Every exponential function will be in the form y = ab^{x}
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 = ab^{x}
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 = ab^{0} 3 = a(1) a = 3 |
Applying (1, 15) 15 = ab^{1} 15 = (3)b b = 15/3 b = 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 = ab^{x}
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 = ab^{0} 5 = a(1) a = 5 |
Applying (3, 40) 40 = ab^{3} 40 = (5)b^{3} b^{3 }= 40/5 b^{3} = 8 b = 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 = ab^{x}
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 = ab^{0} 64 = a(1) a = 64 |
Applying (2, 4) 4 = ab^{2} 4 = 64b^{2} b^{2 }= 4/64 b^{2} = 1/16 b = 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 = ab^{x}
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 = ab^{0} 80 = a(1) a = 80 |
Applying (4, 5) 5 = 80b^{4} 5/80 = b^{4} b^{4 }= 1/16 b = 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 = ab^{x}
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 = ab^{2} |
Applying (4, 40) 40 = ab^{2 } b^{2} 40 = 10 b^{2} 4 = b^{2} b = 2 |
Applying the value of b^{2}, 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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