Doubling time for quantity increasing by 8.6 per year

[Angela123]

Assume a quantity is increasing by 8.6 % per year. What is the doubling time in years?

I tried using the equation: A=Pe^rt and then got .914=e^(.086t) and ended up with an answer of 10.627 which was wrong. Am I using the wrong equation?

 

[Deleted member 4993]

Re: Doubling time

Assume a quantity is increasing by 8.6 % per year.

What is the doubling time in years?

I tried using the equation: A=Pe^rt <<< This formula is not useful here - exponential form is used for continuous increase.
and then got .914=e^(.086t) and ended up with an answer of 10.627 which was wrong. Am I using the wrong equation?

Formula to use

A = P * (1 + r)[sup:1phl36dm]t[/sup:1phl36dm]

 

[stapel]

Am I using the wrong equation?

With the increase being defined as "per year", rather than "continuously", it would probably be better to use the compound-interest formula with r = 0.086, n = 1, P = 1, and A = 2. :wink:

Eliz.