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.
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There are no discrepancies. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * ln(2/3) = ln(2) - ln(3) = a - b. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Looks like I got both of them wrong. But I do understand the concept. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 6.5.26 / 7th edition 5.5.26. ln(0.5) in terms of a and b. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 6.5.52 / 7th edition 5.5.52. log{base 3}(u^2) – log{base 3}(v) as a single log. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * log{base 3}(u^2) – log{base 3}(v) = log{base 3}(u^2 / v). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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
.............................................
Given Solution: * * We get log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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! ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 6.5.82 / 7th edition 5.5.80. Express y as a function of x if ln y = ln(x + C). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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)
.............................................
Given Solution: * * a = ln(b) means e^a = b, so y = ln(x+c) is translated to exponential form as (x+c) = e^y. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t even know how to evaluate mine from the given solution. Obviously, I got it wrong. ------------------------------------------------ Self-critique Rating: 0 "