course Mth 163 I just wanted to let you know that i did take my test last week and turn it into the learning lab and if you havent gotten it please let me know }????????|?i?assignment #018018. `query 18
......!!!!!!!!...................................
20:09:50 query Logarithms, Logarithmic Functions, Logarithmic Equations 1. For what value of x will the function y = log{base 2}(x) first reach 4
......!!!!!!!!...................................
RESPONSE --> 2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:10:48 ** ln(x) = 4 translates to x = e^4, which occurs at x = 55 approx. ln(x) = 2 translates to x = e^2, which occurs at x = 7.4 approx. ln(x) = 3 translates to x = e^3, which occurs at x = 20 approx. **
......!!!!!!!!...................................
RESPONSE --> I understand the first one but I dont understand where the ln(x)=3 came from? self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:11:43 for what value of x will the function y = ln(x) first reach y = 4?
......!!!!!!!!...................................
RESPONSE --> 55 approxamitly i took the inverse of it which is e^4 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:11:59 y = ln(x) means that e^y = x. The function y = ln(x) will first reach y = 4 when x = e^4 = 54.6 approx. **
......!!!!!!!!...................................
RESPONSE --> ok i have the hang of this now self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:12:58 3. Explain why the negative y axis is an asymptote for a log{base b}(x) function explain why this is so only if b > 1
......!!!!!!!!...................................
RESPONSE --> The reason is because it never actually crosses the axis it curves away once it gets close confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:13:17 ** the log{base b} function is the inverse of the y = b^x function. Assuming b > 1, large negative values of x lead to positive b^x values near zero, resulting in a horizontal asymptotes along the negative x axis. When the columns are reversed we end up with small positive values of x associated with large negative y, resulting in a vertical asymptote along the negative y axis. You can take a negative power of any positive b, greater than 1 or not. For b > 1, larger negative powers make the result smaller, and the negative x axis is an asymptote. For b < 1, larger positive powers make the result smaller, and the positive x axis becomes the asymptote. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:19:55 5. What are your estimates for the values of b for the two exponential functions on the given graph?
......!!!!!!!!...................................
RESPONSE --> I have tried multiple ways to figure out how to solve this and i understand how but its just that i dont understand where the y coordinate should come from and i have no way of solving y=b^-3 so i am anxious to see how this one is thought out confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:20:32 ** The exponential function y = A b^x has y intercept (0, A) and basic point (1, A * b). Both graphs pass thru (0, 1) so A = 1. The x = 1 points are (1, 3.5) and (1, 7.3) approx.. Thus for the first A * b = 3.5; since A = 1 we have b = 3.5. For the second we similarly conclude that b = 7.3. So the functions are y = 3.5^x and y = 7.3^x, approx.. **
......!!!!!!!!...................................
RESPONSE --> ok i still dont completely comprehend this one self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:23:13 At what points will each of the logarithmic functions reach the values y = 2, y = 3 and y = 4?
......!!!!!!!!...................................
RESPONSE --> (.69,2) (1.09,3) (1.386,4) I took the natural log of these to get the answer confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:24:20 ** Using trial and error we find that for y = 3.5^x, the x values .55, .88 and 1.11 give us y = 2, 3 and 4. We find that for y = 7.3^x, the x values .35, .55 and .70 give us y = 2, 3 and 4. **
......!!!!!!!!...................................
RESPONSE --> I thought that i was doing something wrong but now that I see the answers I figured out the right way to do it self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:25:29 7. What is the decibel level of a sound which is 10,000 times as loud as hearing threshold intensity?
......!!!!!!!!...................................
RESPONSE --> 4 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:25:52 dB = 10 log(I / I0). If the sound is 10,000 times as loud as hearing threshold intensity then I / I0 = 10,000. log(10,000) = 4, since 10^4 = 10,000. So dB = 10 log(I / I0) = 10 * 4 = 40.
......!!!!!!!!...................................
RESPONSE --> I forgot to multiply by ten self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:26:40 What are the decibel levels of sounds which are 100, 10,000,000 and 1,000,000,000 times louder than threshold intensity?
......!!!!!!!!...................................
RESPONSE --> 20, 70, and 90 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:26:52 10 log(100) = 10 * 2 = 20, so a sound which is 100 times hearing threshold intensity is a 20 dB sound. 10 log(10,000,000) = 10 * 7 = 70, so a sound which is 100 times hearing threshold intensity is a 70 dB sound. 10 log(1,000,000,000) = 10 * 9 = 90, so a sound which is 100 times hearing threshold intensity is a 90 dB sound.
......!!!!!!!!...................................
RESPONSE --> yay i finally understand self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:27:25 how can you easily find these decibel levels without using a calculator?
......!!!!!!!!...................................
RESPONSE --> count the number of zeros after the 1 and then multiply by ten confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:27:36 Since 10^1, 10^2, 10^3, ... are 10, 100, 1000, ..., the power of 10 is the number of zeros in the result. Since the log of a number is the power to which 10 must be raised to get this number, the log of one of these numbers is equal to the number of zeros.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:28:34 What are the decibel levels of sounds which are 500, 30,000,000 and 7,000,000,000 times louder than threshold intensity?
......!!!!!!!!...................................
RESPONSE --> about 27, about 75 and about 99 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:28:42 10 log(500) = 10 * 2.699 = 26.99 so this is a 26.99 dB sound. 10 log(30,000,000) = 10 * 7.477 = 74.77 so this is a 74.77 dB sound. 10 log(7,000,000,000) = 10 * 9.845 = 98.45 so this is a 98.45 dB sound.
......!!!!!!!!...................................
RESPONSE --> i just rounded up self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:28:59 8. If a sound measures 40 decibels, then what is the intensity of the sound, as a multiple of the hearing threshold intensity?
......!!!!!!!!...................................
RESPONSE --> 10000 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:29:07 ** Let x be the ratio I / I0. Then we solve the equation 40= 10*log(x). You solve this by dividing by 10 to get log(x) = 4 then translating this to exponential form x = 10^4 = 10,000. The sound is 10,000 times the hearing threshold intensity, so I = 10,000 I0. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:31:59 Answer the same question for sounds measuring 20, 50, 80 and 100 decibels.
......!!!!!!!!...................................
RESPONSE --> 100, 10000, 100000000, 10000000000 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:32:11 ** since dB = 10 log(I / I0) we have log(I/I0) = dB / 10. Translating to exponential form this tells us that I / I0 = 10^(dB/10) wo that I = I0 * 10^(dB/10). For a 20 dB sound this gives us I = I0 * 10^(20/10) = I0 * 10^2 = 100 I0. The sound is 100 times the intensity of the hearing threshold sound. For a 50 dB sound this gives us I = I0 * 10^(50/10) = I0 * 10^5 = 100,000 I0. The sound is 100,000 times the intensity of the hearing threshold sound. For an 80 dB sound this gives us }I = I0 * 10^(80/10) = I0 * 10^8 = 100,000,000 I0. The sound is 100,000,000 times the intensity of the hearing threshold sound. For a 100 dB sound this gives us I = I0 * 10^(100/10) = I0 * 10^10 = 10,000,000,000 I0. The sound is 10,000,000,000 times the intensity of the hearing threshold sound. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:34:22 What equation you would solve to find the intensity for decibel levels of 35, 83 and 117 dB.
......!!!!!!!!...................................
RESPONSE --> I have an idea how to solve these but i cannot get a good answer and im kindof stuck confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:35:23 ** the equation to find I is dB = 10 log(I / I0) so the equations would be 35 = 10 log(I / I0) 83 = 10 log(I / I0) 117 = 10 log(I / I0). The solution for I in the equation dB = 10 log(I / I0) is I = I0 * 10^(dB/10). For the given values we would get solutions 10^(35/10) I0 = 3162.3 I0 10^(83/10) I0 = 199526231.5 I0 10^(117/10) I0 = 501187233627 I0 **
......!!!!!!!!...................................
RESPONSE --> oh ok this is actually really simple now that i see it layed out like this self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:36:19 9. is log(x^y) = x log(y) valid? If so why, and if not why not?
......!!!!!!!!...................................
RESPONSE --> no because the log(x^y) is not correct confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:36:32 ** log(a^b) = b log a so log(x^y) should be y log (x), not x log(y). **
......!!!!!!!!...................................
RESPONSE --> oh self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:36:51 is log(x/y) = log(x) - log(y) valid. If so why, and if not why not?
......!!!!!!!!...................................
RESPONSE --> yes it is because its in the right order confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:36:57 Yes , this is valid. It is the inverse of the exponential law a^x / a^y = a^(x-y).
......!!!!!!!!...................................
RESPONSE --> yay self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:37:18 is log (x * y) = log(x) * log(y) valid. If so why, and if not why not?
......!!!!!!!!...................................
RESPONSE --> yes either way could be valid really confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:37:30 No. log(x * y) = log(x) + log(y)
......!!!!!!!!...................................
RESPONSE --> oh geeze that was tricky self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:38:02 is 2 log(x) = log(2x) valid. If so why, and if not why not?
......!!!!!!!!...................................
RESPONSE --> no becaue you are doing a different order of operations it would change the answer confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:38:12 ** log(a^b) = b log a so 2 log(x) = log(x^2), not log(2x). **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:38:25 is log(x + y) = log(x) + log(y) valid. If so why, and if not why not?
......!!!!!!!!...................................
RESPONSE --> yes it is valid confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:38:45 ** log(x) + log(y) = log(xy), not log(x+y). **
......!!!!!!!!...................................
RESPONSE --> gosh i hate the multiplying and adding ones i need to review this self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:38:55 is log(x) + log(y) = log(xy) valid. If so why, and if not why not?
......!!!!!!!!...................................
RESPONSE --> yes it is confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:39:09 This is value. It is inverse to the law of exponents a^x*a^y = a^(x+y)
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:39:21 is log(x^y) = (log(x)) ^ y valid. If so why, and if not why not?
......!!!!!!!!...................................
RESPONSE --> no confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:39:28 No. log(x^y) = y log (x). This is the invers of the law (x^a)^b = x^(ab)
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:40:07 No. log(x-y) = log x/ log y
......!!!!!!!!...................................
RESPONSE --> i accidentally hit the button but no its not valid because it should be division self critique assessment:
.................................................
......!!!!!!!!...................................
20:40:17 is 3 log(x) = log(x^3) valid. If so why, and if not why not?
......!!!!!!!!...................................
RESPONSE --> yes confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:40:25 Yes. log(x^a) = a log(x).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:40:46 is log(x^y) = y + log(x) valid. If so why, and if not why not?
......!!!!!!!!...................................
RESPONSE --> no because you have to take the log of both confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:40:53 No. log(x^y) = y log(x).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:41:09 is log(x/y) = log(x) / log(y) valid. If so why, and if not why not?
......!!!!!!!!...................................
RESPONSE --> no it should be subtraction confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:41:14 No. log(x/y) = log(x) - log(y).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:41:31 is log(x^y) = y log(x) valid. If so why, and if not why not?
......!!!!!!!!...................................
RESPONSE --> yes confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:41:38 This is valid.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:42:19 10. what do you get when you simplify log {base 8} (1024)? If it can be evaluated exactly, what is the result and how did you get it?
......!!!!!!!!...................................
RESPONSE --> 3.333 i put it into my calculator confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:42:57 COMMON ERROR: log {base 8} (1024) = Log (1024) / Log (8) = 3.33333 EXPLANATION: log {base 8} (1024) = Log (1024) / Log (8) is correct, but 3.33333 is not an exact answer. log {base 8 } (1024) = log {base 8 } (2^10). Since 8 = 2^3, 2^10 = 2^(3 * 10/3) = (2^3)^(10/3) = 8^(10/3). Thus log {base 8} 1024 = log{base 8} 8^(10/3) = 10/3. Note that 10/3 is not exactly equal to 3.33333. You need to give exact answers where possible.
......!!!!!!!!...................................
RESPONSE --> wow this is confusing self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:44:06 what do you get when you simplify log {base 2} (4 * 32)? If it can be evaluated exactly, what is the result and how did you get it?
......!!!!!!!!...................................
RESPONSE --> 7 i think this one can be solved exactly because its a base of 2 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:44:15 ** log{base 2}(4*32) = log{bse 2}(2^2 * 2^5) = log{base 2}(2^7) = 7. **
......!!!!!!!!...................................
RESPONSE --> alright self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:45:26 what do you get when you simplify log (1000)? If it can be evaluated exactly, what is the result and how did you get it?
......!!!!!!!!...................................
RESPONSE --> would it be log (10)- log(3)???which is .5 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:45:40 Since 10^3 = 1000, we have log (1000) = 3
......!!!!!!!!...................................
RESPONSE --> ohhh i get it now okk self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:46:25 what do you get when you simplify ln(3xy)? If it can be evaluated exactly, what is the result and how did you get it?
......!!!!!!!!...................................
RESPONSE --> 10^3xy confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:46:35 ln(3xy) = ln (3) + ln(x) + ln(y) = 1.0986 + ln(x) + ln(y)
......!!!!!!!!...................................
RESPONSE --> oh ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:46:57 what do you get when you simplify log(3) + log(7) + log(41)? If it can be evaluated exactly, what is the result and how did you get it?
......!!!!!!!!...................................
RESPONSE --> 2.93 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:47:24 log(3) + log(7) + log(41) = log (3*7*41). 3 * 7 * 41 is not a rational-number power of 10 so this can't be evaluated exactly.
......!!!!!!!!...................................
RESPONSE --> that was what i thought was that it could not be evaluated exactly self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:47:51 11. Show how you used the given values to find the logarithm of 12. Explain why the given values don't help much if you want the log of 17.
......!!!!!!!!...................................
RESPONSE --> what given values confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:48:00 ** The problem was as follows: Given that log(2) = .301, log(3) = .477, log(5) = .699, log(7) = .845, find log(4), log(6), log(8), and log(9). Use the given values to find the logarithm of every possible integer between 11 and 20. To get log(12), given the logs of 2, 3, 5 and 7, you have to break 12 down into a product of these numbers. Since 12 = 2 * 2 * 3 we have log(12) = log(2) + log(2) + log(3) = .301 + .301 + .477 = 1.079. Your calculator will confirm this result. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:49:25 12. What do you get when you solve 3 ^ (2x) = 7 ^ (x-4), and how did you solve the equation?
......!!!!!!!!...................................
RESPONSE --> take the log of each side and get log(base3)2x=log(base7)x-4 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:49:41 ** log[3^(2x)]= log [7^(x-4)]. Using the laws of logarithms we get 2xlog(3)= (x-4) log(7). The distributive law gives us 2xlog(3)= xlog(7)- 4log(7). Rearranging to get all x terms on one side we get 2xlog(3)- xlog(7)= -4log(7). Factor x out of the left-hand side to get x ( 2 log(3) - log(7) ) = -4 log(7) so that x = -4 log(7) / [ 2 log(3) - log(7) ]. Evaluating this we get x = -31, approx. **
......!!!!!!!!...................................
RESPONSE --> i was lost on solving that one self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:51:29 What do you get when you solve 2^(3x) + 2^(4x) = 9, and how did you solve the equation?
......!!!!!!!!...................................
RESPONSE --> i got 4.5 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:51:56 COMMON ERROR: 3xlog(2) + 4xlog(2) = 9 Explanation: Your equation would require that log( 2^(3x) + 2^(4x) ) = log(2^(3x) ) + log(2^(4x)). This isn't the case. log(a + b) is not equal to log(a) + log(b). log(a) + log(b) = log(a * b), not log(a + b). If this step was valid you would have a good solution. However it turns out that this equation cannot be solved exactly for x. The best we can do is certain sophisticated forms of trial and error. **
......!!!!!!!!...................................
RESPONSE --> i dont think i quite did it this way self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:52:40 What do you get when you solve 3^(2x-1) * 3^(3x+2) = 12, and how did you solve the equation?
......!!!!!!!!...................................
RESPONSE --> i need to see another example confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:53:19 ** 3^(2x-1) * 3^(3x+2) = 12. Take log of both sides: log{3} [3^(2x-1) * 3^(3x+2)] = log{3} 12. Use log(a*b) = log(a) + log(b): log{3}(3^(2x-1)) + log{3}(3^(3x+2) = log{3} 12. Use laws of logs: (2x-1) + (3x+2) = log{3} 12. Rearrange the left-hand side: 5x + 1 = log{3}12. Subtract 1 from both sides then divide both sides by 5: x = (log {3}(12) -1)/ 5. Evaluate using calculator: x = .2524 **
......!!!!!!!!...................................
RESPONSE --> oh that was alot easier than i tried to make it self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:01:21 query fitting exponential functions to data 1. what is the exponential function of form A (2^(k1 t) ) such that the graph passes thru points (-4,3) and (7,2), and what equations did you solve to obtain your result?
......!!!!!!!!...................................
RESPONSE --> 3=A*(2^k1*-4) 2=A*(2^k1*7) k=0.05 and then A=1.55 confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:02:08 ** Substituting data points into the form y = A * 2^(kx) we get 3= A * 2^(-4k) and 2= A * 2^(7k) Dividing the first equation by the second we get 1.5= 2^(-4k)/ 2^(7k)= 2^(-4-7k)= 2^(-11k) so that log(2^(-11k)) = log(1.5) and -11 k * log(2) = log 1.5 so that k= log(1.5) / (-11log(2)). Evaluating with a calculator: k= -.053 From the first equation A = 3 / (2 ^(-4k) ). Substituting k = -.053 we get A= 3/ 1.158 = 2.591. So our form y = A * 2^(kx) gives us y= 2.591(2^-.053t). **
......!!!!!!!!...................................
RESPONSE --> i messed up on math somewhere and forgot a neegative but i do understand the overall concept of solving these self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:07:38 what is the exponential function of form A e^(k2 t) such that the graph passes thru points thru points (-4,3) and (7,2) and how did you solve the equations to find this function?
......!!!!!!!!...................................
RESPONSE --> I solved them the same way i solved the other one except i used the natural log y=1.1 e^(-.0368t) confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:08:08 ** Substituting data points into the form y = A * e^(kx) we get 3= A * e^(-4k) and 2= A * e^(7k) Dividing the first equation by the second we get 1.5= e^(-4k)/ e^(7k)= e^(-4-7k)= 2^(-11k) so that ln(e^(-11k)) = ln(1.5) and -11 k = 1.5 so that k= ln(1.5) / (-11). Evaluating with a calculator: k= -.037 approx. From the first equation A = 3 / (e ^(-4k) ). Substituting k = -.037 we get A= 3/ 1.158 = 2.591. So our form y = A * e^(kx) gives us y= 2.591(e^-.039 t). **
......!!!!!!!!...................................
RESPONSE --> my answer was close to thsi i dont know where i went wrong self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:11:52 what is the exponential function of form A b^t such that the graph passes thru points thru points (-4,3) and (7,2).
......!!!!!!!!...................................
RESPONSE --> i cannot get a good starting point for this one confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:14:18 ** Our equations are 3= Ab^-4 2= Ab^7 3/2= Ab^-4/Ab^7 1.5= b^-11 b= .96 3= A * .96 ^ -4 3= A * 1.177 2.549= A y= 2.549 * .96^t **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:21:05 2. Find the exponential function corresponding to the points (5,3) and (10,2).
......!!!!!!!!...................................
RESPONSE --> y=96(0.5^t) confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:21:51 ** Using y = A b^t we get equations 3= Ab^5 2= Ab^10 Dividing first by second: 3/2= Ab^5/(Ab^10.) 1.5= b^-5 b= .922 Now A = 3 / b^5 = 3 / .922^5 = 4.5. Our model is y = 4.5 * .922^t. **
......!!!!!!!!...................................
RESPONSE --> I really dont understand how to get from 1.5=Ab^-5 to .922 self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:22:37 What are k1 and k2 such that b = e^k2 = 2^k1?
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:22:51 ** .922 = e^k2 is directly solved by taking the natural log of both sides to get k2 = ln(.922) = -.081. .922= 2^k1 is solved as follows: log(.922) = log(2) k1 k1 = log(.922) / log(2) = -.117 approx.. Using these values for k1 and k2 we get }g(x) = A * 2^(k1 t) = 4.5 * 2^(-.117 t) and h(x) = A e^(k2 t) = 4.5 e^(-.081 t). ****
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:23:33 3. earthquakes measure R1 = 7.4 and R2 = 8.2. What is the ratio I2 / I1 of intensity and how did you find it?
......!!!!!!!!...................................
RESPONSE --> 1.1 confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:23:44 ** R1 = log(I1 / I0) and R2 = log(I2 / I0) so I1/I0 = 10^R1 and I1 = 10^R1 * I0 and I2/I0 = 10^R2 and I2 = 10^R2 * I0 so I2 / I1 = (I0 * 10^R2) / (I0 * 10^R1) = 10^R2 / 10^R1 = 10^(R2-R1). So if R2 = 8.2 and R1 = 7.4 we have I2 / I1 = 10^(R2 - R1) = 10^(8.2 - 7.4) = 10^.8 = 6.3 approx. An earthquake with R = 8.2 is about 6.3 times as intense as an earthquake with R = 7.4. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:24:28 4. I2 / I1 ratios If one earthquake as an R value 1.6 higher than another, what is the ratio I2 / I1?
......!!!!!!!!...................................
RESPONSE --> 9.6/8 confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:24:36 ** As before I2 / I1 = 10^(R2-R1). If R2 is 1.6 greater than R1 we have R2 - R1 = 1.6 and I2 / I1 = 10^1.6 = 40 approx. An earthquake with R value 1.6 higher than another is 40 times as intense. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:24:42 If one earthquake as an R value `dR higher than another, what is the ratio I2 / I1?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
21:24:46 ** As before I2 / I1 = 10^(R2-R1). If R2 is `dR greater than R1 we have R2 - R1 = `dR and I2 / I1 = 10^`dR. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................