document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1193774>Question 1193774</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Moisha is saving up money for a down payment on a house.  She currently has $5579, but knows she can get a loan at a lower interest rate if she can put down $6160.  If she invests the $5579 in an account that earns 5.2% annually, compounded monthly, how long will it take Moisha to accumulate the $6160?  Round your answer to two decimal places, if necessary. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #825820 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=825820\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> formula to use is f = p * (1+r) ^ n<BR>\n" );
document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "1 + r is the growth factor per time period.<BR>\n" );
document.write( "n is the number of time periods.\r<BR>\n" );
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document.write( "your interest rate per year is 5.2% / 100 = .052<BR>\n" );
document.write( "your interest rate per month is .052/12 = .00433333333<BR>\n" );
document.write( "your growth factor per month is 1.00433333333\r<BR>\n" );
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document.write( "the number of time periods is the number of months.\r<BR>\n" );
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document.write( "formula becomes:<BR>\n" );
document.write( "6160 = 5579 * 1.00433333333 ^ n<BR>\n" );
document.write( "divide both sides of the equation by 5579 to get:<BR>\n" );
document.write( "6160/5579 = 1.00433333333 ^ n<BR>\n" );
document.write( "take the log of both sides of the equation to get:<BR>\n" );
document.write( "log(6160/5579) = log(1.00433333333 ^ n)<BR>\n" );
document.write( "by one of the rules of logs, this becomes:<BR>\n" );
document.write( "log(6160/5579) = n * log(1.00433333333)<BR>\n" );
document.write( "divide both sides by log(1.00433333333) to get:<BR>\n" );
document.write( "log(6160/5579) / log(1.00433333333) = n<BR>\n" );
document.write( "solve for n to get:<BR>\n" );
document.write( "n = 22.91116607 months.<BR>\n" );
document.write( "confirm by replacing n in the original equation to get:<BR>\n" );
document.write( "f = 5579 * 1.00433333333 ^ 22.91116607 = 6160.<BR>\n" );
document.write( "your solution is that 5579 will grow to 6160 in 22.91116607 months at the rate of .00433333333 per month.\r<BR>\n" );
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document.write( "the annual rate if 5.2%.<BR>\n" );
document.write( "divide that by 12 to get a monthly rate of 5.2%/12.<BR>\n" );
document.write( "since 5.2% = .052, then you get a monthly rate of .052/12 = .00433333333<BR>\n" );
document.write( "the growth factor per month is 1 plus that = 1.00433333333.\r<BR>\n" );
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document.write( "here are the rules of log arithmetic.<BR>\n" );
document.write( "<a  rel=nofollow HREF= \"https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/\" target = \"_blank\">https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/</a><BR>\n" );
document.write( "the one used in this problem is rule 3.\r<BR>\n" );
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document.write( "  </SPAN>\n" );
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