Question 84434
Given f(-8) = 200 and f(30) = 580
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Find an exponential formula that fits these given data.
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The formula you will need to find is of the form {{{ y = A*e^(b*x)}}}
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So what you need to do is to solve for A and b using the given data and appropriate
math processes.
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Start with the given that y = 200 when x = -8. Plug these values into the formula to get:
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{{{200 = A*e^(b*(-8))}}}
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Take the natural logarithm of both sides:
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{{{ln(200) = ln(A*e^(-8*b)) = ln(A) -8*b*ln(e)}}}
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But the natural log of e is 1. Substitute that and you get:
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{{{ln(200) = ln(A)-8*b }}}
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Solve this for ln(A) by adding 8*b to both sides to get:
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{{{ln(A) = ln(200)+ 8*b}}}
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Next follow the identical process for the second data point in which y = 580 and x = 30.
Using this data in the formula and solving for ln(A) results in:
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{{{ln(A) = ln(580) - 30*b}}}
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Now you can set the right sides of the two equations you have for ln(A) equal and solve
for b. You should be solving:
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{{{ln(200) + 8*b = ln(580) - 30*b}}}
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When you solve this you should find that b is somewhere around 0.028018703 or whatever
you choose to round it to.
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Now that you know b you can return to either of the two equations for the data points and
solve for A.  For example, return to:
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{{{y = A*(e^(b*x))}}}
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Substitute 200 for y, -8 for x, and 0.028018703 for b to get:
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{{{200 = A*(e^(0.028018703*(-8)))= A*e^(-0.224149628)}}}
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Solve for A by dividing both sides by {{{e^(-0.224149628)}}}
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{{{A = 200/(e^(-0.224149628))}}}
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But a quantity in the denominator that has a negative exponent can be brought into the 
numerator with the same exponent only positive.  This means that A becomes equal to:
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{{{A = 200*e^(0.224149628)}}}
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and by a calculator {{{e^(0.224149628) = 1.25125823}}}. Substituting this into the equation 
for A results in:
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{{{A = 200*1.25125823 = 250.2516459}}}
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Now put this value of A and also the known value of b into the equation and you should
end up with something around:
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{{{y = (250.2516459)*e^(0.028018703*x)}}}
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as the equation you are looking for.
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Hope you can track your way through this.  You can check by letting x = 30 and solving
to see if y = 580. You should find that y computes to be 579.9999895 or so. Close 
enough to assume this works. Cheers and good luck!