115.3 Solutions to some odd numbered problems from section 1.3
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1 . Solutions to some odd numbered problems from section.- y. Solutions to some odd numbered problems from section..- Sketch the graph of y = x + without using a graphing calculator. Starting with the graph of y = x in black, and take that graph shifted up unit in red: y = x +..- Sketch the graph of y = exp(x ) = e x without using a graphing calculator. Starting with the graph of y = e x in black, we get that shifted units to the right in red: y = e x x 00 Doug MacLean y x
2 . Solutions to some odd numbered problems from section.-.- Find the following numbers on a number line that is on a logarithmic scale (base 0): 0.000,0.0,,,0,00,000, 000, and Doug MacLean
3 . Solutions to some odd numbered problems from section.-.- 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. 00 Doug MacLean (x,y ) = (0, ), (x,y ) = (, ) 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 (x,y ) = (0, ) we have log = m(0) + b, sob = log. At (x,y ) = (, ) we have log = m() + b, so0= m + log, and thus m = log 0. and we have the equation Y = log x + log. Exponentiating, we get ) y = 0 Y = 0 log y = 0 log x+log = 0 log (0 log x (0.) x Y=log y 0 x
4 . Solutions to some odd numbered problems from section.-.- 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. 00 Doug MacLean (x,y ) = (, ), (x,y ) = (, ) 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 (x,y ) = (, ) we have log = m( ) + b, and at (x,y ) = (, ) we have log = m() + b, or0= m + b, sob = m. Substituting this in the first equation, we get log = m m = m, som = log and b = log and we have the equation Y = log x + log. Exponentiating, we get y = 0 Y = 0 log y = 0 log log x+ = 0 log ) (0 log x ) = ( x Y=log y x
5 . Solutions to some odd numbered problems from section.-.- Use a logarithmic transformation to find a linear relationship between the given quantities and graph the resulting linear relationship on a log-linear plot: y = 0 x. We have Y = log( 0 x ) = log + x Y=log y 00 Doug MacLean - 0 x
6 . Solutions to some odd numbered problems from section.-.- 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.) (x,y ) = (, ), (x,y ) = (, ) 00 Doug MacLean We have Y = log y = mlogx + b, so when (x,y ) = (, ), we have log = m log + b = m(0) + b = b, sob = log, and when (x,y ) = (, ), we have log = m log + log, so 0 = m log + log and thus m = log log, so we have the equation Y = log X + log. log Exponentiating, we have 0 Y = 0 + log y = y = 0 log log X+log = 0 log 0 log log log x = x log log. y 0
7 . Solutions to some odd numbered problems from section.-.- Use a logarithmic transformation to find a linear relationship between the given quantities and graph the resulting linear relationships on a log-log plot. 00 Doug MacLean y = x Y = log y = log(x ) = log + log x = log + X, where X = log x. y 0
8 . Solutions to some odd numbered problems from section.-.- 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 00 Doug MacLean f(x) = x. ( F = log f(x) = log x.) ( = log + log x.) = log +. log x = log +.X which we plot on log-log paper: y 0
9 . Solutions to some odd numbered problems from section.-.- 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 00 Doug MacLean N(t) = 0.t ( log N(t) = log 0.t) ( = log 0 + log.t) = log 0 + (. log )t which we plot on log-linear paper: log L 00-0 c
10 . Solutions to some odd numbered problems from section.-0.- 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 x 0 0 x and y y log-log paper: First we graph the data on semi-log and Y=log y Y=log y 00 Doug MacLean x 0 X=log x We conclude that the data displayed on loglog paper is closest to being a straight line, so we take a line Y = mx + b, orlogy = m log x + b, through the points corresponding to (,.) and (0,.): At (,.) we have Y = log y = log. = m log + b = m(0) + b = b, sob = log.. At (0,.) we have Y = log. = m log 0 + b = m log 0 + log., so m =. log. Thus we have Y = log 0 X + log. 0.X + log. Exponentiating, we get 0 Y log.. = y = 0 log 0 X+log. = 0 log. log.. 0 log 0 X =. (0 log.. X) log 0. x 0. log. log. log 0 = log.. log 0.
11 . Solutions to some odd numbered problems from section.-.- 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 x x and y We only graph the data on semi-log, y because negative values cannot be represented on log-log paper: Y 00 Doug MacLean
12 . Solutions to some odd numbered problems from section.- We conclude that the data displayed on semilog paper is close to being a straight line, so we take a line Y = mx + b, orlogy = mx + b, through the points corresponding to (-,0.) and (,0.): At (-,0.) we have Y = log y = log 0. = m( ) + b = m + b, and at (,0.) we have Y = log 0. = m() + b = m + b, so adding the two equations, we get b = log 0. + log 0. = log(0. 0.) = log., and subtracting the first equation ( from the ) second gives 0. m = log 0. log 0. = log = log 00., 0. ( ) So m = log and b = log(0. 0.) log. 0.0 ( ) Thus we have Y = log x + log(0. 0.) 0.x Exponentiating, we get ( 0 Y = y = 0 log ( (0. 0.) ) ) x = ( x+ log(0. 0.) = 0 log(0. 0.) 0 log ( 0 log 0. )) x ( ) x 0. = x Doug MacLean
13 . Solutions to some odd numbered problems from section.-.- 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 x x and y y log-log paper: First we graph the data on semi-log and 00 Doug MacLean Y
14 . Solutions to some odd numbered problems from section.- We conclude that the data displayed on loglog paper is close to being a straight line, so we take a line Y = mx + b, orlogy = m log x + b, through the points corresponding to (0.,0.0) and (,.): At (0.,0.0) we have Y = log y = log 0.0 = m log(0.) + b = m( ) + b = m + b, and 00 Doug MacLean at (,.) we have Y = log. = m(log ) + b = (log )m + b, so on subtracting the second equation from the first, we get log 0. log. = ( + log )m, log. log 0.0 so m = +log. and b = m + log (.) = 0. Thus we have Y =.X + 0. Exponentiating, we get 0 Y = y = 0.X+0. = log x =. x.
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