document.write( "Question 963867: A principal of
\n" ); document.write( "$1300
\n" ); document.write( " is invested at
\n" ); document.write( "8.75%
\n" ); document.write( " interest, compounded annually. How many years will it take to accumulate
\n" ); document.write( "$4000
\n" ); document.write( " or more in the account?
\n" ); document.write( "

Algebra.Com's Answer #588842 by nerdybill(7384)\"\" \"About 
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A principal of
\n" ); document.write( "$1300
\n" ); document.write( " is invested at
\n" ); document.write( "8.75%
\n" ); document.write( " interest, compounded annually. How many years will it take to accumulate
\n" ); document.write( "$4000
\n" ); document.write( " or more in the account?
\n" ); document.write( ".
\n" ); document.write( "Use:
\n" ); document.write( "A = P(1+i)^t
\n" ); document.write( "where
\n" ); document.write( "A is amount after time t (4000)
\n" ); document.write( "P is principal (1300)
\n" ); document.write( "i is interest (.075)
\n" ); document.write( "t is time (years) -- this is what we're looking for
\n" ); document.write( ".
\n" ); document.write( "4000 = 1300(1+.075)^t
\n" ); document.write( "4000 = 1300(1.075)^t
\n" ); document.write( "40 = 13(1.075)^t
\n" ); document.write( "40/13 = (1.075)^t
\n" ); document.write( "\"log%281.075%2C%2840%2F13%29%29\" = t
\n" ); document.write( "15.54 years = t
\n" ); document.write( "answer: approximately 16 years
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