MTH 163
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Express the function y=7*2^(.97x) in the form y=Ab^x and also y=Ae^(kx)
For the first function i Have y=7(1.95884)^x
I cannot figure our how to come up with the second function notation.
Can you give me an example of how to work through this...
Thanks
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The first notation would be y = 7 * (2^.97) ^ x = 7 * 1.95884 ^ x, as you say.
The second form y = A e^(k x) can similarly be rearranged to the for y = A (e^k)^x.
This is equal to the given function for A = 7, provided e^k = 2^.97.
This equation can be solved for k by taking the natural log of both sides:
ln ( e^k ) = ln ( 2 ^.97).
ln(e^k) is just k, since the natural log and exponential functions are inverse functions.
ln(2^.97) = .97 ln(2), by the laws of logarithms, so the equation becomes
k = .97 ln(2), which is easily evaluted. We get
k = .67, approx..
Thus the function is
y = 7 e^(.67 k).