#$&* course Mth 158 5/6/13 Around 1:30 PM 038. * 38
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Given Solution: * * log{base 3}(8) * log{base 8}(9) = log 8 / log 3 * log 9 / log 8 = log 9 / log 3 = log{base 3}(9). log{base 3}(9) is the power to which 3 must be raised to get 9, and is therefore equal to 2. Thus log{base 3}(8) * log{base 8}(9) = 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 6.5.30 / 6.5.24 / 7th edition 5.5.24. ln(2) = a, ln(3) = b. What is ln(2/3) in terms of a and b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ln(2/3) = ln(2) - ln(3) a - b confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ln(2/3) = ln(2) - ln(3) = a - b. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 6.5.32 / 6.5.26 / 7th edition 5.5.26. ln(0.5) in terms of a and b. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am not sure. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * Since ln(2) = a, and since ln(1) = 0, we have ln(.5) = ln(1/2) = ln(1) - ln(2) = 0 - a = -a. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn't think of turning 0.5 into 1/2. I totally see how it works now. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * Extra problem / 6.5.52 / 7th edition 5.5.52. log{base 3}(u^2) - log{base 3}(v) as a single log. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base 3}(u^2) - log{base 3}(v) log{base 3}(u^2 / v) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * log{base 3}(u^2) - log{base 3}(v) = log{base 3}(u^2 / v). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * Extra Problem / 6.5.58 / 7th edition 5.5.68. Using a calculator express log{base1 / 2}(15) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base1 / 2}(15) = -3.88 approximate. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * We get log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 6.5.88 / 6.5.82 / 7th edition 5.5.80. Express y as a function of x if ln y = ln(x + C). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = x + C confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I'm not sure I understand the given solution. Could you explain please???? ------------------------------------------------ Self-critique Rating: 1