document.write( "Question 1080926: how much must you deposit in an account that pays 8% interest, compounded monthly, to have a balance of $1000 after three years \n" ); document.write( "

Algebra.Com's Answer #695010 by jorel1380(3719) $\"\"$ $\"About$

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In order to have a balance of $1000, after three years of 8% interest, compounded monthly, we have:
\n" ); document.write( "1000=N(1 + .08/12)^36 where N is the initial deposit. So
\n" ); document.write( "1000=N(1.00666667)^36
\n" ); document.write( "N=1000/(1.00666667)^36
\n" ); document.write( "ln N=ln 1000/(1.00666667)^36=ln 1000-36 ln (1.00666667)
\n" ); document.write( "ln N=6.6685517411100682158933244766027
\n" ); document.write( "N=e^6.6685517411100682158933244766027=$787.25463 as the initial deposit. ☺☺☺☺ \n" ); document.write( "

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