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course Mth 163
4/1/13 around 5:30 p.m.
019. `query 19*********************************************
Question: `qexplain the steps in fitting an exponential function to data
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Your solution:
If you have two points you can solve the simultaneous equations by either:
• Substitute the coordinates into the form y = A b^x and solve the two resulting equations for A and b.
• Use the form y = A * 2^(k x) or y = A * e^(k x), in which case you would solve for A and k.
You may also use transformations if you have more data.
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.
Since the graph is a straight line, you can find its equation using either slope and y- intercept, or two points on the line. If the slope of a y vs. x graph is m and the y-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.
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 can 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.
Therefore y = 10^(mx) * 10^b, where m and b are just the numbers (slope and y-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.
confidence rating #$&*: 3
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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. **
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Self-critique (if necessary): ok
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Self-critique rating: 3
If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.
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Question: The graph of log(y) vs. log(x) has slope 2.5 and vertical-axis intercept 4.
What is the equation relating log(y) to log(x)?
What is the equation relating y to x?
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Your solution:
The equation relating log(y) to log(x) would be log (y) = 2.5log(x) + 4
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This is correct.
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The equation relating y to x would be y = 2.5x + 4
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The equation log(y) = 2.5 log(x) + 4 must be solved by valid methods, performing a series of identical operations on both sides of the equation, as covered in the worksheets.
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confidence rating #$&*:3
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Question: The graph of log(y) vs. x has slope 2.5 and vertical-axis intercept 4.
What is the equation relating log(y) to x?
What is the equation relating y to x?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
The equation relating log(y) to log(x) would be log (y) = 2.5x + 4
Then we use transformation using the inverse function getting 10^log(y) = 10^(2.5x+4)
We then would get y = 10^(2.5*4) * 10^4
Therefore we get y = 10^(2.5x) * 10^4
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Good.
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Self-critique rating: 3
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Good overall, but you need to know how to solve the equation
log (y) = 2.5log(x) + 4.
The method is similar to the method you used on the last problem, but there are some differences in the details.
I suggest you submit a revised solution to that problem.
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