Assign 38

course mth 158

I wish every chapter had been as much fun as this one. I loved working with logarithmic functions (even though I didn't get them all right) it was challenging. This section was more like logic math instead of numbers.

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.

038. * 38

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Question: * 6.5.18 / 7th edition 5.5.18. Exact value of log{base 3}{(8) * log{base 8}(9).

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Your solution:

log{base 3}(8) * log {base 8} (9)

ln8/ln3 * ln9/ln8

= ln9/ln3 = log {base 3} (9)

To find the value of y you take

log {base 3} (9) =

3^y = 3^2

y = 2

log {base 3} (9) = 2

confidence rating: 3

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Given Solution:

* * log{base 3}(8) * log{base 8}(9) = log 8 / log 3 * log 9 / log 8 = log 9 / log 3 = log{base 3}(9).

log{base 3}(9) is the power to which 3 must be raised to get 9, and is therefore equal to 2.

Thus log{base 3}(8) * log{base 8}(9) = 2.

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Self-critique (if necessary):

There are no discrepancies.

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Self-critique Rating: 3

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Question: * 6.5.24 / 7th edition 5.5.24. ln(2) = a, ln(3) = b. What is ln(2/3) in terms of a and b?

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Your solution:

ln(2/3) where 2 = a and 3 = b

ln2^a - ln3^b

= a ln 2 – b ln 3

I’m not sure, I may have that backwards. It may be 2ln(a) – 3ln(b). The book had an example of one like ln x^2/(x-1)^3 where ln x^2 – ln(x – 1)^3 and the answer was 2 ln x – 3 ln (x – 1). I tried to follow along with that example.

confidence rating: 2

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Given Solution:

* * ln(2/3) = ln(2) - ln(3) = a - b.

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Self-critique (if necessary):

Looks like I got both of them wrong. But I do understand the concept.

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Self-critique Rating: 3

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Question: * 6.5.26 / 7th edition 5.5.26. ln(0.5) in terms of a and b.

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Your solution:

This one is a little bit more difficult. It seems a little tricky since there are no 2 or 3 values. The value .5 is ½, but I’m not sure how that works out in this solution either.

confidence rating: 0

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Given Solution:

* * Since ln(2) = a, and since ln(1) = 0, we have

ln(.5) = ln(1/2) = ln(1) - ln(2) = 0 - a = -a.

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Self-critique (if necessary):

Well, I would have never guessed how to solve that one. I see how you did it though. I was thinking that you had to use the 3 = b somewhere in the solution.

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Self-critique Rating: 3

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Question: * 6.5.52 / 7th edition 5.5.52. log{base 3}(u^2) – log{base 3}(v) as a single log.

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Your solution:

2 log{base 3}(U) – log {base 3}(V)

log{base 3} (U)^2 - log{base 3}(V)

= log{base3}(U^2/V)

confidence rating: 3

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Given Solution:

* * log{base 3}(u^2) – log{base 3}(v) = log{base 3}(u^2 / v).

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Self-critique (if necessary):

I think I should have put the U^2 inside my parenthesis, instead I put it (U)^2 although I don’t really see a difference here since there is nothing else inside the ( ) but it would make a difference if it were something like (2 + U)^2

The question that arises is whether

log{base 3} (U)^2 means

log{base 3} (U^2) or

(log{base 3} (U)) ^ 2.

Order of operations is very tricky here unless you are specific with your grouping. It's always best to be clear about the grouping.

It turns out that the order of operations applies the log function before the exponentiation, so log{base 3} (U)^2 would be interpreted as (log{base 3} (U)) ^ 2. This would not give the correct result for the problem.

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Self-critique Rating: 3

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Question: * 6.5.58 / 7th edition 5.5.68. Using a calculator express log{base1 / 2}(15)

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Your solution:

-3.907

confidence rating: 3

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Given Solution:

* * We get log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595.

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Self-critique (if necessary):

I didn’t work it out on paper. I just used my calculator, but the answers are the same. I truncated mine to 3 places or if I had used floor(log {base ½} (15)) it would have given me 3…..just had to throw that in for fun!! I’m really not even sure that’s right, it might have given me 4!

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Self-critique Rating: 3

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Question: * 6.5.82 / 7th edition 5.5.80. Express y as a function of x if ln y = ln(x + C).

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Your solution:

I’m not sure about this one, but I’ll try it like this:

y = ln e(x+c) =

e^y = x + c =

ln e^y = ln(x+c)

you've gone around in a circle; ln(e^y) is just y so this equation is y = ln(x + c), which is the equation you started with.

y ln e = ln(x+c)

y = ln(x+c) / ln e

That’s probably not right.

confidence rating: 0

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Given Solution:

* * a = ln(b) means e^a = b, so y = ln(x+c) is translated to exponential form as

(x+c) = e^y.

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Self-critique (if necessary):

I don’t even know how to evaluate mine from the given solution. Obviously, I got it wrong.

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Self-critique Rating: 0

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