#$&* course Mth 158 4/30 9:45p 038. * 38
.............................................
Given Solution: * * log{base 3}(8) * log{base 8}(9) = log 8 / log 3 * log 9 / log 8 = log 9 / log 3 = log{base 3}(9). log{base 3}(9) is the power to which 3 must be raised to get 9, and is therefore equal to 2. Thus log{base 3}(8) * log{base 8}(9) = 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 6.5.30 / 6.5.24 / 7th edition 5.5.24. ln(2) = a, ln(3) = b. What is ln(2/3) in terms of a and b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ln(2/3) = ln(2) - ln(3) = a -b confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * ln(2/3) = ln(2) - ln(3) = a - b. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 6.5.32 / 6.5.26 / 7th edition 5.5.26. ln(0.5) in terms of a and b. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ln (2) =a and ln(1) = 0, so Ln(.5) = ln(1/2) = ln (1) - ln(2) = 0 - a = -a confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * Since ln(2) = a, and since ln(1) = 0, we have ln(.5) = ln(1/2) = ln(1) - ln(2) = 0 - a = -a. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * Extra problem / 6.5.52 / 7th edition 5.5.52. log{base 3}(u^2) - log{base 3}(v) as a single log. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Log{base 3}(u^2 / v) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * log{base 3}(u^2) - log{base 3}(v) = log{base 3}(u^2 / v). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * Extra Problem / 6.5.58 / 7th edition 5.5.68. Using a calculator express log{base1 / 2}(15) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * We get log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I do not have a scientific calculator, so I just took notes and read the solution. ------------------------------------------------ Self-critique Rating:1 ********************************************* Question: * 6.5.88 / 6.5.82 / 7th edition 5.5.80. Express y as a function of x if ln y = ln(x + C). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A = ln means e^a, so the exponential form of y = ln(x + c) is (x + c) = e^y confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):ok ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 158 4/30 9:45p 038. * 38
.............................................
Given Solution: * * log{base 3}(8) * log{base 8}(9) = log 8 / log 3 * log 9 / log 8 = log 9 / log 3 = log{base 3}(9). log{base 3}(9) is the power to which 3 must be raised to get 9, and is therefore equal to 2. Thus log{base 3}(8) * log{base 8}(9) = 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 6.5.30 / 6.5.24 / 7th edition 5.5.24. ln(2) = a, ln(3) = b. What is ln(2/3) in terms of a and b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ln(2/3) = ln(2) - ln(3) = a -b confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * ln(2/3) = ln(2) - ln(3) = a - b. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 6.5.32 / 6.5.26 / 7th edition 5.5.26. ln(0.5) in terms of a and b. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ln (2) =a and ln(1) = 0, so Ln(.5) = ln(1/2) = ln (1) - ln(2) = 0 - a = -a confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * Since ln(2) = a, and since ln(1) = 0, we have ln(.5) = ln(1/2) = ln(1) - ln(2) = 0 - a = -a. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * Extra problem / 6.5.52 / 7th edition 5.5.52. log{base 3}(u^2) - log{base 3}(v) as a single log. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Log{base 3}(u^2 / v) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * log{base 3}(u^2) - log{base 3}(v) = log{base 3}(u^2 / v). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * Extra Problem / 6.5.58 / 7th edition 5.5.68. Using a calculator express log{base1 / 2}(15) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * We get log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I do not have a scientific calculator, so I just took notes and read the solution. ------------------------------------------------ Self-critique Rating:1 ********************************************* Question: * 6.5.88 / 6.5.82 / 7th edition 5.5.80. Express y as a function of x if ln y = ln(x + C). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A = ln means e^a, so the exponential form of y = ln(x + c) is (x + c) = e^y confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):ok ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course Mth 158 4/30 9:45p 038. * 38
.............................................
Given Solution: * * log{base 3}(8) * log{base 8}(9) = log 8 / log 3 * log 9 / log 8 = log 9 / log 3 = log{base 3}(9). log{base 3}(9) is the power to which 3 must be raised to get 9, and is therefore equal to 2. Thus log{base 3}(8) * log{base 8}(9) = 2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 6.5.30 / 6.5.24 / 7th edition 5.5.24. ln(2) = a, ln(3) = b. What is ln(2/3) in terms of a and b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ln(2/3) = ln(2) - ln(3) = a -b confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * ln(2/3) = ln(2) - ln(3) = a - b. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 6.5.32 / 6.5.26 / 7th edition 5.5.26. ln(0.5) in terms of a and b. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ln (2) =a and ln(1) = 0, so Ln(.5) = ln(1/2) = ln (1) - ln(2) = 0 - a = -a confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * Since ln(2) = a, and since ln(1) = 0, we have ln(.5) = ln(1/2) = ln(1) - ln(2) = 0 - a = -a. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * Extra problem / 6.5.52 / 7th edition 5.5.52. log{base 3}(u^2) - log{base 3}(v) as a single log. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Log{base 3}(u^2 / v) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * log{base 3}(u^2) - log{base 3}(v) = log{base 3}(u^2 / v). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * Extra Problem / 6.5.58 / 7th edition 5.5.68. Using a calculator express log{base1 / 2}(15) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: * * We get log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I do not have a scientific calculator, so I just took notes and read the solution. ------------------------------------------------ Self-critique Rating:1 ********************************************* Question: * 6.5.88 / 6.5.82 / 7th edition 5.5.80. Express y as a function of x if ln y = ln(x + C). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A = ln means e^a, so the exponential form of y = ln(x + c) is (x + c) = e^y confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):ok ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: