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February 1, 2015

February 1, 2015

Posted by **Jim** on Tuesday, December 1, 2009 at 12:47am.

- Math -
**Reiny**, Tuesday, December 1, 2009 at 8:54amI don't know at what level you are working at, but any exponential of the form y = p^x can be written as

y = a(e^kx) where a and k are constants.

So let your y=2^(x+1) + 3 be

y = a(e^kx) + 3

let's use 2 points from your first equation, say (0,5) and (1,7)

in the new one:

for (0,5) ,

5 = a(e^0) + 3

a = 2

for (1,7)

7 = 2(e^k) + 3

2 = e^k

k = ln2

so a second equation would be

y = 2(e^[(ln2)x)] + 3

check it by using (2,11) from the first equation and trying it in our new one.

LS = 11

RS = 2(e^2ln2) + 3

= 2(e^1.386294361) + 3

= 2(4) + 3

= 11

= RS

- Math -
**Jim**, Tuesday, December 1, 2009 at 1:23pmThank you so much for your detailed reply. My problem is that we haven't done logs or "e" yet. Is there any other way arrive at an answer without knowing these? I really appreciate your help.

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