#$&* course MTH 272 10/1 about 4:45 pm 002. `query 2
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Given Solution: `a y = ln x is the same as e^y = x, so in exponential form the equation should read e^-2.8824 = .056 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q 4.4.4 (previously 4.4.8 (was 4.3 #8)) write e^(.25) = 1.2840 as a logarithmic equation YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ln 1.2840 = 0.25 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a e^x = y is the same as x = ln(y) so the equation is .25 = ln(1.2840). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q4.4.5 (previously 4.4.16 (was 4.3 #16)) Sketch the graph of y = 5 + ln x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I did plug in values and sketch a graph on paper prior to looking at the given solution. I did see that the asymptote was the negative y axis, that it passed through the point (0,5), and is concave down. It increases at a decreasing rate as the x values increase. I did know that the logarithmic function is an inverse to the exponential function. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Did not initially analyze the function with respect to the exponential function, but did understand the concepts. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q 4.4.8 (previously 4.4.22 (was 4.3 #22)) Show e^(x/3) and ln(x^3) inverse functions 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I understand the problem but could not figure out another way to show it. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q4.4.9 (previously 4.4.46 (was query 4.3 #44)) simplify 1/3 [ 2 ln(x+3) + ln x - ln(x^2-1) ] YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1/3(2ln(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)] confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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) ] ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q4.4.21 (previously 4.4.58 (was 4.3 #58)) solve 400 e^(-.0174 t) = 1000. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 400 e^(-0.0174 * t) = 1000 e^(-0.0174 * t)=1000/400=2.5 ln 2.5 = -0.0174 * t t = ln(2.5) / -0.0174 = -52.66 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q4.4.25 (previously 4.4.72 (was 4.3 #68)) p = 250 - .8 e^(.005x), price and demand; find demand for price $200 and $125 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, solve the equation for x. p = 250 - .8*e^(0.005*x) p-250 = -.8*e^(0.005*x) (p-250)/-.8 = e^(0.005*x) -1.25p + 312.5 = e^(0.005*x) ln(-1.25p + 312.5) = 0.005x x = 200 ln(-1.25p + 312.5) Then plug in the dollar amounts and solve. x = 200 ln((-1.25*200) + 312.5) = 827.033 x = 200 ln((-1.25*125) + 312.5) = 1010.29 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!