#$&* course Mth 163 19 Query*********************************************
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Given Solution: `a** If you have two points you can solve the simultaneous equations: Substitute the coordinates into the form y = A b^x and solve the two resulting equations for A and b.You could alternatively use the form y = A * 2^(k x) or y = A * e^(k x), in which case you would solve for A and k. If you have a more extensive data set you can use transformations. For exponential data you plot log(y) vs. x. If the graph is well approximated by a straight line then you get an exponential function. Then since the graph is a straight line, you can find its equation using using either slope and vertical intercept, or two points on the line. If the slope of a y vs. x graph is m and the vertical intercept is b then the function is y = m x + b. However in this case the graph is not of y vs. x, but of log(y) vs. x. So if the slope of your graph is m and the y intercept is b, the function is log(y) = m x + b. This equation needs to be solved for y: You invert the transformation using the inverse function 10^x, obtaining 10^log(y) = 10^(mx+b). 10^log(y) = y, by the definition of the logarithm, and 10^(mx + b) = 10^(mx) * 10^b, by the laws of exponents. Thus y = 10^(mx) * 10^b, where m and b are just the numbers (slope and vertical intercept) that you determined from your graph. Note that if a power function fits the data then log y vs. log x will give a straight line so that log y = m log x + b. In this case our solution will be y = 10^b * x^m, a power function rather than an exponential function. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course Mth 163 19 Query*********************************************
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Given Solution: `a** If you have two points you can solve the simultaneous equations: Substitute the coordinates into the form y = A b^x and solve the two resulting equations for A and b.You could alternatively use the form y = A * 2^(k x) or y = A * e^(k x), in which case you would solve for A and k. If you have a more extensive data set you can use transformations. For exponential data you plot log(y) vs. x. If the graph is well approximated by a straight line then you get an exponential function. Then since the graph is a straight line, you can find its equation using using either slope and vertical intercept, or two points on the line. If the slope of a y vs. x graph is m and the vertical intercept is b then the function is y = m x + b. However in this case the graph is not of y vs. x, but of log(y) vs. x. So if the slope of your graph is m and the y intercept is b, the function is log(y) = m x + b. This equation needs to be solved for y: You invert the transformation using the inverse function 10^x, obtaining 10^log(y) = 10^(mx+b). 10^log(y) = y, by the definition of the logarithm, and 10^(mx + b) = 10^(mx) * 10^b, by the laws of exponents. Thus y = 10^(mx) * 10^b, where m and b are just the numbers (slope and vertical intercept) that you determined from your graph. Note that if a power function fits the data then log y vs. log x will give a straight line so that log y = m log x + b. In this case our solution will be y = 10^b * x^m, a power function rather than an exponential function. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 163 19 Query*********************************************
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Given Solution: `a** If you have two points you can solve the simultaneous equations: Substitute the coordinates into the form y = A b^x and solve the two resulting equations for A and b.You could alternatively use the form y = A * 2^(k x) or y = A * e^(k x), in which case you would solve for A and k. If you have a more extensive data set you can use transformations. For exponential data you plot log(y) vs. x. If the graph is well approximated by a straight line then you get an exponential function. Then since the graph is a straight line, you can find its equation using using either slope and vertical intercept, or two points on the line. If the slope of a y vs. x graph is m and the vertical intercept is b then the function is y = m x + b. However in this case the graph is not of y vs. x, but of log(y) vs. x. So if the slope of your graph is m and the y intercept is b, the function is log(y) = m x + b. This equation needs to be solved for y: You invert the transformation using the inverse function 10^x, obtaining 10^log(y) = 10^(mx+b). 10^log(y) = y, by the definition of the logarithm, and 10^(mx + b) = 10^(mx) * 10^b, by the laws of exponents. Thus y = 10^(mx) * 10^b, where m and b are just the numbers (slope and vertical intercept) that you determined from your graph. Note that if a power function fits the data then log y vs. log x will give a straight line so that log y = m log x + b. In this case our solution will be y = 10^b * x^m, a power function rather than an exponential function. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!