Assignment - 18

course Mth163

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

018. `query 18

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Question: `qquery 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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Your solution:

to find the x value of this function that reaches 4 you would have to do the inversion of this function:

x = e^4….. x = 55 approx.

Confidence rating: positive

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Given Solution:

`a** 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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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qfor what value of x will the function y = ln(x) first reach y = 4?

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Your solution:

to solve for x you would use the same function as above:

x = e^4…x = 55 approx.

Confidence rating: positive

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Given Solution:

`ay = 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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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `q3. 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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Your solution:

log{base b}(x) is the inverse function of y = b^x. Since x is the negative number and the squaring of b, laws of exponents tells us we have to convert to a fraction giving you a positive exponent, and thus a positive answer. The result will result in an asymptote along the negative axis.

Confidence rating: mostly

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Given Solution:

`a** 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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Self-critique (if necessary):

b>1 = larger neg. powers will make result smaller giving a neg. asymptote of x axis.

b< 1 = larger positive powers making result smaller giving pos. asymptote of x axis.

Self-critique rating:

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Question: `q5. What are your estimates for the values of b for the two exponential functions on the given graph?

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Your solution:

using the function y = A b^x we can see the y intercept and basic point are (0, A) and (1, A*b) We can see that a would be 1.

If x = 1 looking at our graph we have coordinates (1,3.3) and (1, 7.2) To solve for b we have A * b = 3.3…. b = 3.3 and A * b = 7.2…..b = 7.2 so for function y = b^x you would have these functions:

y = 3.3^x and y = 7.2^x

Confidence rating: positive

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Given Solution:

`a** 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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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qAt what points will each of the logarithmic functions reach the values y = 2, y = 3 and y = 4?

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Your solution:

using our functions from above and trying x values until we have a fit we get:

y = 3.3^x… y = 2, 3, 4… x = .58, .919, 1.348

y = 7.2^x…y = 2, 3, 4…x =.35, .56, .7

Confidence rating: positive

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Given Solution:

`a** 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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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `q7. What is the decibel level of a sound which is 10,000 times as loud as hearing threshold intensity?

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Your solution:

Your decibel scale is defined as dB = 10log(I/Io) if it is 10,000 times as loud as threshold intensity then (I/Io) = 10,000 so:

log(10,000) = 4, if you calculate that on your calculator so:

dB = 10 * 4 = 40

Confidence rating: mostly

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Given Solution:

`adB = 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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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qWhat 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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Your solution:

log(100 = 10 * 2 = 20

log(10,000,000) = 10 * 7 = 70

log(1,000,000,000) = 10 * 9 = 90

Confidence rating: positive

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Given Solution:

`a10 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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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qhow can you easily find these decibel levels without using a calculator?

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Your solution:

You just use your powers of 10; say we have 1,000 then you would take 10 ^ how many zeros you have: which is 10^3 or 10 * 3 = 10

Confidence rating: positive

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Given Solution:

`aSince 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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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qWhat 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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Your solution:

you would do your function as before:

10 log(500) = 10 * 2.699 = 26.99

10 log(30,000,000) = 10 * 7.477 = 74.77

10 log(7,000,000,000) = 10 * 9.845 = 98.45

Confidence rating: positive

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Given Solution:

`a10 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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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `q8. 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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Your solution:

you would have 40 = 10 log(I/Io) try to solve for (I/Io)

40 / 10 =log( I/Io)……10 ^4 = (I/Io)……which is 10,000 hearing threshhold

Confidence rating: mostly

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Given Solution:

`a** 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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Self-critique (if necessary):

could use x as I/Io and answer will be I = 10,000 Io

Self-critique rating:

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Question: `qAnswer the same question for sounds measuring 20, 50, 80 and 100 decibels.

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Your solution:

I don’t know using the steps as before when given your decibels my calculator will not function it.

Confidence rating:

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Given Solution:

`a** 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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Self-critique (if necessary):

So we take our equation db= 10 log(I/Io) and solve for I

log(I/Io) = dB/10

I/Io = 10^(dB/10)

I = Io * 10^(dB/10)

using decibels of 20: I = Io * 10^(20/10)…..I = Io * 10^10….I = 100 Io

Self-critique rating:

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Question: `qWhat equation you would solve to find the intensity for decibel levels of 35, 83 and 117 dB.

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Your solution:

using the equation from above: I = Io * 10^(dB/10)

for decibel 35: I = Io * 10^(35/10) ….I = Io * 10 ^ 3.5….I = approx. 3162 Io

for decibel 83: I = Io * 10^(83/10) …I = Io * 10 ^ 8.3 … I = approx. 199526232 Io

for decibel 117: I = Io * 10^(117/10) … I= Io * 10 ^ 11.7 … I = approx. 5.012

Confidence rating: positive

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Given Solution:

`a** 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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Self-critique (if necessary):

???? Mr. Smith, I got everything the same as you did except for the last one and I keep getting 5.011872336 E11, I don’t understand what our differences is> Could you please explain ????

5.011872336 E11 on your calculator means 5.011872336 * 10^11

Actually both answers should be rounded to 2 significant figures, as either 500 000 000 000 I0 or 5.0 * 10^11 I0.

Self-critique rating:

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Question: `q9. is log(x^y) = x log(y) valid? If so why, and if not why not?

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Your solution:

It is not valid because the properties of logarithms says that log(x^y) is equal to y * log(x) , the problem states log is of x to the power of y not log of y to the power of x.

Confidence rating: positive

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Given Solution:

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

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qis log(x/y) = log(x) - log(y) valid. If so why, and if not why not?

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Your solution:

Yes this is valid because just like you exponential rules of x^a / x^b = x(a-b), so are your log. properties so log(x/y) would be equal to log(x) – log(y)

Confidence rating: positive

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Given Solution:

`aYes , this is valid. It is the inverse of the exponential law a^x / a^y = a^(x-y).

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qis log (x * y) = log(x) * log(y) valid. If so why, and if not why not?

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Your solution:

No it is not valid: in your exponential rules if you have x^b * x^a you would have x^(b+a) same with your log. properties: log(x * y) = log(a) + log(b)

Confidence rating: positive

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Given Solution:

`aNo. log(x * y) = log(x) + log(y)

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qis 2 log(x) = log(2x) valid. If so why, and if not why not?

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Your solution:

no because you are saying 2 times the log to the x so you would say log(x^2)

Confidence rating: positive

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Given Solution:

`a** log(a^b) = b log a so 2 log(x) = log(x^2), not log(2x). **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qis log(x + y) = log(x) + log(y) valid. If so why, and if not why not?

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Your solution:

This is valid because it fits with the proper way to distribute your problem so:

log(x+y) is the same as log(x) + log(y)

Confidence rating: positive

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Given Solution:

`a** log(x) + log(y) = log(xy), not log(x+y). **

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Self-critique (if necessary):

ok, I see were I misunderstood the problem and I also remember if you have log(a*b) it is the same as Log(a) + log(b)

Self-critique rating:

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Question: `qis log(x) + log(y) = log(xy) valid. If so why, and if not why not?

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Your solution:

yes because as stated before when you add the two logs together it is the same as multiplying them.

Confidence rating: positive

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Given Solution:

`aThis is value. It is inverse to the law of exponents a^x*a^y = a^(x+y)

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qis log(x^y) = (log(x)) ^ y valid. If so why, and if not why not?

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Your solution:

no because log(x^y) is the same as y * log(x)

Confidence rating: positive

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Given Solution:

`aNo. log(x^y) = y log (x).

This is the invers of the law (x^a)^b = x^(ab)

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qis log(x - y) = log(x) - log(y) valid. If so why, and if not why not?

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Your solution:

no because log(x/y) = log(x) – log(y)

Confidence rating: positive

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Given Solution:

`aNo. log(x-y) = log x/ log y

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qis 3 log(x) = log(x^3) valid. If so why, and if not why not?

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Your solution:

yes because b * log(a) = log(a^b)

Confidence rating: positive

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Given Solution:

`aYes. log(x^a) = a log(x).

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qis log(x^y) = y + log(x) valid. If so why, and if not why not?

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Your solution:

No because log(x^y) is the same as y*log(x)not y + log(x)

Confidence rating: positive

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Given Solution:

`aNo. log(x^y) = y log(x).

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qis log(x/y) = log(x) / log(y) valid. If so why, and if not why not?

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Your solution:

No because log(x/y) is the same as log(x) – log(y) not log(x) / log(y)

Confidence rating: positive

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Given Solution:

`aNo. log(x/y) = log(x) - log(y).

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qis log(x^y) = y log(x) valid. If so why, and if not why not?

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Your solution:

yes because log(a^b) = b * log(a)

Confidence rating: positive

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Given Solution:

`aThis is valid.

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Self-critique (if necessary):

Self-critique rating:

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Question: `q10. 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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Your solution:

log{base8}(1024) = log(1024) / log(8) = 3.333333333

Confidence rating: positive

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Given Solution:

`aCOMMON 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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Self-critique (if necessary):

** Why would that be a common error, my calculator shows the same answer when you take 10/3 ?

Self-critique rating:

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Question: `qwhat 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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Your solution:

log{base 2}(4 * 32)… note that both 4 and 32 are powers of 2.

log{base2}(2^2 +* 2^5)

log{base 2}(2^7) = 7

Confidence rating: positive

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Given Solution:

`a** log{base 2}(4*32) = log{bse 2}(2^2 * 2^5) = log{base 2}(2^7) = 7. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qwhat 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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Your solution:

log(1000)….recalling that log without a base means – 10 logarithm

so 10^3 = 1000 …..log(1000) = 3

Confidence rating: positive

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Given Solution:

`aSince 10^3 = 1000, we have log (1000) = 3

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qwhat 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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Your solution:

ln(3xy) can be rewritten as:

ln(3) + ln(x) + ln(y)…… to simplify it:

1.0986 + ln(x) + ln(y)

Confidence rating: positive

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Given Solution:

`a ln(3xy) = ln (3) + ln(x) + ln(y) = 1.0986 + ln(x) + ln(y)

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qwhat 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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Your solution:

with the properties of logarithm you would get log(3 * 7 * 41)

you cannot solve this because it is not a power of 10

Confidence rating: mostly

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Given Solution:

`a 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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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `q11. 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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Your solution:

Given that log(2) = .301, log(3)= .477. log(5) = .699, log(7) = .845, use the given values to find the logarithm of every possible integer between 11 and 20.

To get the log(12), you will have to find it with the numbers given, 2, 3, 5, 7

You can eliminate 5 and 7 because they don’t multiply into 12, so that leaves 2 and 3:

2 * 2 *3 = 12…so…log(2) + log(2) + log(3) …. .301 + .301 + .477 = 1.079 = log(12)

Confidence rating: positive

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Given Solution:

`a** 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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Self-critique (if necessary):

ok

Self-critique rating:

*********************************************

Question: `q12. What do you get when you solve 3 ^ (2x) = 7 ^ (x-4), and how did you solve the equation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

take log[3^(2x)] = log[7^(x-4)] because of the properties of logarithm you:

2x log(3) = (x-4)log(7)……. 2x log(3) = x log(7) – 4 log(7) .. Need to put x on one side.

2x log(3) – x log(7) = -4 log(7)..factoring out the x you get:

x(2 log(3) – log(7) = -4 log(7)…solving for x we get:

x = -4 log(7) / [2 log(3) – log(7) = approx -31

Confidence rating: mostly

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a ** 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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Self-critique (if necessary):

ok

Self-critique rating:

*********************************************

Question: `qWhat do you get when you solve 2^(3x) + 2^(4x) = 9, and how did you solve the equation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

log[2^(3x *4x)] = log(9)But I don’t know how to solve this from here.

Confidence rating: unsure

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Given Solution:

`aCOMMON 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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Self-critique (if necessary):

ok

Self-critique rating:

*********************************************

Question: `qWhat do you get when you solve 3^(2x-1) * 3^(3x+2) = 12, and how did you solve the equation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

first we take logs of both sides:

log{3}[3^(2x-1) * 3^(3x +2)] = log{3}(12)

log{3}(3^(2x – 1)) + log{3}(3^(3x+2)) = log{3}(12)

(2x-1) + (3x+2) = log{3}(12)

5x + 1 = log{3}(12)

5x = log{3}(12) – 1

x =[ log{3}(12) - 1] / 5

x = .9451 approx.

Confidence rating: mostly

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** 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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Self-critique (if necessary):

???? Mr. Smith will you tell me the order in which I would put that in the calculator TI-83 Plus , I am getting a different answer. I figure it is because I’m not putting it in exactly right. ????

log{base 3}(x) = log(x) / log(3).

So

(log {3}(12) -1)/ 5 means

(log(12) / log(3) - 1) / 5

which evaluates to the given result.

If you don't get that, let me know how you are entering the expression.

Self-critique rating:

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Question: `qquery 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

putting in the data points into our function we get:

3 = A * (2^(-4k) and 2 = A * (2^(7k)…..divide the first equation by the second we get:

1.5 = (2^(-4k)) / (2^(7k)) ….. with the laws of distribution we have:

1.5 = 2^(-4k – 7k)…… 1.5 = 2^(-11k)…now we have:

log(1.5) = log(2^(-11k))…which gives:

log(1.5) = -11k * log(2)…so:

k = log(1.5) / (-11k * log(2))…….k = -.0532

substituting k value into the first equation we get:

3 = A * (2^(-4 * -.0532)……3 = A * (2^(.2128)…..3 = A * 1.159

A = 3 / 1.159….A =2.59

So for the equation A * (2^(kt)) we have 2.59 * (2^(-.0532)

Confidence rating: mostly

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** 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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Self-critique (if necessary):

ok

Self-critique rating:

*********************************************

Question: `qwhat is the exponential function of form A e^(kt) 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

in the form of y = A e^(kt) using the coordinates (-4,3) and ( 7,2)

3 = A * e^(-4k)

2 = A * e^(7k) divide the first problem by the second and get:

1.5 = e^(-4k) / e^(7k)……we can distribute this like:

1.5 = e^(-4k – 7k)…..1.5 = e^(-11)….ln is inversion of e^ so we have:

ln(1.5) = ln(e^(-11k))……ln(1.5) = -11k…. k = ln(1.5) / -11…..k = -.0369

To solve for A we have to rearrange the first function 3 = A * e^(-4k) to A = 3/ (e^(-4k))

Now we plug in k value into A equation and solve for A:

A = 3 / (e^(-4 * -.0369))….A = 3 /1.159……a = 2.589

So in the form A * e^(kt) we get:

2.589 * e^(-.0369)

Confidence rating: mostly

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** 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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Self-critique (if necessary):

ok

Self-critique rating:

*********************************************

Question: `qwhat is the exponential function of form A b^t such that the graph passes thru points thru points

(-4,3) and (7,2).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

for A * b^t we can form this functions from the given points:

3 = A * b^-4….and….2 = A * b^7

dividing the first one by the second one we get:

1.5 = (b^-4) / (b^7)….which is the same as 1.5 = b^(-4 – 7) ,,,,1.5 = b^(-11)

b = 1.5 ^(1/-11)…. b = .964… rearrange first equation in order to solve for A and plug in b:

A = 3 / (b^-4)….a = 3 / (.964^-4)…A = 3 / 1.158…A = 2.59

In the form of A * b^t we have y = 2.59 * .964^t

Confidence rating: positive

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Given Solution:

`a** 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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Self-critique (if necessary):

ok

Self-critique rating:

*********************************************

Question: `q2. Find the exponential function corresponding to the points (5,3) and (10,2).

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Your solution:

We can use the same function as above A * b^t and we have:

3 = A * b^5 and 2 = A * b^10 divide the first by the second:

1.5 = b^(5 – 10)…1.5 = b^-5…b = .922

in A = 3 / (b^5) we have A = 3 / (.922^5)…A = 3 / .666…A = 4.50

giving us the function: y = 4.50 * .922^t

Confidence rating: positive

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** 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. **

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Self-critique (if necessary):

ok

Self-critique rating:

*********************************************

Question: `qWhat are k1 and k2 such that b = e^k2 = 2^k1?

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Your solution:

.922 = e^k2…we can solve by the inversion of e^ which is ln:

ln(.922) = ln(e^kt)…..-.081 which is you k2

so if e^k2 = 2^k1…then….. .922 = 2^k1

we can solve for k1 by taking the log of both sides and putting k1 to one side:

k1 = log(.922) / log(2) = -.117

now we would have these functions:

A * 2^(-.117t) …. and …. A * e^(-.081t)

Confidence rating: mostly

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** .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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Self-critique (if necessary):

ok, but I should have put the value of A into the equation: 4.5

Self-critique rating:

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Question: `q3. 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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Your solution:

Using R = log(I/Io) we will get R1 = log(I/Io) and R2 = log(I/Io)

being the same as base – 10 logs we now can solve to get:

I1/Io = 10^R1…and…I2/Io = 10^R2 we can further solve it to:

I1 = 10^R1 * Io and I2 = 10^ R2 * Io dividing:

I2/ I1 = 10^(R2 – R1)…R2 = 8.2 and R1 = 7.4

I2 / I1 = 10^(8.2 – 7.4) = 6.31 approx.

R2 has a 6.31 intensity more than R1

Confidence rating: mostly

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** 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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Self-critique (if necessary):

ok

Self-critique rating:

*********************************************

Question: `q4. 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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Your solution:

using I2/I1 = 10^(R2 – R1) and the intensity from R1 to R2 is 1.6 then we would have:

I2/I1 = 10^1.6 = 39.81 more as intense than the other R value

Confidence rating: mostly

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** 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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Self-critique (if necessary):

ok, I could have rounded my value up but I was trying to be as accurate as possible.

Self-critique rating:

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Question: `qIf one earthquake as an R value `dR higher than another, what is the ratio I2 / I1?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

if the intensity is `dR then you would use I2/I1 = 10^(R2-R1) = 10^`dR

Confidence rating: positive

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** 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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Self-critique (if necessary):

ok

Self-critique rating:

"

&#This looks good. See my notes. Let me know if you have any questions. &#

This was received on or about July 16. Responses were composed at that time but the work did not get posted. It is being posted on July 21.