Question 4186
I was thinking to use the formula for interest compounded continuously

             {{{A = P* e^(rt)}}} 

In this formula, A=$30,000,  P=$10,000, r=.12, and of course t is the unknown in years.

Substituting, this gives you:

             {{{ 30000 = 10000 e^(.12t) }}}


Divide both sides by 10000 to obtain

             {{{  3 = e^(.12t) }}}

Take the ln of both sides


             {{{ ln 3 = ln e^(.12t) }}}
             {{{ ln 3 = .12t }}}


Divide both sides by .12, and the answer I get is {{{ ln 3/.12}}}

I don't have my calculator with me, but I think that is approximately 9.185 years, which rounds off to the nearest year of 9.



What do you think?

R^2
Seminole Community College
Sanford, Florida