document.write( "Question 1197706: I have invested $1200 in a bank acount with an interest rate 4.5% compounded monthly. After how many years will my account pass $2100 \n" ); document.write( "
Algebra.Com's Answer #831102 by Theo(13342)\"\" \"About 
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f = p * (1 + r) ^ n
\n" ); document.write( "f is the future value
\n" ); document.write( "p is the present value
\n" ); document.write( "r is the interest rate per time period.
\n" ); document.write( "1 + r is the growth factor per time period.
\n" ); document.write( "n is the number of time periods.\r
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\n" ); document.write( "\n" ); document.write( "in your problem, the equation become:
\n" ); document.write( "2100 = 1200 * (1 + .045/12) ^ n
\n" ); document.write( "this becomes 2100 = 1200 * 1.00375 ^ n
\n" ); document.write( "divide both sides of the equation by 1200 to get:
\n" ); document.write( "2100/1200 = 1.00375 ^ n
\n" ); document.write( "this becomes 1.75 = 1.00375 ^ n
\n" ); document.write( "take the log of both sides of the equation to get:
\n" ); document.write( "log(1.75) = n * log(1.00375)
\n" ); document.write( "solve for n to get:
\n" ); document.write( "n = log(1.75) / log(1.00375) = 149.5105101 months.\r
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\n" ); document.write( "\n" ); document.write( "confirm by replacing n with that to get:
\n" ); document.write( "f = 1200 * 1.00375 ^ 149.5105101 = 2100.
\n" ); document.write( "this confirms the value of n is correct.\r
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\n" ); document.write( "\n" ); document.write( "149.5105101 / 12 = 12.45920918 years.
\n" ); document.write( "the account will pass 2100 some time in the 13th year.\r
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