course MTH 272 01/31, 20:00 assignment #002002. `query 2
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16:35:14 4.4.4 (was 4.3 #40 write ln(.056) = -2.8824 as an exponential equation
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RESPONSE --> ln(.056)=-2.8824 e^(ln .056)=e^(-2.8824) .056=e^(-2.8824) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:35:22 y = ln x is the same as e^y = x, so in exponential form the equation should read e^-2.8824 = .056 **
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RESPONSE --> ------------------------------------------------ Self-critique rating #$&*:ent: 3
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16:37:06 4.4.8 (was 4.3 #8) write e^(.25) = 1.2840 as a logarithmic equation
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RESPONSE --> e^.25 = 1.2840 ln e^.25 = ln 1.2840 .25=ln 1.2840 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:37:13 e^x = y is the same as x = ln(y) so the equation is .25 = ln(1.2840). **
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RESPONSE --> ------------------------------------------------ Self-critique rating #$&*:ent: 3
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16:40:24 4.4.16 (was 4.3 #16) Sketch the graph of y = 5 + ln x.
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RESPONSE --> The graph of y=5 + ln x is increasing at a decreasing rate and is concave downward. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:42:07 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 --> The answer given is much more detailed than my own answer. But, generally, seems to describe the same graph. I likely would have gotten a better sense of the graph by forming a table for various x values and the resulting value of the function (y-value). ------------------------------------------------ Self-critique rating #$&*:ent: 3
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16:55:17 4.4.22 (was 4.3 #22) Show e^(x/3) and ln(x^3) inverse functions
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RESPONSE --> When graphing the functions e^(x/3) and ln (x^3), the resulting lines mirror one another. The graph of e^(x/3) is concave upward and is increasing at an increasing rate. The ln(x^3) graph is concave downward and is increasing at a decreasing rate. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:03:17 4.4.46 (was query 4.3 #44) simplify 1/3 [ 2 ln(x+3) + ln x - ln(x^2-1) ]
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RESPONSE --> 1/3 [ 2 ln(x + 3) + ln x - ln(x^2 - 1) ] 1/3 [ (ln(x+3)^2 + ln x)/ln(x^2 -1) ] 1/3 ln [(x(x+3)^2)/(x^2-1)] ln [(x(x+3)^2)/(x^2-1)]^(1/3) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:03:59 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 --> ------------------------------------------------ Self-critique rating #$&*:ent: 3
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17:07:22 4.4.58 (was 4.3 #58) solve 400 e^(-.0174 t) = 1000.
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RESPONSE --> 400e^(-.0174t)=1000 e^(-.0174t)=2.5 ln e^(-.0174t)=ln(2.5) -.0174t = .916 t=-52.66 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:08:44 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 --> My answer differs from the one given by being a negative (-52.66) versus a positive 52.7. ------------------------------------------------ Self-critique rating #$&*:ent: 2
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17:16:12 4.4.72 (was 4.3 #68) p = 250 - .8 e^(.005x), price and demand; find demand for price $200 and $125
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RESPONSE --> p = 250 - .8 e^(.005x) p=200 200 = 250 - .8 e^(.005x) -50 = - .8e^(.005x) 62.5 = e^(.005x) ln(62.5)= ln e^(.005x) 4.14 =.005x 827=x p=125 125=250 - .8e^(.005x) -125=-.8e^(.005x) 100=e^(.005x) ln(100)=ln (e^(.005x)) 4.61=.005x 921=x confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:16:24 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 --> ------------------------------------------------ Self-critique rating #$&*:ent: 3
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