Given an initial value of 17 g, which is halving once each 32 hours, an exponential function in term of hour t which models this situation is ...?
Value= 17(1/2)^time/32 where time is in hours.
or...
Value= 17 e-.692t
To find an exponential function that models the given situation, let's break it down step by step:
1. We know that the initial value is 17 g, so the exponential function can be written in the form f(t) = a * (b^t), where "a" represents the initial value.
2. The given problem states that the value is halved once every 32 hours. In exponential form, this means that the base "b" is 1/2, representing the halving factor.
3. We need to find the value of "b" that corresponds to the halving every 32 hours. Since the value is halved once every 32 hours, we have b = 1/2^(1/32), which is the inverse of the halving factor raised to the power of the time interval.
4. Plugging in the values, the exponential function can be expressed as f(t) = 17 * (1/2^(1/32))^t.
Therefore, the exponential function in terms of hour "t" that models this situation is f(t) = 17 * (1/2^(1/32))^t.