document.write( "Question 294557: Suppose $1000 dollars is invested at an interest rate r. compounded continuously. If the balance grows to $1144.54 in 3 years. Find the time it will take for the balamce to grow to $2500. \n" ); document.write( "
Algebra.Com's Answer #212516 by nerdybill(7384)\"\" \"About 
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Suppose $1000 dollars is invested at an interest rate r. compounded continuously. If the balance grows to $1144.54 in 3 years. Find the time it will take for the balance to grow to $2500.
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\n" ); document.write( "Continuous Compound Interest Formula
\n" ); document.write( "A = Pe^(rt)
\n" ); document.write( " where, P = principal amount (initial investment)
\n" ); document.write( "r = annual interest rate (as a decimal)
\n" ); document.write( "t = number of years
\n" ); document.write( "A = amount after time t
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\n" ); document.write( "First, find the interest rate:
\n" ); document.write( "A = Pe^(rt)
\n" ); document.write( "1144.54 = 1000e^(r(3))
\n" ); document.write( "1144.54/1000 = e^(3r)
\n" ); document.write( "ln(1144.54/1000) = 3r
\n" ); document.write( "ln(1144.54/1000)/3 = r
\n" ); document.write( ".045 = r
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\n" ); document.write( "Now, you can answer:
\n" ); document.write( "Find the time it will take for the balance to grow to $2500.
\n" ); document.write( "A = Pe^(rt)
\n" ); document.write( "2500 = 1000e^(.045t)
\n" ); document.write( "2500/1000 = e^(.045t)
\n" ); document.write( "ln(2500/1000) = .045t
\n" ); document.write( "ln(2500/1000)/.045 = t
\n" ); document.write( "20.36 years = t\r
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