Yahoo Answers: Answers and Comments for Does anyone know how to solve this? [Mathematics]
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From Bob
enCA
Mon, 30 Mar 2020 00:33:35 +0000
3
Yahoo Answers: Answers and Comments for Does anyone know how to solve this? [Mathematics]
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From coreyA: You have y = a*b^x and you also know that (0,...
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Mon, 30 Mar 2020 00:56:11 +0000
You have y = a*b^x and you also know that (0,1) and (2,.25) are also on the curve.
Start with (0,1)
1 = a*b^0
1 = a*1
a = 1
y = (1)b^x, use (2,.25) to find b:
.25 = (1)*b^2
b^2 = .25
b = .5

From Philip: y = a(b^x).
(0,1)........on y > 1 = a(...
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Mon, 30 Mar 2020 02:21:40 +0000
y = a(b^x).
(0,1)........on y > 1 = a(b^0) = a, ie., 1 = a(1) = a. Then a = 1
(2,0.25)...on y > 0.25 = (b^2). Then b^2 = (1/4) and b = (+/) (1/2).
Now (2,4) on the graph is shown to be on y. Therefore 4=[(1/2)^(2)]...(1). or
4 = [(1/2)^(2)]...(2).
For (1) holding, 4 = (1/2)^(2) = [2^(1)]^(2)] = 2^[(1)(2)] = 2^(2) = 4 which is not
true. Therefore b =/= (1/2).
For (2) holding, 4 = (1/2)^(2) = [2^(1)]^(2) = 2^[(1)(2)] = 2^(2) = 4 which is true.
Therefore b = (1/2).