Question

1. rewrite each of the equations in the indicated form: a. rewrite in logarithmic form: 7^3 = 343 b. rewrite in exponential form: log 0.01 = - 2 9. solve each of the equations. round answers to 3 decimal places as needed. a. 5e^x + 4 = 39 b. log_2(3x + 1) = 6 10. based in data from 1965 to 2002, the number of local municipal governments in the u.s. can be modeled by the function f(t)=17,658.6 + 490.8ln(t) where t is time in years since 1965. a. use the given function to evaluate f(20) and explain its practical meaning in the context of the problem. round to the nearest whole number. f(20)=17658.6 + 490.8ln(t) 0.001132 + 490.8ln(t) b. according to the given model, in what year were there about 19,360 local municipal governments in the u.s.?

Explanation:

8.

a.

Step1: Recall log - exp relationship

The general form is \(a^b = c\) can be written as \(\log_a c=b\). For \(7^3 = 343\), \(a = 7\), \(b = 3\) and \(c = 343\).
So, \(\log_7 343=3\)

b.

Step1: Recall log - exp relationship

The general form \(\log_a c=b\) can be written as \(a^b = c\). For \(\log 0.01=-2\), since the base of the common - log is 10 (\(\log=\log_{10}\)), we have \(a = 10\), \(b=-2\) and \(c = 0.01\).
So, \(10^{-2}=0.01\)

9.

a.

Step1: Isolate the exponential term

Given \(5e^x + 4=39\), first subtract 4 from both sides: \(5e^x=39 - 4=35\). Then divide both sides by 5: \(e^x=\frac{35}{5}=7\).

Step2: Take the natural - log of both sides

Since \(y = e^x\) and \(y=\ln^{-1}(x)\) are inverse functions, \(\ln(e^x)=\ln(7)\). By the property \(\ln(e^x)=x\), we get \(x=\ln(7)\approx1.946\)

b.

Answer:

1. a. \(\log_7 343 = 3\)

b. \(10^{-2}=0.01\)

1. a. \(x\approx1.946\)

b. \(x = 21\)

1. a. \(f(20)\approx19129\), meaning there were approximately 19129 local municipal governments in the U.S. in 1985.

b. 1996