SOLUTION: Find equation of an exponential curve passing through the given pair of points and graph these exponents. Is there any other exponent of the same kind passing through these two po

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Question 117703: Find equation of an exponential curve passing through the given pair of points and graph these exponents. Is there any other exponent of the same kind passing through these two points?
A.)(2,12)and(5,96)
B.)(-1,-2/3)and(3,-54)

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
): Find equation of an exponential curve passing through the given pair of points and graph these exponents. Is there any other exponent of the same kind passing through these two points?
A.)(2,12)and(5,96)
I'll just do the first one. If I have time, I'll come back and
do the second one too.

The standard exponential function is easy to remember,
because the letters on the right spell the word "pert".

y = Pe%5Ert 

For the point (2, 12), plug in t = 2 and y = 12

12 = Pe%5E%28r%282%29%29

12 = Pe%5E%282t%29

y = Pe%5Ert

For the point (5, 96), plug in t = 5 and y = 96

96 = Pe%5E%28r%285%29%29

96 = Pe%5E%285r%29

Now we have the system of equations:

96 = Pe%5E%285r%29
12 = Pe%5E%282r%29

Write these as

96 = P%28e%5Er%29%5E5
12 = P%28e%5Er%29%5E2

Since equals divided by equals give equals,
the quotient of the left sides equals the
quotient of the right sides:

96%2F12 = P%28e%5Er%29%5E5%2FP%28e%5Er%29%5E2
 
Simplify by dividing on the left, canceling P's
on the right and subtracting exponents on the right:

          8 = %28e%5Er%29%5E3

Take cube roots of both sides:

          root%283%2C8%29 = root%283%2C%28e%5Er%29%5E3%29

                        2 = e%5Er

Take the natural logs of both sides:

                    ln(2) = r

Substitute ln(2) for r in

                       12 = Pe%5E%282r%29

                       12 = Pe%5E%282%2Aln%282%29%29

                       12 = Pe%5E%28ln%282%5E2%29%29

                       12 = Pe%5E%28ln%284%29%29

                       12 = P%284%29

                       12 = 4P

                        3 = P

So substituting P = 3 and r = ln(2) into

y = Pe%5Ert

y = 3e%5E%28ln%282%29t%29

y = 3%28++e%5Eln%282%29++%29%5Et++%29

y = 3%2A2%5Et


B.)(-1,-2/3)and(3,-54)

Edwin