Query 19

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course Mth 163

12/14 11

019. `query 19*********************************************

Question: `qexplain the steps in fitting an exponential function to data

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Your solution:

If you have an exponential function, you will use the y = A b^x function and solve for a and b. If you have data points to use, do a t vs. y, log(y) vs. x, y vs. log(x) and log(y) vs. log (x) to see if the table yields a linear graph.

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: ok

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:

y = 2.5x + 4

@&

It's the graph of log(y) vs. log(x), not the graph of y vs. x, that has slope 2.5 and vertical intercept 4.

So the equation would be

log(y) = 2.5 log(x) + 4.

You treated the equation as such in your steps below, but you should be careful to write it correctly in the beginning.

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10^log(y) = l0^2.5x + 4

y = 10^2.5x + 4

@&

You need to be careful with your signs of grouping.

The preceding two steps should read

10^log(y) = l0^(2.5x + 4)

y = 10^(2.5x + 4)

*@

@&

From this point on you have a good solution.

*@

y = 10^2.5x * 10^4

y = 10^2.5)^x * 10^4

y = 316.23^x * 10,000

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:

y = 2.5x + 4

10^log(y) = l0^2.5x + 4

y = 10^2.5x + 4

y = 10^2.5x * 10^4

y = 10^2.5)^x * 10^4

y = 316.23^x * 10,000

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Self-critique rating:

&#Your work looks good. See my notes. Let me know if you have any questions. &#