course Mth 272 When you check these assignments and make assessments, where are they posted. Are they sent in an email? Thanks. ?|????????|?????assignment #002002. `query 2
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00:42:54 4.4.4 (was 4.3 #40 write ln(.056) = -2.8824 as an exponential equation
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RESPONSE --> lnx=b is equivalent to e^b=x so this is rewritten as e^-2.8824=.056 confidence assessment: 3
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00:42:59 y = ln x is the same as e^y = x, so in exponential form the equation should read e^-2.8824 = .056 **
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RESPONSE --> ok self critique assessment: 3
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00:43:09 4.4.8 (was 4.3 #8) write e^(.25) = 1.2840 as a logarithmic equation
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RESPONSE --> this is the same type of problem just reversed. it is rewritten as ln(1.2840)=.25 confidence assessment: 3
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00:43:14 e^x = y is the same as x = ln(y) so the equation is .25 = ln(1.2840). **
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RESPONSE --> ok self critique assessment: 3
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00:43:26 4.4.16 (was 4.3 #16) Sketch the graph of y = 5 + ln x.
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RESPONSE --> The graph of this function shows that it is discontinuous from 0 to negative infinite. The function also never reaches the negative y axis either, because it is nearing positive infinite as y as x approaches infinite. The graph is concave down, but at an increasing rate. When x is 1, y is 5, because of the 5 added to the lnx. This is the reason for the graph keeping out of the negative y region. confidence assessment: 3
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00:43:30 Plugging in values is a good start but we want to explain the graph and construct it without having to resort to much of that. The logarithmic function is the inverse of the exponential function, which is concave up, has asymptote at the negative x axis and passes thru (0, 1). Reversing the coordinates of the exponential graph gives you a graph which is concave down, asymptotic to the negative y axis and passes thru (1, 0). Adding 5 to get y = 5 + ln x raises this graph 5 units so that the graph passes thru (1, 5) and remains asymptotic to the negative y axis and also remains concave down. The graph is asymptotic to the negative y axis, and passes through (1,5). The function is increasing at a decreasing rate; another way of saying this is that it is increasing and concave down. STUDENT COMMENT: I had the calculator construct the graph. I can do it by hand but the calculator is much faster. INSTRUCTOR RESPONSE: The calculator is faster but you need to understand how different graphs are related, how each is constructed from one of a few basic functions, and how analysis reveals the shapes of graphs. The calculator doesn't teach you that, though it can be a nice reinforcing tool and it does give you details more precise than those you can imagine. Ideally you should be able to visualize these graphs without the use of the calculator. For example the logarithmic function is the inverse of the exponential function, which is concave up, has asymptote at the negative x axis and passes thru (0, 1). Reversing the coordinates of the exponential graph gives you a graph which is concave down, asymptotic to the negative y axis and passes thru (1, 0). Adding 5 to get y = 5 + ln x raises this graph 5 units so that the graph passes thru (1, 5) and remains asymptotic to the negative y axis and also remains concave down. **
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RESPONSE --> ok self critique assessment: 3
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00:43:41 4.4.22 (was 4.3 #22) Show e^(x/3) and ln(x^3) inverse functions
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RESPONSE --> These functions are inverses of each other. y=e^(x/3) x=e^(y/3) y=lnx^3 opposite y=lnx^3 x=lny^3 y=e^(x/3) confidence assessment: 3
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00:43:46 GOOD STUDENT RESPONSE: Natural logarithmic functions and natural exponential functions are inverses of each other. f(x) = e^(x/3) y = e^(x/3) x = e^(y/3) y = lnx^3 f(x) = lnx^3 y = ln x^3 x = lny^3 y = e^(x/3) INSTRUCTOR RESPONSE: Good. f(x) = e^(x/3) so f(ln(x^3)) = e^( ln(x^3) / 3) = e^(3 ln(x) / 3) = e^(ln x) = x would also answer the question MORE ELABORATION You have to show that applying one function to the other gives the identity function. If f(x) = e^(x/3) and g(x) = ln(x^3) then f(g(x)) = e^(ln(x^3) / 3) = e^( 3 ln(x) / 3) = e^(ln(x)) = x. **
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RESPONSE --> ok confidence assessment: 3
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00:43:55 4.4.46 (was query 4.3 #44) simplify 1/3 [ 2 ln(x+3) + ln x - ln(x^2-1) ]
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RESPONSE --> 1/3[2ln(x+3)+lnx-ln(x^2-1)] first distribute the 1/3 2/3ln(x+3)+1/3lnx-1/3ln(x^2+1) then reverse order of the addition and subtraction rule of logarithms can be used ln(x+3)^(2/3)+ln(x^(1/3))- ln((x^2-1)^(1/3)) then the rules backwards ln [(x+3)^(2/3)(x^(1/3)/(x^2-1)^(1/3)] confidence assessment: 3
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00:43:59 1/3 [ 2 ln(x+3) + ln x - ln(x^2-1) ] = 2/3 ln(x+3) + 1/3 ln x - 1/3 ln(x^2-1) = ln(x+3)^(2/3) + ln(x^(1/3)) - ln((x^2-1)^(1/3)) = ln [ (x+3)^(2/3) (x^(1/3) / (x^2-1)^(1/3) ] = ln [ {(x+3)^2 * x / (x^2-1)}^(1/3) ] **
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RESPONSE --> ok self critique assessment: 3
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00:44:10 4.4.58 (was 4.3 #58) solve 400 e^(-.0174 t) = 1000.
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RESPONSE --> 400e^(-.0174t)= 1000 first divide both sides by 400 to make things more manageable e^(-.0174t)=2.5 then log form ln(2.5)= -.0174t then divide to find t t= ln(2.5)/-.0174 t= -52.66 confidence assessment: 3
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00:44:16 The equation can easily be arranged to the form e^(-.0174) = 2.5 We can convert the equation to logarithmic form: ln(2.5) = -.0174t. Thus t = ln(2.5) / -.0174 = 52.7 approx.. **
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RESPONSE --> ok self critique assessment: 3
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00:44:29 4.4.72 (was 4.3 #68) p = 250 - .8 e^(.005x), price and demand; find demand for price $200 and $125
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RESPONSE --> p= 250-.8e^(.005x) x needs to be isolated to find out how many were sold p-250=-.8e^(.005x) subtraction, then division p-250/-.8 =e^(.005x) .005x=ln(312.5-1.25p) x=200ln(312-1.25p) then substitute 200 so x=200ln(312.5-1.25*200)= 827.033 then substitute 125 to get x= 1010.29 confidence assessment: 3
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00:44:34 p = 250 - .8 e^(.005x) so p - 250 = - .8 e^(.005x) so e^(.005 x) = (p - 250) / (-.8) so e^(.005 x) = 312.5 - 1.25 p so .005 x = ln(312.5 - 1.25 p) and x = 200 ln(312.5 - 1.25 p) If p = 200 then x = 200 ln(312.5 - 1.25 * 200) = 200 ln(62.5) = 827.033. For p=125 the expression is easily evaluated to give x = 1010.29. **
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RESPONSE --> ok self critique assessment: 3
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