Question 753000
Under the onslaught of the college algebra second period class, a pile of homework problems decreased exponentially. It decreased form 900 to 500 problems in only 40 minutes. How long would it take until only 300 problems remained?
--------
y = ab^x
-------
You have 2 points (0,900) and (40,500)
---------------------
Using (0,900) 900 = ab^0
900 = a
-------
y = 900*b^x
-----
Using (40,500) solve for "b":
500 = 900*b^40
b^40 = 5/9
------
Equation:
f(x) = 900(5/9)^(x/40)
-------------------
How long would it take until only 300 problems remained?
Solve: 300 = 900(5/9)^(x/40)
(5/9)^(x/40) = 1/3
----
(x/40) = log(1/3)/log(5/9)
x/40 = 1.87
x = 75 minutes
======================
Cheers,
Stan H.
==================