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This case one of the set is x is -1 and y is 3 the other pair is x is 1 and y is 75 How do I find the multiplier of the 2 to make an exponential function?
Wiki User
∙ 8y ago
It sounds like you want the form y = A*2^(b*x). Where A and b are constants to be found?
If you plug in the numbers: 3 = A*2^(-b), and then 75 = A*2^b. So take that one and you have A = 75 / (2^b), but 1/(2^b) is the same as 2^(-b), so then we have A = 75*2^(-b). plug this into the first one: 3 = A*2^(-b) which equals (75*2^(-b))*2^(-b). So you have 3/75 = 2^(-2*b). Take reciprocal of both sides: 75/3 = 2^(2*b). And 75/3 is 25. So take square root of both sides: 5 = 2^b. Take log[base2] of both sides: and we have log[base2](5) = b.
Wiki User
∙ 8y ago
Wiki User
∙ 8y agoThe answer depends on whether x is the base of the exponential function or the power.
Suppose y = AB^xThen 3 = A*B^(-1)
and
75 = A*B^1
Dividing the second by the first gives 25 = B^2 so that B = +/- 5
Substituting for B in the second result, 75 = A*(+/-5) = A = -/+ 15.
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