document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1161154>Question 1161154</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1)Find the time required for an investment of 7,000 dollars to grow to 11,000 dollars at an interest rate of 6% per year, compounded monthly. Give your answer accurate to 2 decimal places.<BR>\n" );
document.write( "2) Joy invests $19,000 at 14% simple interest for 1 year. How much is in the account at the end of the 1 year period?<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #784609 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=784609\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 1)<BR>\n" );
document.write( "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( "r is the interest rate per time period<BR>\n" );
document.write( "n is the number of time periods<BR>\n" );
document.write( "your equation is:<BR>\n" );
document.write( "11000 = 7000 * (1+.06/12)^n<BR>\n" );
document.write( "your time period is in months.<BR>\n" );
document.write( "the interest rate per month is .06/12<BR>\n" );
document.write( "the number of months is n.<BR>\n" );
document.write( "divide both sides of that equation by 7000 to get:<BR>\n" );
document.write( "11000/7000 = (1+.06/12)^n<BR>\n" );
document.write( "take the log of both sides of this equation to get:<BR>\n" );
document.write( "log(11000/7000) = log((1+.06/12)^n)<BR>\n" );
document.write( "since log(x^a) = a*log(x), this becomes:<BR>\n" );
document.write( "log(11000/7000) = n*log((1+.06/12)<BR>\n" );
document.write( "divide both sides of this equation by log(1+.06/12)) and solve for n to get:<BR>\n" );
document.write( "n = log(11000/7000)/log(1+.06/12) = 90.62282945.<BR>\n" );
document.write( "this says that 7000 will become 11000 in 90.62282945 months.<BR>\n" );
document.write( "to confirm, replace n in the original equation to get:<BR>\n" );
document.write( "f = 7000*(1+.06/12)^90.6228i2945.<BR>\n" );
document.write( "solve for f to get:<BR>\n" );
document.write( "f = 11000<BR>\n" );
document.write( "this confirms the solution is correct.<BR>\n" );
document.write( "90.622812945 months divided by 12 is equivalent to 7.551902454 years.<BR>\n" );
document.write( "2)<BR>\n" );
document.write( "19000 * 1.14 = 21660.<BR>\n" );
document.write( "that's what's in the account at the end of the year.<BR>\n" );
document.write( "simple interest formula is f = p*p+r*n which can be simplified to f = p*(1+r*n)<BR>\n" );
document.write( "in this problem:<BR>\n" );
document.write( "p = 19000<BR>\n" );
document.write( "r=.14 per year<BR>\n" );
document.write( "n = 1 year.<BR>\n" );
document.write( "formula becomes f = 19000*(1+.14) = 19000 * 1.14 = 21660.\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );