How long will it take for $ 2000 to grow to $ 46,800 at an interest rate of 10.1% if the interest is compounded continuously?

Question

Round the number of years to the nearest hundredth.

Answer 1

P(t) = P₀e^rt  where:

• P(t) = the value of the investment at time t = $46,800
• P₀ = the initial investment = $2000
• r = the interest rate expressed as a decimal = 10.1% = 0.101
• t = time = what you're trying to find

 

P(t) = P₀e^rt

$46,800 = $2000e^0.101t

46,800/2000 = e^0.101t

ln(46800/2000) = 0.101t

ln(46800/2000)/0.101 = t

 

Plug the numbers into your calculator and compute the answer