course Mth 272 Some of these problems did not matchup with the textbook I have. I have the latest edition. I feel this will be an ongoing problem. I did every problem assigned using my textbook, which is the latest edition.
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15:20:27 4.4.4 (was 4.3 #40 write ln(.056) = -2.8824 as an exponential equation
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RESPONSE --> This problem was not one that matches up with my textbook. confidence assessment: 3
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15:20:45 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 --> I understand how to solve these problems. self critique assessment: 3
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16:21:26 4.4.8 (was 4.3 #8) write e^(.25) = 1.2840 as a logarithmic equation
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RESPONSE --> ln1.2840=0.25 confidence assessment: 3
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15:23:23 e^x = y is the same as x = ln(y) so the equation is .25 = ln(1.2840). **
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RESPONSE --> understand self critique assessment: 3
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15:25:34 4.4.16 (was 4.3 #16) Sketch the graph of y = 5 + ln x.
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RESPONSE --> a graph that starts at (0,5) and arches up somewhat. Its next point is (2, 5.6931), the next point after that is (3, 6.0986) confidence assessment: 3
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15:25:58 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 --> I know how to do the problem. I hope I explained it well enough. self critique assessment: 3
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15:26:37 4.4.22 (was 4.3 #22) Show e^(x/3) and ln(x^3) inverse functions
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RESPONSE --> y=3(ln(x)) confidence assessment: 3
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15:27:15 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 --> I needed to show more steps of the process. Im glad this was an in depth response to my answer. confidence assessment: 3
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15:29:23 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+4)] ln cubert x(x+3)2-(x^2-1) confidence assessment: 3
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15:29:35 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 --> self critique assessment:
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15:30:17 4.4.58 (was 4.3 #58) solve 400 e^(-.0174 t) = 1000.
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RESPONSE --> This does not matchup with a problem in my textbook. confidence assessment: 3
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15:30:54 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 --> I know the process of how to solve these problems. The problems themselves did not match up. self critique assessment: 3
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15:31:43 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 --> This problem does not match up with my textbook. confidence assessment: 3
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15:32:13 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 --> I know how to do these problems. The problems in my textbook do not matchup with what I was assigned. self critique assessment: 3
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