SOLUTION: write an exponential function in {{{y=ab^x}}} passing through the points (2,1)(5,16)

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: write an exponential function in {{{y=ab^x}}} passing through the points (2,1)(5,16)      Log On


   

Question 961247: write an exponential function in y=ab%5Ex passing through the points (2,1)(5,16)
Found 2 solutions by MathLover1, stanbon:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
write an exponential function in y=ab%5Ex passing through the points (2,1) and (5,16)



y=ab%5Ex=> use (2,1), substitute x and y
1=ab%5E2...solve for a
a=1%2Fb%5E2........eq.1

now use (5,16)
y=ab%5Ex
16=ab%5E5...solve for a
a=16%2Fb%5E5......eq.2

from eq.1 and eq.2 we have
1%2Fb%5E2=16%2Fb%5E5........both sides multiply by b%5E5
b%5E5%2Fb%5E2=16b%5E5%2Fb%5E5
b%5E%285-2%29=16
b%5E3=16
b=root%283%2C2%2A2%5E3%29
b+=+2%2Aroot%283%2C2%29
find a
a=1%2Fb%5E2........eq.1
a=1%2F%282%2Aroot%283%2C2%29%29%5E2
a=1%2F%284%28root%283%2C2%29%5E2%29%29

substitute a and b in y=ab%5Ex and you have:
y=%281%2F%284%28root%283%2C2%29%5E2%29%29%29%282%2Aroot%283%2C2%29%29%5Ex
y=%28+2%2Aroot%283%2C2%29%29%5Ex%2F%284%28root%283%2C2%29%29%5E2%29

highlight%28+y+=+2%5E%28%284%28x-2%29%29%2F3%29+%29







Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
write an exponential function in y=ab%5Ex passing through the points (2,1)(5,16)
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Solve for "a" and "b"::
y = ab^x
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Using (2,1) you get: 1 = ab^2
Using (5,16) you get: 16 = ab^5
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Dividing the bottom eq. by the top eq. you get:
16 = b^3
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Then b = 2^4/3
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Solve for "a" by substituting for "b"::
1 = a*(2^(4/3)]^2
1 = a*(2^(8/3))
a = 2^(-8/3)
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Ans: y = (2^(-8/3))*(2^(4x/3))
y = 2^[(-8/3)+(4x/3)]
y = 2^((4x-8)/3)
y = 2^[4(x-2)/3]
y = 16^[(x-2)/3]
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Cheers,
Stan H.