Ask your own question, for FREE!
Mathematics 9 Online
OpenStudy (anonymous):

Given an exponential function for compounding interest, A(x) = P(.77)x, what is the rate of change?

OpenStudy (anonymous):

@paki

OpenStudy (paki):

To find rate we subtract the rate in this case is .77 or 77% - 1 or 100% and that would give you the rate in this case its decreasing by an 18%

OpenStudy (freckles):

I thought rate of change was derivative...

OpenStudy (anonymous):

I dont know:(

OpenStudy (freckles):

rate of change is derivative

OpenStudy (anonymous):

What did you get as an answer?

OpenStudy (freckles):

I think he thought you were looking for the interest rate

OpenStudy (freckles):

\[A=P(.77)^x \\ \frac{A}{P}=(.77)^x \] P is a constant take ln( ) of both sides

OpenStudy (freckles):

this will allow you to bring down that power of x

OpenStudy (freckles):

and then differentiate both sides

OpenStudy (anonymous):

ok:)

OpenStudy (anonymous):

I got 23% as an answer?!

OpenStudy (freckles):

wait? does that mean you aren't looking for rate of change and you are looking for interest rate?

OpenStudy (anonymous):

no! I think i am looking for rate of change. Why dont we just try out both?

OpenStudy (anonymous):

−0.13% −13% 0.23% −23% these are the answer choics

OpenStudy (freckles):

none of those are the rate of change so you must be looking for interest rate

OpenStudy (anonymous):

yes :)

OpenStudy (freckles):

@paki gave you the way to calculate the interest rate above just do 77%-100%

OpenStudy (anonymous):

-23?

OpenStudy (anonymous):

@paki

OpenStudy (freckles):

yes 77-100 is -23

OpenStudy (freckles):

so 77%-100% is -23%

OpenStudy (anonymous):

Thanks :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!