course Mth 272
......!!!!!!!!...................................
13:10:45 4.4.4 (was 4.3 #40 write ln(.056) - -2.8824 as an exponential equation
......!!!!!!!!...................................
RESPONSE --> e^-2.8824 = .056
.................................................
......!!!!!!!!...................................
13:12:50 ** y = ln x is the same as e^y = x, so in exponential form the equation should read e^-2.8824 = .056 **
......!!!!!!!!...................................
RESPONSE --> ok i understand
.................................................
......!!!!!!!!...................................
13:13:06 4.4.8 (was 4.3 #8) write e^(.25) = 1.2840 as a logarithmic equation
......!!!!!!!!...................................
RESPONSE --> .25 = ln(1.2840)
.................................................
......!!!!!!!!...................................
13:14:18 ** e^x = y is the same as x = ln(y) so the equation is .25 = ln(1.2840). **
......!!!!!!!!...................................
RESPONSE --> ok i understand
.................................................
......!!!!!!!!...................................
13:14:50 4.4.16 (was 4.3 #16) Sketch the graph of y = 5 + ln x.
......!!!!!!!!...................................
RESPONSE --> My graph is increasing at a decreasing rate. It is asymptotic to the negative y axis. It passes through the points (1,5)
.................................................
......!!!!!!!!...................................
13:16:14 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok i understand that the logarithmic function is the inverst of the exponential function. This is concave up and has a asymptote at the negative x axis. It passes through the (0,1). So if the coordinates was revereesed the exponential graph gives you a graph which is concave down and asymptotice to the negatvie y axis. This passes through the points (1,0).
.................................................
......!!!!!!!!...................................
13:19:05 4.4.22 (was 4.3 #22) Show e^(x/3) and ln(x^3) inverse functions
......!!!!!!!!...................................
RESPONSE --> f(x) = e^(x/3) so f(ln(x^3)) = e^( ln(x^3) / 3) = e^(3 ln(x) / 3)=e^(ln x)=x
.................................................
......!!!!!!!!...................................
13:21:48 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. **
......!!!!!!!!...................................
RESPONSE --> I understand that applying one functions to the other gives the idenity function.
.................................................
......!!!!!!!!...................................
13:26:39 4.4.46 (was query 4.3 #44) simplify 1/3 [ 2 ln(x+3) + ln x - ln(x^2-1) ]
......!!!!!!!!...................................
RESPONSE --> =ln[(x+3)^2 * x / (x^2-1)}^(1/3)]
.................................................
......!!!!!!!!...................................
13:28:06 ** 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) ] **
......!!!!!!!!...................................
RESPONSE --> ok i understand the how to simplify.
.................................................
......!!!!!!!!...................................
13:30:34 4.4.58 (was 4.3 #58) solve e^(-.0174 t) = .5
......!!!!!!!!...................................
RESPONSE --> t = ln(.5) / -.0174 = 39
.................................................
......!!!!!!!!...................................
13:32:46 ** We can convert the equation to logarithmic form: ln(0.5) = -.0174t. Thus t = ln(.5) / -.0174 = 40 approx.. **
......!!!!!!!!...................................
RESPONSE --> ok i understand
.................................................
......!!!!!!!!...................................
13:35:05 4.4.72 (was 4.3 #68) p = 250 - .8 e^(.005x), price and demand; find demand for price $200 and $125
......!!!!!!!!...................................
RESPONSE --> p=200 then x = 200 ln(312.5 - 1.25 * 200) = 200 ln(62.5) = 827 p=125 then x= 125 ln(312.5-1.25*125)=125 ln(62.5)=1010
.................................................
......!!!!!!!!...................................
13:38:22 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok i understand how to evaluate the expression"