course MTH272 Ï€ªœEyÔŽ€â±[ÉN‹¦R¹{°íÑǦassignment #002
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19:52:59 4.4.4 (was 4.3 #40 write ln(.056) = -2.8824 as an exponential equation
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RESPONSE --> lnx=y e^y=x e^(-2.8824)=.056 confidence assessment: 3
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19:53:03 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 assessment:
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19:56:07 4.4.8 (was 4.3 #8) write e^(.25) = 1.2840 as a logarithmic equation
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RESPONSE --> lnx=y e^y=x .25=ln(1.2840) confidence assessment: 3
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19:56:15 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 assessment: 3
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19:57:16 4.4.16 (was 4.3 #16) Sketch the graph of y = 5 + ln x.
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RESPONSE --> always increasing with an asymptote that is the y-axis.concave down confidence assessment: 3
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19:58:09 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 --> self critique assessment: 3
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20:06:02 4.4.22 (was 4.3 #22) Show e^(x/3) and ln(x^3) inverse functions
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RESPONSE --> e^(x/3) lne^(x/3) =x/3 ln(x^3) e^(ln(x^3) = x^3 confidence assessment: 2
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20:09:31 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 wasnt really sure what the question was asking.Simple confusion. it looks like to prove the inverse you substituted g(x) inside f(x) to form f(g(x)). I understand the concept. There was no examples in the book. confidence assessment: 2
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20:20:04 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[lnx^2 + 2ln3 + lnx - lnx^2 + ln(1)]
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20:20:12 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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20:21:33 4.4.58 (was 4.3 #58) solve 400 e^(-.0174 t) = 1000.
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RESPONSE --> 400e^(-.0714t)=1000 2.5=e^(.0714t) ln2.5=.0714t .916=.0714t t=.0189 confidence assessment: 3
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20:23:05 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 --> had the right steps, but I did it all on the computer without writing it down and switched my division around -.0174/ln(2.5) when it should be ln2.5/-.0174 self critique assessment: 2
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20:31:55 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 --> the question in the book asks to solve for p, but i think you want us to solve for x. because your asking for the units sold. so that would make the equation: (p-250)/(-.8)= e^(.005x) p=200 x=827 units sold p=125 x= 1010.3 units sold confidence assessment: 3
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20:32:17 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 --> economics is definilty more my forte'. self critique assessment: 3
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