Question

Find the exponential function that satisfies the given conditions: Initial value = 70, decreasing at a rate of 0.43% per week

Answer 1

Answer:

$f(t) \ = \ 70 \ * \ e ^ { \ - 0.0043 \frac{1}{week} \ * \ t}$

Step-by-step explanation:

Exponentials functions are of the form:

$f(t) \ = \ A \ * \ e ^ { \ b \ * \ t}$

where A and b are constants.

Now, the initial value of the exponential function its

$f(0) \ = \ A \ * \ e ^ { \ b \ * \ 0}$

$f(0) \ = \ A \ * \ e ^ { \ 0 \ }$

$f(0) \ = \ A$

If the initial value must be 70, this must means:

$A \ = \ 70$

So

$f(t) \ = \ 70 \ * \ e ^ { \ b \ * \ t}$

We also know that it must decrease at a rate of 0.43 %, this mean that after one week we got:

$100 \ \% - 0.43 \ \% = 99.57 \ \%$

$f(1 week) \ = \ 70 \ * 0.9957 \ = \ 70 \ * \ e ^ { \ b \ * \ 1 \ week}$

This means :

$0.9957 \ = \ e ^ { \ b \ * \ 1 \ week}$

$ln ( 0.9957) \ = \ b \ * \ 1 \ week$

$\ b \ = \frac{ln ( 0.9957)}{ 1 \ week}$

$\ b \ = - 0.0043 \frac{1}{week}$

So, our equation, finally, its:

$f(t) \ = \ 70 \ * \ e ^ { \ - 0.0043 \frac{1}{week} \ * \ t}$

Answer 2

That'd be y = 70* (1-.00043)^t, where the rate of decrease is really 0.43% and t denotes the # of weeks.  

This simplifies to y = 70*(0.00057)^t.