Approximately 13.3 years.
For continuous interest, depending on your class's terminology and notation, you'll see either
#F=Pe^(rt)# where#F# is the future value, #P# is the present value, #r# is the (annual) interest rate, and #t# is time (in years).
or
#A=Pe^(rt)# where #A# is the accumulated amount, #P# is the principal, #r# is the (annual) interest rate, and #t# is time (in years).
In this question, #r=0.0825#, #P=500# and we want #F# (or #A#) to be #3P=1500#
Find #t# so that: #3P = Pe^(0.0825 t)# or #1500=500e^(0.0825 t)#
The first thing we'd like to do, is get #e^"power"# alone on one side of the equation. To do this, divide by #P# or #500#, to get:
#3 = e^(0.0825 t)#
(Notice that it doesn't matter how much Carl invests. If he wants to triple his money, he'll have to solve this equation.)
Solve: #e^(0.0825 t) = 3# by taking the natural logarithm of both sides.
#0.0825 t = ln(3)#
#t= ln(3)/0.0825# years
In order to get a numerical approximation, we'll need #ln(3)~~1.098612#. Divide to get #t~~13.32# years.
Note In this solution, #ln# is used for the natural logarithm. Others use #log# for the natural logarithm.
