(A^B) = B \log A \), and the quotient rule \( \log(A/B) = \log A - \log B \). These properties are essential when transforming exponential functions. For example, using the product rule, you can expand \( \log(10,000 \times 2^x) \) as \( \log(10,000) + \log(2^x) \). Then apply the power rule to simplify \( \log(2^x) \) into \( x \log 2 \). Mastering these properties makes transformations smoother and helps in solving logarithmic equations efficiently.

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Most popular questions from this chapter

For each of the following equations, either prove that it is correct (by using the rules of logarithms and exponents) or else show that it is not correct (by finding numerical values for \(x\) that make the values on the two sides of the equation different). a. \(e^{x+\ln x}=x \cdot e^{x}\) b. \(\ln \left(\frac{(x+1)^{2}}{x}\right)=2 \ln (x+1)-\ln x\) c. \(\ln \left(\frac{x}{x+1}\right)=\frac{\ln x}{\ln (x+1)}\) d. \(\ln \left(x+x^{2}\right)=\ln x+\ln x^{2}\) e. \(\ln \left(x+x^{2}\right)=\ln x+\ln (x+1)\)

(Requires a graphing program.) A woman starts a training program for a marathon. She starts in the first week by doing 10 -mile runs. Each week she increases her run length by \(20 \%\) of the distance for the previous week. a. Write a formula for her run distance, \(D,\) as a function of week, \(W\) b. Use technology to graph your function, and then use the graph to estimate the week in which she will reach a marathon length of approximately 26 miles. c. Now use your formula to calculate the week in which she will start running 26 miles.

Contract, expressing your answer as a single logarithm. a. \(\frac{1}{4} \ln (x+1)+\frac{1}{4} \ln (x-3)\) b. \(3 \ln R-\frac{1}{2} \ln P\) c. \(\ln N-2 \ln N_{0}\)

The half-life of bismuth-214 is about 20 minutes. a. Construct a function to model the decay of bismuth- 214 over time. Be sure to specify your variables and their units. b. For any given sample of bismuth- 214 , how much is left after I hour? c. How long will it take to reduce the sample to \(25 \%\) of its original size? d. How long will it take to reduce the sample to \(10 \%\) of its original size?

For each of the following, find the half-life, then rewrite each function in the form \(P=P_{0} e^{r t}\). Assume \(t\) is measured in years. a. \(P=P_{0}\left(\frac{1}{2}\right)^{t / 10}\) b. \(P=P_{0}\left(\frac{1}{2}\right)^{t / 215}\) c. \(P=P_{0}\left(\frac{1}{2}\right)^{4 t}\)
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