Assignment 18

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22:17:57 query Logarithms, Logarithmic Functions, Logarithmic Equations 1. For what value of x will the function y = log{base 2}(x) first reach 4

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RESPONSE --> When the function y = log{base 2}(x) first reaches 4, the value of x will be 2. ??

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22:18:50 ** 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. **

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RESPONSE --> I'm not sure if I did it correct or not. ??

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22:23:07 for what value of x will the function y = ln(x) first reach y = 4?

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RESPONSE --> The value of x will be approx. 55 when the function y = ln(x) first reaches y = 4 ??

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22:23:54 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. **

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RESPONSE --> I was almost correct. i said 55 and the answer is 54.6

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22:29:03 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

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RESPONSE --> the negative y axis is an asymptote for a log{base b}(x) function because the distance to the curve of a log{base b}(x) function tends to zero, therefore it wouldn't work for negative numbers on the y-axis. If b > 1, then it will work because it won't be a negative. ??

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22:30:23 ** 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. **

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RESPONSE --> i don't think i was correct, but i think i understand what you're trying to say ??

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22:34:48 5. What are your estimates for the values of b for the two exponential functions on the given graph?

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RESPONSE --> my estimates for the values of b for the two exponential functions on the graph are b = 10 and b = .954 ???

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22:35:53 ** 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.. **

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RESPONSE --> i wasn't correct, but i think i understand how you solved for b

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22:41:15 At what points will each of the logarithmic functions reach the values y = 2, y = 3 and y = 4?

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RESPONSE --> the logarithmic functions will reach the values of y = 2, y = 3, and y = 4 at (7.4, 2), (20, 3) and (55, 4) ?? I'm not sure how to solve this one...

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22:42:04 ** 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. **

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RESPONSE --> I was almost correct, although i'm still a little uncertain how to solve it

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22:47:49 7. What is the decibel level of a sound which is 10,000 times as loud as hearing threshold intensity?

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RESPONSE --> the decibel level of a sound which is 10,000 times as loud as hearing threshold intensity is 40 dB ??

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22:48:31 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.

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RESPONSE --> i was right! woo-hoo!

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22:50:20 What are the decibel levels of sounds which are 100, 10,000,000 and 1,000,000,000 times louder than threshold intensity?

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RESPONSE --> the decibel levels of sounds which are 100, 10,000,000 and 1,000,000,000 times louder than threshold intensity is 100 = 20 dB 10,000,000 = 70 dB 1,000,000,000 = 90dB ??

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22:50:54 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.

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RESPONSE --> I was correct...i think i'm getting the hang of this decibel stuff

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22:54:36 how can you easily find these decibel levels without using a calculator?

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RESPONSE --> since dB = 10 log(I / I0) and 1 to 1,000,000,000,000 threshold intensity is the same as saying 0 to 120 dB, then you could just count the number of zeros in the threshold intensity and make it into the hundreths. For example.... 10,000,000,000 has 10 zeros, therefore the dB scale would be 100 dB ??

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22:55:25 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.

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RESPONSE --> i was correct, but i didn't explain it as well as you did....sorry about that. i knew what i wanted to say, but it didn't come out the way i wanted it to.

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23:04:32 What are the decibel levels of sounds which are 500, 30,000,000 and 7,000,000,000 times louder than threshold intensity?

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RESPONSE --> the decibel levels of sound for 500 = 80 dB 30,000,000 = 140 dB 7,000,000,000 = 97 dB ???

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23:05:59 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.

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RESPONSE --> i was close, but not close enough. i understand how to solve for the dB sounds now.

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23:07:16 8. If a sound measures 40 decibels, then what is the intensity of the sound, as a multiple of the hearing threshold intensity?

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RESPONSE --> if a sound measures 40 decibels, then the intensity of the sound as a multiple of the hearing threshold intensity would be 10,000 ??

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23:07:42 ** 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. **

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RESPONSE --> i was correct

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23:08:58 Answer the same question for sounds measuring 20, 50, 80 and 100 decibels.

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RESPONSE --> 20 dB = 100 50 dB = 100,000 80 dB = 100,000,000 100 dB = 10,000,000,000

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23:09:30 ** 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. **

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RESPONSE --> i was correct

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23:17:46 What equation you would solve to find the intensity for decibel levels of 35, 83 and 117 dB.

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RESPONSE --> i'm not sure how to solve this one...i'm a little confused. ???

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23:19:23 ** 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 **

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RESPONSE --> i think i understand how to solve the equation now and what it is.

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23:22:28 9. is log(x^y) = x log(y) valid? If so why, and if not why not?

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RESPONSE --> log(x^y) = x log(y) isn't valid because i think that the first part of the equation is saying that log * x to the y power and the second part of the equation is saying x * log * y ???

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23:23:34 ** log(a^b) = b log a so log(x^y) should be y log (x), not x log(y). **

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RESPONSE --> i was correct in that it wasn't valid, but my reasoning behind it was not. i understand how it's invalid now

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23:25:33 is log(x/y) = log(x) - log(y) valid. If so why, and if not why not?

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RESPONSE --> no, i don't think it's valid, because to undo division, you multiply, not subtract. so log(x/y) should equal log(x) * log(y) ???

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23:27:10 Yes , this is valid. It is the inverse of the exponential law a^x / a^y = a^(x-y).

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RESPONSE --> i was wrong, because i thought it was invalid...how does the subtraction sign in log(x) - log(y) come into play? why wouldn't it be multiplication?

The log is inverse of the exponential function, and b^x / (b^y) = b^(x-y). That's how subtraction comes into it.

However the reasoning can be confusing, so you should be working from the laws of logarithms. One of the laws of logarithms states that log(x / y) = log(x) - log(y).

The laws of logarithms are stated in the worksheet, and you need to memorize them, just as you have memorized the laws of exponents.

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23:28:49 is log (x * y) = log(x) * log(y) valid. If so why, and if not why not?

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RESPONSE --> yes, i think it would be valid, because when you use the distributive property, it works out... ??

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23:29:44 No. log(x * y) = log(x) + log(y)

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RESPONSE --> i must be getting these all mixed up...i need to get on the right page.... ??

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23:30:20 is 2 log(x) = log(2x) valid. If so why, and if not why not?

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RESPONSE --> yes, it's valid

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23:31:12 ** log(a^b) = b log a so 2 log(x) = log(x^2), not log(2x). **

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RESPONSE --> i was incorrect, but i see how in would be invalid. i understand the concept.

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23:32:11 is log(x + y) = log(x) + log(y) valid. If so why, and if not why not?

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RESPONSE --> No, because log(x * y) = log(x) + log(y)

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23:32:50 ** log(x) + log(y) = log(xy), not log(x+y). **

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RESPONSE --> i was correct....i think

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23:34:08 is log(x) + log(y) = log(xy) valid. If so why, and if not why not?

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RESPONSE --> yes, it is valid because log(x) + log(y) = log(xy)

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23:35:02 This is value. It is inverse to the law of exponents a^x*a^y = a^(x+y)

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RESPONSE --> i was correct in that it was valid

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23:36:47 is log(x^y) = (log(x)) ^ y valid. If so why, and if not why not?

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RESPONSE --> no, because log(x^y) = y log x

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23:37:12 No. log(x^y) = y log (x). This is the invers of the law (x^a)^b = x^(ab)

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RESPONSE --> i was correct

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23:38:46 is log(x - y) = log(x) - log(y) valid. If so why, and if not why not?

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RESPONSE --> no, because log(x/y) = log(x) - log(y), not log(x - y)

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23:39:04 No. log(x-y) = log x/ log y

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RESPONSE --> i was correct

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23:39:44 is 3 log(x) = log(x^3) valid. If so why, and if not why not?

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RESPONSE --> yes, it is valid

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23:39:58 Yes. log(x^a) = a log(x).

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RESPONSE --> i was correct

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23:40:54 is log(x^y) = y + log(x) valid. If so why, and if not why not?

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RESPONSE --> yes, because it is the inverse of the law (x^a)^b = x^(ab)

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23:42:19 No. log(x^y) = y log(x).

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RESPONSE --> i see what i did wrong...i didn't see the + sign between y and log(x) i thought it read log(x^y) = ylog(x), in which case it would've been correct.

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23:43:27 is log(x/y) = log(x) / log(y) valid. If so why, and if not why not?

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RESPONSE --> no, it should be log(x/y) = log(x) - log(y)

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23:43:42 No. log(x/y) = log(x) - log(y).

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RESPONSE --> i was correct

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23:44:10 is log(x^y) = y log(x) valid. If so why, and if not why not?

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RESPONSE --> yes

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23:44:27 This is valid.

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RESPONSE --> correct

You did very well once you understood that you need to apply the laws of logarithms.

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23:46:35 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?

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RESPONSE --> you would get 8 ^ x = 1024 ???

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23:47:44 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.

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RESPONSE --> i was incorrect, but i think i understand how to solve it now

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23:48:56 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?

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RESPONSE --> i'm still a little confused when it comes to working with log and base...i'm not sure

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23:49:51 ** log{base 2}(4*32) = log{bse 2}(2^2 * 2^5) = log{base 2}(2^7) = 7. **

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RESPONSE --> that helps me to understand it jsut a little, but i'm still confused

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23:50:27 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?

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RESPONSE --> log(1000) = 3

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23:50:43 Since 10^3 = 1000, we have log (1000) = 3

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RESPONSE --> correct

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23:51:43 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?

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RESPONSE --> would ln(3xy) = 1.100xy ??

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23:52:20 ln(3xy) = ln (3) + ln(x) + ln(y) = 1.0986 + ln(x) + ln(y)

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RESPONSE --> i was partially correct. i understand now

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23:53:27 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?

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RESPONSE --> log(3) + log(7) + log(41) = 2.93 ??

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23:54:25 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.

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RESPONSE --> why couldn't you just take the log of 3, 4 and 7 and then add them?

You could, but you would have to use approximations. In any case you have to understand that the sum is also equal to log(3 * 7 * 41).

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23:56:03 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.

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RESPONSE --> i was a little confused on getting the values of 12, therefore i'm not sure how to answer the problem ??

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23:56:48 ** 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. **

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RESPONSE --> i'm starting to better understand the problem

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23:59:47 12. What do you get when you solve 3 ^ (2x) = 7 ^ (x-4), and how did you solve the equation?

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RESPONSE --> i'm not sure exactly how to solve it correctly...

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00:00:49 ** 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. **

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RESPONSE --> i'm starting to understand a little more, but i'm still a little uneasy about solving them

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00:01:18 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. **

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RESPONSE --> i'm not sure...

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00:05:04 What do you get when you solve 3^(2x-1) * 3^(3x+2) = 12, and how did you solve the equation?

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RESPONSE --> can you help explain to me how i can better understand how to solve these and the concepts of log and base ??

I need to know the specific details of what you do and do not understand about the given solutions, and I need to see the details of your attempts at solution.

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00:05:36 ** 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 **

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RESPONSE --> ok

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00:16:50 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?

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RESPONSE --> 1.5 = 2^(-4k) / 2^(3k) = 2^(-4k - 3k) = 2^(-7k) so 2^(-7k) = 1.5 k = -.023

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00:18:11 ** 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). **

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RESPONSE --> i think i got the concept of the equation right and how to work it, but i put in the worng values...

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00:18:56 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?

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RESPONSE --> i'm not sure how to correctly solve the problem

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00:19:09 ** 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). **

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RESPONSE --> ok

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00:20:44 what is the exponential function of form A b^t such that the graph passes thru points thru points (-4,3) and (7,2).

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RESPONSE --> i didn't really understand this while i was working on the exercises.

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00:21:07 ** 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 **

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RESPONSE --> ok

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00:37:32 2. Find the exponential function corresponding to the points (5,3) and (10,2).

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RESPONSE --> i can get through the equation until i have to find b, then i get stuck... 3 = Ab^5 2 = Ab^10 3/2 = Ab^5/Ab^10 1.5 = b^-5 b = ???

Good. You have b^(-5) = 1.5. So

(b^(-5) ) ^ (-1/5) = 1.5 ^(-1/5). Since (b^(-5))^(-1/5) = b^(-5 * (-1/5) ) = b^1 = b, we get

b = 1.5^(-1/5), which you can find using your calculator.

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00:39:03 ** 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= Ab^-5 b= .922 Now A = 3 / b^5 = 3 / .922^5 = 4.5. Our model is y = 4.5 * .922^t. **

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RESPONSE --> i still don't understand how to find b, but i understand the rest of the problem and how to work it.

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00:46:08 What are k1 and k2 such that b = e^k2 = 2^k1?

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RESPONSE --> i had a little trouble understanding this when i was doing the exercises, can you explain it to me?

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00:46:48 ** .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). ****

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RESPONSE --> i think i'm starting to understand it...

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00:56:28 3. earthquakes measure R1 = 7.4 and R2 = 8.2. What is the ratio I2 / I1 of intensity and how did you find it?

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RESPONSE --> .045 ??

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00:57:06 ** 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. **

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RESPONSE --> man, why can't i get the hang of this??? grrrr

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01:26:13 4. I2 / I1 ratios If one earthquake as an R value 1.6 higher than another, what is the ratio I2 / I1?

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RESPONSE --> i'm not sure how to solve it

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01:33:35 ** 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. **

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RESPONSE --> ok, i got that

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01:35:52 If one earthquake as an R value `dR higher than another, what is the ratio I2 / I1?

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RESPONSE --> why is this so hard for me??

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01:36:11 ** As before I2 / I1 = 10^(R2-R1). If R2 is `dR greater than R1 we have R2 - R1 = `dR and I2 / I1 = 10^`dR. **

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RESPONSE --> ok

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"

I suggest that you go back, read the worksheet again and try to work the problems again without reference to these solutions.

Another thing you need to do is attempt every solution, and give me detailed self-critiques telling me exactly what you do and do not understand about every step of each of the given solutions.