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21:38:11 behavior and source of exponential functions problem 1, perversions of laws of exponents Why is the following erroneous: a^n * b^m = (ab) ^ (n*m)
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RESPONSE --> a and b are two different variables, the law of exponents applies to the variables that are the same ex. a^n * a^m = a^(n+m) confidence assessment: 3
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21:38:21 STUDENT RESPONSE: my example was (4^2)(5^3) did not equal 20^6 INSTRUCTOR COMMENT ** more generally a^n * b^n = a^(n+m), which usually does not equal (ab) ^ (n * m) **
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RESPONSE --> self critique assessment: 3
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21:39:06 Why is the follow erroneous: a^(-n) = - a^n
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RESPONSE --> law of exponents a^(-n)= 1/(a^n) confidence assessment: 3
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21:39:27 STUDENT RESPONSE: 2^-3 is not equal to -2^3 INSTRUCTOR COMMENT: ** A more general counterexample: a^(-n) = 1 / a^n is positive when a is positive whereas -a^n is negative when a is positive **
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RESPONSE --> self critique assessment: 3
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21:40:38 Why is the following erroneous: a^n + a^m = a^(n+m)
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RESPONSE --> law or exponents rule works when the same variable is eing multiplied not added ex. a^n*a^m=a^(n+m) confidence assessment: 3
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21:40:43 STUDENT RESPONSE: (5^3)+(5^4) is not equal to 5^7 INSTRUCTOR COMMENT: (5^3) * (5^4) = (5^7) since (5*5*5) * (5*5*5*5) = 5*5*5*5*5*5*5 = 5^7. However (5^3) + (5^7) = 5*5*5 + 5*5*5*5*5 = 5*5*5( 1 + 5*5) = 5^3( 1 + 5^2), not 5^7.**
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RESPONSE --> self critique assessment: 3
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21:41:02 STUDENT RESPONSE: Why is the following erroneous: a^0 = 0
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RESPONSE --> law of exponents a^0=1 self critique assessment: 3
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21:41:07 4^0 is not equal to 0 INSTRUCTOR COMMENT: ** a^(-n) * a^n = 0 but neither a^(-n) nor a^n need equal zero **
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RESPONSE --> self critique assessment: 3
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21:41:44 Why is the following erroneous: a^n * a^m = a^(n*m).
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RESPONSE --> law of exponents you add the powers together confidence assessment: 3
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21:41:50 STUDENT RESPONSE: (4^7)(4^2) is not equal to 4^14 INSTRUCTOR COMMENT: Right. Generally a^n * a^m = a^(n+m), not a^(n*m).
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RESPONSE --> self critique assessment: 3
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21:46:18 problem 2. Graph and describe Give basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = 1200 (2^(.12 t) )
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RESPONSE --> y=Ab^x A=1200 b=2^.12 y=1200(2^(.12t)) y=1200(2^(.12(0)))=1200 y=1200(2^(.12(1)))=1304 (0,1200) (1,1304) ratio 2^.12 or 1.09 negative x-axis confidence assessment: 3
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21:46:26 STUDENT RESPONSE (0,1200),(1,1304) negative x-axis ratio=163/150 INSTRUCTOR COMMENT: the precise ratio is 2^.12, which is probably pretty close to 163/150
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RESPONSE --> self critique assessment: 3
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21:47:53 give basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = 400 ( 1.07 ) ^ t
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RESPONSE --> y=Ab^x A=400 b=1.07 y = 400 ( 1.07 ) ^ t y = 400 ( 1.07 ) ^ 0=400 y = 400 ( 1.07 ) ^ 1=428 (0,400) (1,428) ratio 1.07 confidence assessment: 3
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21:48:02 STUDENT RESPONSE (0,400),(1,428) Neg. x-axis 1.07 or 107/100 is ratio INSTRUCTOR COMMENT: that ratio is correct and is of course equal to 1.07
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RESPONSE --> self critique assessment: 3
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21:51:17 give basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = 250 ( 1 - .12 ) ^ t
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RESPONSE --> y=Ab^x A=250 b=(1-.12) or .88 y = 250 ( 1 - .12 ) ^ t y = 250 ( 1 - .12 ) ^ 0=250 y = 250 ( 1 - .12 ) ^ 1=220 (0,250) (1,220) ratio .88 confidence assessment: 3
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21:51:24 STUDENT RESPONSE The basic points are (0,250),(1,220) The positive x-axis is the horizontal asymptote The ratio of y values at the basic points is 220 / 250 = .88. INSTRUCTOR COMMENT: Note also that the ratio .88 is equal to 1-.12; .88 is the growth factor and .12 is the growth rate.
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RESPONSE --> self critique assessment: 3
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22:01:44 give basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = .04 ( .8 ) ^ t
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RESPONSE --> y=Ab^x A=.04 b=.8 y = .04 ( .8 ) ^ t y = .04 ( .8 ) ^ 0=.04 y = .04 ( .8 ) ^ 1=.032 (0,.04) (1,.032) ratio .8 confidence assessment: 3
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22:01:50 STUDENT RESPONSE (0,.04),(1,.032) are the basic points and the asymptote is the positive x-axis. The ratio is .32 / .4 = .8. The pattern is that the ratio is equal to b, where b is the base for the form y = A b ^ x.
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RESPONSE --> self critique assessment: 3
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22:09:24 problem 3. y = f(x) = 5 (1.27^x).
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RESPONSE --> y = f(x) = 5 (1.27^x) y = Ab^x A=5 b=1.27 y=5(1.27^0)=5 y=5(1.27^1)=6.35 (0,5) (1,6.35) ratio 1.27 confidence assessment: 3
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22:09:41 What is the ratio between the y values at x = 0 and at x = 1?
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RESPONSE --> ratio 1.27 confidence assessment: 3
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22:09:47 ** f(1) / f(0) = 5 * 1.27^1 / ( 5 * 1.27^0) = 1.27 **
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RESPONSE --> self critique assessment: 3
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22:11:51 What is the ratio between the y values at to x = 3.4 and x = 4.4?
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RESPONSE --> y = f(x) = 5 (1.27^x) y = f(3.4) = 5 (1.27^3.4)=11.26 y = f(4.4) = 5 (1.27^4.4)=14.31 (3.4,11.26) (4.4,14.31) ratio 1.27 confidence assessment: 3
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22:11:56 ** f(4.4) / f(3.4) = 5 * 1.27^4.4 / ( 5 * 1.27 ^3.4) = 1.27. **
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RESPONSE --> self critique assessment: 3
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22:13:15 Verify that the ratio of y values is again the same for your own points where x differs by 1 unit.
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RESPONSE --> ratio 1.27^1=1.27 confidence assessment: 3
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22:13:23 ** STUDENT RESPONSE: My points were 4.5 and 5.5, and the y ratio was again 1.27 **
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RESPONSE --> self critique assessment: 3
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22:13:44 What is the ratio of y values when x values are separated by two units?
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RESPONSE --> ratio 1.27^2=1.61 confidence assessment: 3
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22:15:02 ** If x values are separated by 2 units then the ratio is 1.27^(x+2) / 1.27^x = 1.27^(x+2 - x) = 1.27^2 = 1.61 approx. **
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RESPONSE --> self critique assessment: 3
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22:33:14 ** If y = A b^x then the value at x1 is A b^x1 and the value at x1 + 1 is A b ^(x1 + 1). The ratio of these values is A b^(x1+1) / A b^x1 = b^(x1 + 1 - x1) = b^1 = b. The ratio should have been b, where b is the base in the form y = A b^x. This is the same as in all previous examples, which shows that there is no dependence on x1. **
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RESPONSE --> okay note taken! self critique assessment: 3
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22:37:57 problem 5. y = 3 (2 ^ (.3 x) ). What is the ratio of the two basic-point y values?
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RESPONSE --> y=Ab^x A=3 b=2^.3 y=3(2^(.3(0)))=3 y=3(2^(.3(1)))=3.69 (0,3) (1,3.69) ratio 2^.3=1.23 confidence assessment: 3
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22:38:03 ** The basic points are the x = 0 and x = 1 points. The corresponding y values are 3(2^.(3*0) ) = 3 and 3(2^(.3 *1) ) = 3 * 2^.3 = 3.69 approx. The ratio of these values is 3.69 / 3 = 1.23. **
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RESPONSE --> self critique assessment: 3
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22:41:50 What is the y = A b^x form of this function?
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RESPONSE --> y=3(1.23)^x or y=3*2^.3x confidence assessment: 3
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22:42:09 ** 3 (2 ^ (.3 x) ) = 3 (2^.3)x = 3 * 1.23^x, approx. This is in the form y = A b^x for A = 3 and b = 1.23. **
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RESPONSE --> which I shown in previous answer self critique assessment: 3
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22:42:39 What does the value of 2 ^ .3 have to do with this situation?
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RESPONSE --> 2^.3=1.23 which is b in the equation and the ratio confidence assessment: 3
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22:43:07 ** CORRECT STUDENT RESPONSE: this is the b value in the form y = A b^x. **
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RESPONSE --> self critique assessment: 3
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22:46:35 problem 6 P(n+1) = (1+r) P(n), with r = .1 and P(0) = $1000. What are P(1), P(2), ..., P(5)?
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RESPONSE --> n=0 P0=1000 r =.1 P(n+1)=P0(1+r)^n P(1)=1000(1.1)^1=1100 P(2)=1000(1.1)^1=1210 P(3)=1000(1.1)^1=1331 P(4)=1000(1.1)^1=1464.10 P(5)=1000(1.1)^1=1610.51 confidence assessment: 3
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22:50:53 ** If n = 0 we get P(0 + 1) = (1 + .1) * P(0), or P(1) = 1.1 * P(0). Since P(0) = $1000 we have P(1) = 1.1 * $1000 = $1100. If n = 1 we get P(1 + 1) = (1 + .1) * P(1), or P(2) = 1.1 * P(1). Since P(1) = $1000 we have P(2) = 1.1 * $1100 = $1210. If n = 2 we get P(2 + 1) = (1 + .1) * P(2), or P(3) = 1.1 * P(2). Since P(2) = $1000 we have P(3) = 1.1 * $1210 = $1331. If n = 3 we get P(3 + 1) = (1 + .1) * P(3), or P(4) = 1.1 * P(3). Since P(3) = $1000 we have P(4) = 1.1 * $1331 = $1464/1. If n = 4 we get P(4 + 1) = (1 + .1) * P(4), or P(5) = 1.1 * P(4). Since P(4) = $1000 we have P(5) = 1.1 * $1464.1 = $1610.51. **
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RESPONSE --> okay note taken on the method self critique assessment: 2
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23:01:33 problem 8. Q(n+1) = .85 Q(n), Q(0) = 400. What are Q(n) for n = 1, 2, 3 and 4 /
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RESPONSE --> Q(n+1) = .85 Q(n), Q(0) = 400 n=0 Q(0+1) = .85*Q0= Q(1)=.85*Q0=340 n=1 Q(1+1) = .85*Q1= Q(2)=.85*Q1=289 n=2 Q(2+1) = .85*Q2= Q(3)=.85*Q2=245.63 n=3 Q(3+1) = .85*Q3= Q(4)=.85*Q3=208.80 n=4 Q(4+1) = .85*Q4= Q(5)=.85*Q4=177.48 confidence assessment: 3
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23:01:41 ** For n = 0 we have Q(0 + 1) = .85 Q(0) so Q(1) = .85 * 400 = 340. For n = 1 we have Q(1 + 1) = .85 Q(1) so Q(2) = .85 * 340 = 289. For n = 2 we have Q(2 + 1) = .85 Q(2) so Q(3) = .85 * 400 = 245.65. For n = 3 we have Q(3 + 1) = .85 Q(3) so Q(4) = .85 * 400 = 208.803. **
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RESPONSE --> self critique assessment: 3
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23:01:59 What is the growth rate for this equation?
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RESPONSE --> growth rate is .15 confidence assessment: 3
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23:02:04 ** The growth factor is .85, which is 1 + growth rate. It follows that the growth rate is .85 - 1 = .15 **
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RESPONSE --> self critique assessment: 3
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23:03:41 problem 9. interest rate 12%, initial principle $2000. What is your difference equation?
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RESPONSE --> P0=2000 r=.12 P(n+1)=(1+r)*P(n) confidence assessment: 2
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23:03:59 ** The growth rate is 12% = .12 The growth factor is therefore 1 + .12 and the difference equation is P(n+1)=(1+.12)P(n), P(0)=2000. **
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RESPONSE --> self critique assessment: 3
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23:06:25 How did you use your difference equation to find the principle after 1, 2, 3 and 4 years and what did you get?
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RESPONSE --> P0=2000 r=.12 P(n+1)=1.12*P(n) P1=2240 P2=2508.8 P3=2809.86 P4=3147.04 confidence assessment: 3
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23:06:36 ** STUDENT RESPONSE P(0+1)=(1+.12)2000 and so on up to P(4) was found. P1=2240 P2=2508.8 P3=2809.856 P4=3147.03872 **
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RESPONSE --> self critique assessment: 3
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23:14:00 problem 11. Texcess(t) = 50 (.97 ^ t). What is your estimate of the time required to fall to 1/8 of the original value?
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RESPONSE --> Texcess(t) = 50 (.97 ^ t) Texcess(0) = 50 (.97 ^ 0)=50 Texcess(1) = 50 (.97 ^ 1)=48.5 (0,50) (1,48.5) 50*(1/8)=6.25 confidence assessment: 1
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23:14:17 ** The original value takes place at t = 0 and is Texcess(0) = 50 * .97^0 = 50. 1/8 of the original value is therefore 1/8 * 50 = 6.25. You have to find t such that 50 * .97^t = 6.25. Dividing both sides by 50 you get .97^t = 6.25 / 50 or .97^t = .125. Use trial and error to find t: Try t = 10: .97^10 = .74 approx. That's too high. Try t = 100: .97^100 = .04 approx. That's too low. So try a number between 10 and 100, probably closer to 100. Try 70: .97^70 = .118. Lucky guess. That's close to .125 but a little low. {Try 65: .97^65 = .138. Too high. Try a number between 65 and 70, closer to 70 but not too much closer. Try 68: .97^68 = .126. That's good to the nearest whole number. The process could be continued and refined to get more accurate values of t. **
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RESPONSE --> okay note taken self critique assessment: 3
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23:16:38 Identify the values of A, b and c in the generalized form y = A b^x + c.
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RESPONSE --> y=Ab^x+c A=50 b=.97 c=75 confidence assessment: 3
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23:16:43 ** Since the function is Temp(t) = 50(.97^t)+75 , the y = A b^x + c has y = Temp(t), A = 50, b = .97 and c = 75. **
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RESPONSE --> self critique assessment: 3
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23:21:14 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> This assignment was very useful and I learned a lot from it...some of these same problems were in a link in the assignments work..this help me understand the problems I couldn't work out. self critique assessment: 2
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