Find a formula for the exponential function that satisfies h(2)=22 and h(4)=14.08
Done
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To find a formula for the exponential function that satisfies h(2) = 22 and h(4) = 14.08, we can start by assuming the exponential function has the form h(x) = a * b^x.
We can plug in the values from the given data points to form a system of equations:
h(2) = 22
22 = a * b^2 --- Equation (1)
h(4) = 14.08
14.08 = a * b^4 --- Equation (2)
Now, we can solve this system of equations to find the values of a and b.
Dividing Equation (2) by Equation (1), we get:
(14.08 / 22) = (a * b^4) / (a * b^2)
0.64 = b^2
Taking the square root of both sides, we have:
b = ± sqrt(0.64)
b = ± 0.8
Since exponential functions cannot have negative bases, we can ignore the negative value. Thus, b = 0.8.
Plugging this value of b into Equation (1), we can solve for a:
22 = a * (0.8)^2
22 = a * 0.64
a = 22 / 0.64
a ≈ 34.375
Therefore, the formula for the exponential function that satisfies h(2) = 22 and h(4) = 14.08 is:
h(x) = 34.375 * 0.8^x
To find a formula for the exponential function that satisfies the given conditions, we can start by assuming that the exponential function has the form h(t) = a * b^t, where 'a' and 'b' are constants that we need to determine.
We are given two conditions: h(2) = 22 and h(4) = 14.08. Plugging these values into the assumed formula, we get two equations:
1) h(2) = a * b^2 = 22
2) h(4) = a * b^4 = 14.08
Now we have a system of two equations with two unknowns (a and b). We can solve this system to find the values of a and b.
Dividing equation (2) by equation (1), we get:
(b^4) / (b^2) = 14.08 / 22
b^2 = 14.08/22
b^2 ≈ 0.640
Taking the square root of both sides, we get:
b ≈ √(0.640)
b ≈ 0.8
Substituting the value of b back into equation (1), we can solve for a:
a * (0.8)^2 = 22
0.64a = 22
a ≈ 22/0.64
a ≈ 34.375
Therefore, we have determined the values of a and b. The formula for the exponential function that satisfies the given conditions is:
h(t) ≈ 34.375 * 0.8^t