# FROM THE GIVEN INPUT AND OUTPUT FIND THE EXPONENTIAL FUNCTION

Every exponential function will be in the form y = abx

• If b > 1, then it must be exponential growth function.
• If 0 < b < 1, then it must be exponential decay function.

Find an exponential function having the given values.

Problem 1 :

f(0) = 3 and f(1) = 15

Solution :

Exponential function will be in the form

y = abx

From the given inputs and outputs, we can understand that the points (0, 3) and (1, 15) on the exponential function.

 Applying (0, 3)3 = ab03 = a(1)a = 3 Applying (1, 15)15 = ab115 = (3)bb = 15/3b = 5

Applying the values of a and b, we get the required exponential function

y = 3(5)x

Problem 2 :

f(0) = 5 and f(3) = 40

Solution :

Exponential function will be in the form

y = abx

From the given inputs and outputs, we can understand that the points (0, 5) and (3, 40) on the exponential function.

 Applying (0, 5)5 = ab05 = a(1)a = 5 Applying (3, 40)40 = ab340 = (5)b3b3 = 40/5b3 = 8b = 2

Applying the values of a and b, we get the required exponential function

y = 5(2)x

Problem 3 :

f(0) = 64 and f(2) = 4

Solution :

Exponential function will be in the form

y = abx

From the given inputs and outputs, we can understand that the points (0, 64) and (2, 4) on the exponential function.

 Applying (0, 64)64 = ab064 = a(1)a = 64 Applying (2, 4)4 = ab24 = 64b2b2 = 4/64b2 = 1/16b = 1/4

Applying the values of a and b, we get the required exponential function

y = 64(1/4)x

Problem 4 :

f(0) = 80 and f(4) = 5

Solution :

Exponential function will be in the form

y = abx

From the given inputs and outputs, we can understand that the points (0, 80) and (4, 5) on the exponential function.

 Applying (0, 80)80 = ab080 = a(1)a = 80 Applying (4, 5)5 = 80b45/80 = b4 b4 = 1/16b = 1/2

Applying the values of a and b, we get the required exponential function

y = 80(1/2)x

Problem 5 :

f(2) = 10 and f(4) = 40

Solution :

Exponential function will be in the form

y = abx

From the given inputs and outputs, we can understand that the points (2, 10) and (4, 40) on the exponential function.

 Applying (2, 10)10 = ab2 Applying (4, 40)40 = ab2  b240 = 10 b24 = b2b = 2

Applying the value of b2, we get

10 = 4a

a = 10/4

a = 5/2

Applying the values of a and b, we get

y = (5/2)(2)x

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