document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=937996>Question 937996</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A loan of $40,000 is made at 4.75% interest , compounded annually after how many years will the amount due reach 80,000 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #571375 by </B><A HREF=/tutors/aboutme.mpl?userid=lwsshak3><B>lwsshak3(11628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=lwsshak3><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=571375\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A loan of $40,000 is made at 4.75% interest , compounded annually after how many years will the amount due reach 80,000<BR>\n" );
document.write( "compound interest formula: A=P(1+r/n)^nt, P=initial investment, r=interest rate, n=number of compounding periods per year, A=amount after t-years.<BR>\n" );
document.write( "..<BR>\n" );
document.write( "For given problem:<BR>\n" );
document.write( "P=40000<BR>\n" );
document.write( "r=.0475<BR>\n" );
document.write( "n=1<BR>\n" );
document.write( "..<BR>\n" );
document.write( "80000=40000(1+.0475/1)^1*t<BR>\n" );
document.write( "80000/40000=2=(1.0475)^t<BR>\n" );
document.write( "(1.0475)^t=2<BR>\n" );
document.write( "take log of both sides<BR>\n" );
document.write( "tlog(1.0475)=log(2)<BR>\n" );
document.write( "t=log(2)/log(1.0475)<BR>\n" );
document.write( "t&#8776;14.9<BR>\n" );
document.write( "after how many years will the amount due reach 80,000? approximately 15 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );