if g(x) varies exponentially with x, and g(0) =37 and g(7) =5940 find the particular equation...
g = ka^x
37 = ka^0 = k
so, g = 37a^x
5940 = 37a^7
160.54 = a^7
a = 2.066
g = 37*2.066^x
To find the particular equation for the function g(x) that varies exponentially with x, we need to determine the base of the exponential function. The general form of an exponential function is g(x) = a * b^x, where a is the initial value and b is the base.
Given that g(0) = 37, we know that when x = 0, g(x) = 37.
Therefore, we have g(0) = a * b^0 = a * 1 = a = 37.
Next, we are given that g(7) = 5940. Plugging this into the general form of the exponential function, we have:
g(7) = 37 * b^7 = 5940.
Now, we can solve for the base, b:
37 * b^7 = 5940.
Dividing both sides of the equation by 37:
b^7 = 5940 / 37.
b^7 = 160.
To find the value of b, we need to take the 7th root of both sides:
b = (160)^(1/7).
Now, we can finalize the particular equation by plugging in the values we've found:
g(x) = 37 * [(160)^(1/7)]^x.
Therefore, the particular equation for g(x) is: g(x) = 37 * (160)^(x/7).