115.3 Solutions to some odd numbered problems from section 1.3-1 y 1.3-1 Sketch the graph of y = x + 1 without using a graphing calculator. 2 Solution: Starting with the graph of y = x 2 in black, and take that graph shifted up 1 unit in red: y = x 2 + 1. 1.3-11 Sketch the graph of y = exp(x − 2) = ex−2 without using a graphing calculator. Starting with the graph of y = ex in black, we get that shifted 2 units to the right in red: y = e−x . Solution: y 10 9 8 7 6 5 4 3 2 1 0 -4-3-2-10 1 2 3 4 x 10 9 8 7 6 5 4 3 2 1 0 -4-3-2-10 1 2 3 4 Sa 115.3 Solutions to some odd numbered problems from section 1.3 sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean x 115.3 Solutions to some odd numbered problems from section 1.3-2 Solution: 0.0002 1 -4 10 2 3 4 5 67891 -3 10 2 3 4 5 67891 0.02 -2 10 2 1 3 4 5 67891 -1 10 2 3 4 5 67891 0 10 5 2 1 3 4 5 6789 1 10 Sa 1.3-33 Find the following numbers on a number line that is on a logarithmic scale (base 10): 0.0002,0.02,1,5,50,100,1000, 8000, and 20000. sk 3 4 5 67891 2 10 PAT- ET RIÆ atc h e w ane n 2003 Doug MacLean 50 100 2 DEO sis iversitas Un 1000 2 3 4 5 67891 3 10 8000 20000 2 3 4 5 67891 4 10 2 3 4 5 6789 5 10 115.3 Solutions to some odd numbered problems from section 1.3-3 (x1 , y1 ) = (0, 5), Sa 1.3-43 When log y is graphed as a function of x on log-linear paper, a straight line results. Graph straight lines, each given by two points, on a log-linear plot, and determine the functional relationship. sk Solution: Since the graph is a straight line on log-linear graph paper, we have Y = mx + b for constants m and b which we must find. At (x1 , y1 ) = (0, 5) we have log 5 = m(0) + b, so b = log 5. have the equation Y = − log 5 3 x y = 10Y = 10log y = 10− log 5 −0.23 and we 3 + log 5. Exponentiating, we get log 5 3 x+log 5 log 5 x = 10log 5 × 10− 3 5 × (0.58)x Y=log y 9 8 7 6 5 4 3 2 1 0 1 2 3 4 PAT- ET RIÆ atc h e w ane n 2003 Doug MacLean (x2 , y2 ) = (3, 1) At (x2 , y2 ) = (3, 1) we have log 1 = m(3) + b, so 0 = 3m + log 5, and thus m = − DEO sis iversitas Un x 115.3 Solutions to some odd numbered problems from section 1.3-4 (x1 , y1 ) = (−2, 3), Sa 1.3-45 When log y is graphed as a function of x on log-linear paper, a straight line results. Graph straight lines, each given by two points, on a log-linear plot, and determine the functional relationship. sk (x2 , y2 ) = (1, 1) Since the graph is a straight line on log-linear graph paper, we have Y = mx + b for constants m and b which we must find. At (x1 , y1 ) = (−2, 3) we have log 3 = m(−2) + b, and at (x2 , y2 ) = (1, 1) we have log 1 = m(1) + b, or 0 = m + b, so b = −m. Substituting this in the first log 3 log 3 equation, we get log 3 = −2m − m = −3m, so m = − and b = 3 3 log 3 log 3 and we have the equation Y = − x+ . 3 3 Exponentiating, we get 1 x log 3 log 3 log 3 log 3 x 1 y = 10Y = 10log y = 10− 3 x+ 3 = 10 3 × 10− 3 = 3 3 × 3− 3 Y=log y 9 8 7 6 5 4 3 2 1 -1 0 1 2 PAT- ET RIÆ atc h e w ane n 2003 Doug MacLean Solution: -2 DEO sis iversitas Un 3 x 115.3 Solutions to some odd numbered problems from section 1.3-5 y = 2 × 10−2x . Solution: Sa 1.3-47 Use a logarithmic transformation to find a linear relationship between the given quantities and graph the resulting linear relationship on a log-linear plot: sk We have Y = log(4 × 10 ) = log 4 + 5x 9 8 7 6 5 4 3 2 1 -1 0 PAT- ET RIÆ atc h e w ane n 2003 Doug MacLean Y=log y 5x DEO sis iversitas Un 1 x 115.3 Solutions to some odd numbered problems from section 1.3-6 Solution: We have Y = log y = mlogx + b, so when (x1 , y1 ) = (1, 2), we have log 2 = m log 1 + b = m(0) + b = b, so b = log 2, and when (x2 , y2 ) = (5, 1), we have log 1 = m log 5 + log 2, so 0 = m log 5 + log 2 and thus m = − so we have the equation Y = − log 2 X + log 2. log 5 Exponentiating, we have 10Y = 10 + log y = y = 10 log 2 − log 5 X+log 2 = 10log 2 10 log 2 − log 5 log x = 2×x log 2 − log 5 . y 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 log 2 , log 5 Sa 1.3-55 When log y is graphed as a function of log x a straight line results. Graph straight lines, each given by two points on a log-log plot, and determine the functional relationship. (The original x-y coordinates are given.) (x1 , y1 ) = (1, 2), (x2 , y2 ) = (5, 1) sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean 115.3 Solutions to some odd numbered problems from section 1.3-7 y = 2x 5 Solution: Y = log y = log(2x 5 ) = log 2 + 5 log x = log 2 + 5X, where X = log x. y 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 Sa 1.3-59 Use a logarithmic transformation to find a linear relationship between the given quantities and graph the resulting linear relationships on a log-log plot. sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean 115.3 Solutions to some odd numbered problems from section 1.3-8 f (x) = 3x 1.7 Solution: F = log f (x) = log 3x 1.7 = log 3 + log x 1.7 = log 3 + 1.7 log x = log 3 + 1.7X which we plot on log-log paper: y 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 Sa 1.3-67 Use a logarithmic transformation to find a linear relationship between the given quantities and determine whether a log-log or a log-linear plot should be used to graph the resulting relationship sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean 115.3 Solutions to some odd numbered problems from section 1.3-9 N(t) = 130 × 21.2t Solution: log N(t) = log 130 × 21.2t = log 130 + log 21.2t = log 130 + (1.2 log 2)t which we plot on log-linear paper: log L 900 800 700 600 500 400 300 200 -1 100 0 1 c Sa 1.3-69 Use a logarithmic transformation to find a linear relationship between the given quantities and determine whether a log-log or a log-linear plot should be used to graph the resulting relationship sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean 115.3 Solutions to some odd numbered problems from section 1.3-10 x and y x 1 2 4 10 20 y 1.8 2.07 2.38 2.85 3.28 Solution: Sa 1.3-73 The following table is based on a functional relationship between x and y, which is either an exponential or a power function. Use an appropriate logarithmic transformation and a graph to decide whether the table comes from a power function or an exponential function and find the functional relationship between sk log-log paper: Y=log y 9 8 7 6 5 4 3 9 8 7 6 5 4 3 2 2 x 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 1 2 3 4 5 67891 2 3 4 5 678910 We conclude that the data displayed on loglog paper is closest to being a straight line, so we take a line Y = mX + b, or log y = m log x + b, through the points corresponding to (1,1.8) and (20,3.28): At (1,1.8) we have Y = log y = log 1.8 = m log 1 + b = m(0) + b = b, so b = log 1.8. log 3.28 log 3.28 − log 1.8 1.8 At (20,3.28) we have Y = log 3.28 = m log 20 + b = m log 20 + log 1.8, so m = = . log 20 log 20 Thus we have Y = log 3.28 1.8 log 20 X + log 1.8 0.2X + log 1.8 Exponentiating, we get 10Y = y = 10 log 3.28 1.8 log 20 X+log 1.8 = 10log 1.8 10 log 3.28 1.8 log 20 X = 1.8 × 10X 3.28 1.8 log 20 log 1.8 × x 0.2 PAT- ET RIÆ atc h e w ane n 2003 Doug MacLean First we graph the data on semi-log and Y=log y DEO sis iversitas Un X=log x 115.3 Solutions to some odd numbered problems from section 1.3-11 x and y x −1 −0.5 0 0.5 1 y 0.398 1.26 4 12.68 40.18 Solution: We only graph the data on semi-log, because negative values cannot be represented on log-log paper: Y 9 8 7 6 5 4 3 2 1 9 8 7 6 5 4 3 2 -2.0 -1.5 -1.0 -0.5 1 9 8 7 6 5 4 3 2 1 0.0 0.5 1.0 1.5 2.0 Sa 1.3-75 The following table is based on a functional relationship between x and y, which is either an exponential or a power function. Use an appropriate logarithmic transformation and a graph to decide whether the table comes from a power function or an exponential function and find the functional relationship between sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean 115.3 Solutions to some odd numbered problems from section 1.3-12 At (-1,0.398) we have Y = log y = log 0.398 = m(−1) + b = −m + b, and at (1,40.18) we have Y = log 40.18 = m(1) + b = m + b, so adding the two equations, we get 2b = log 0.398 + log 40.18 = log(0.398 × 40.18) = log 15.99, and subtracting the first equationfrom the second gives 40.18 2m = log 40.18 − log 0.398 = log = log 100.95, 0.398 1 1 So m = 12 log 40.18 0.398 and b = 2 log(0.398 × 40.18) 2 log 15.99 0.60 40.18 1 Thus we have Y = 2 log 0.398 x + 12 log(0.398 × 40.18) 50.43x + 0.60 Exponentiating, we get Y 10 = y = 10 1 2 log 40.18 0.398 1 x+ 2 log(0.398×40.18) = 10 1 2 log(0.398×40.18) 1 2 log 40.18 0.398 x 10 = 1 x x 1 1 1 40.18 2 log 40.18 0.398 = 15.99164 2 × 4 × 10x (0.398 × 40.18) 2 × 10 2 0.398 Sa We conclude that the data displayed on semilog paper is close to being a straight line, so we take a line Y = mx + b, or log y = mx + b, through the points corresponding to (-1,0.398) and (1,40.18): sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean 115.3 Solutions to some odd numbered problems from section 1.3-13 x and y x 0.1 0.5 1 1.5 2 y 0.045 1.33 5.7 13.36 24.44 Solution: First we graph the data on semi-log and log-log paper: Y 9 8 7 6 5 4 3 2 9 8 7 6 5 4 3 2 1 9 8 7 6 5 4 3 2 1 9 8 7 6 5 4 3 2 1 9 8 7 6 5 4 3 2 1 0.0 0.5 1.0 1.5 2.0 1 1 9 8 2 3 45677 891 6 5 4 3 2 1 9 8 7 6 5 4 3 2 1 2 3 45678910 Sa 1.3-77 The following table is based on a functional relationship between x and y, which is either an exponential or a power function. Use an appropriate logarithmic transformation and a graph to decide whether the table comes from a power function or an exponential function and find the functional relationship between sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean 115.3 Solutions to some odd numbered problems from section 1.3-14 At (0.1,0.045) we have Y = log y = log 0.045 = m log(0.1) + b = m(−1) + b = −m + b, and at (2,24.44) we have Y = log 24.44 = m(log 2) + b = (log 2)m + b, so on subtracting the second equation from the first, we get log 0.45 − log 24.44 = −(1 + log 2)m, so m = log 24.44−log 0.045 1+log 2 2.1 and b = m + log 0.045 2.1 + (−1.346) = 0.76 Thus we have Y = 2.1X + 0.76 Exponentiating, we get 10Y = y = 102.1X+0.76 = 100.76 102.1 log x = 5.7 × x 2.1 Sa We conclude that the data displayed on loglog paper is close to being a straight line, so we take a line Y = mX + b, or log y = m log x + b, through the points corresponding to (0.1,0.045) and (2,24.44): sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean