course mth 272 §”±û‰ÅýàÜ^å¾³C̘áœòœéôÎñassignment #002 002. `query 2 Applied Calculus II 06-19-2008
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07:46:22 4.4.4 (was 4.3 #40 write ln(.056) = -2.8824 as an exponential equation
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RESPONSE --> e^(.056)=-2.8824 i know that in one example I have where y =ln x is written as e^x =x. there is no y in the equation above, so this is my best guess. confidence assessment: 0
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07:50:07 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, i understand that because y= ln x is the same as saying e^y =x, and in the equation the x=.056 and y=-2.8824. then plug into e^y=x self critique assessment: 2
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07:52:52 4.4.8 (was 4.3 #8) write e^(.25) = 1.2840 as a logarithmic equation
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RESPONSE --> because y=ln x is the same as e^y=x this equation should be written out as ln(1.2840) =.25 confidence assessment: 2
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07:53:03 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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08:01:04 4.4.16 (was 4.3 #16) Sketch the graph of y = 5 + ln x.
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RESPONSE --> we need to differentiate this first. y= 5 + ln x y'= 0 +(1/x) y'= 1/x the graph is concave down from (0,4) and decreasingly increasing towards the positive x axis confidence assessment: 1
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08:05:56 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 I understand the explanation given that the log function is the inverse so we reverse what we get from that function, then add 5 to raise the graph. self critique assessment: 2
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08:10:31 4.4.22 (was 4.3 #22) Show e^(x/3) and ln(x^3) inverse functions
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RESPONSE --> i think ln(x^3) would be written as 1/x^3. then the inverse of that would be x^3. not sure about e^(x/3) confidence assessment: 1
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08:17:17 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. nat logs and nat exponential functions are inverses of each other. to illustrate this if you sat up e^(x/3) and lnx^3 to each other you would get x. these two inverses are each other. confidence assessment: 2
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08:19:48 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 --> confidence assessment: 0
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08:23:16 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. i u follow all the steps. self critique assessment: 3
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08:28:17 4.4.58 (was 4.3 #58) solve 400 e^(-.0174 t) = 1000.
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RESPONSE --> 400e^(-.0174 t) = 1000 e^(-.0174 t) =2.5 divide 1000 by 400 (-.0174 t) = e^2.5 =12.18249396
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08:30:03 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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08:36:54 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 --> for $200 200=250-.8e^(.005x) 62.5=e(.005x) ln(62.5)=.005x x=ln(2.5)/.005 x= 183.258 do the same plugging in $125 confidence assessment: 2
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08:43:23 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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