Suppose f(t) = a * b^t. If f(5) = 1000 and f(8)= 3000, then find the exact value of f(20)
thanks
$\displaystyle f(5) = ab^5 = 1000$
and
$\displaystyle f(8) = ab^8 = 3000$
So
$\displaystyle \frac{f(8)}{f(5)} = \frac{ab^8}{ab^5} = \frac{3000}{1000}$
$\displaystyle b^3 = 3$
$\displaystyle b = 3^{1/3}$
Thus
$\displaystyle f(8) = a(3^{1/3})^8 = 3000$
$\displaystyle a3^{8/3} = 3000$
$\displaystyle a = \frac{3000}{3^{8/3}} = 3000 \cdot 3^{-8/3}$
If you want you can write this as
$\displaystyle a = 1000 \cdot 3 \cdot 3^{-8/3} = 1000 \cdot 3^{-5/3}$
which is what you would have gotten directly using f(5).
So
$\displaystyle f(x) = 1000 \cdot 3^{-5x/3}$
So
$\displaystyle f(20) = 1000 \cdot 3^{5 \cdot 20/3}$
$\displaystyle f(20) = 1000 \cdot 3^{100/3}$
$\displaystyle f(20) = 1000 \cdot 3^{99/3} \cdot 3^{1/3}$
$\displaystyle f(20) = 1000 \cdot 3^{33} \cdot 3^{1/3}$
or you could write
$\displaystyle f(20) = 1000 \cdot 3^{33} \sqrt[3]{3}$
-Dan
You're correct.
topsquark made a few typing boo boo's in his formula for f(t), which led to an accumulation of arithmetic boo boo's in his answer for f(20). Boo boo's that you should easily have seen and taken into account if you read through his post carefully, I might add:
1. He forgot to include the b^t bit in his formula, that is, $\displaystyle b^t = (3^{1/3})^t = 3^{t/3}$.
2. Consequently he had the t (he used x) in the wrong spot - in a rather than b.
2. He lost the negative in his exponent.
But the important stuff (the stuff needed to make the problem totally routine if tedious) - the values for a and b - was spot on.
But topsquark's BIGGEST MISTAKE, his GREATEST BLUNDER, his FAUX PAS GRANDE, ( ) was in continuing the solution after finding the correct values of a and b ...... he should have left the tedious arithmetic for you!!!! Way too nice a guy and he paid the price.