document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1204592>Question 1204592</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An amount of money is invested at 12%/a interest. Write an equation and solve it to determine the length of time it will take for each of the following events to occur. Round answers to the nearest tenth of a year.<BR>\n" );
document.write( "- the original amount to double if interest is compounded annually<BR>\n" );
document.write( "- the original amount to triple if interest is compounded monthly </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #840931 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=840931\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> An amount of money is invested at 12%/a interest. Write an equation and solve it to determine the length of time it will take for each of the following events to occur. Round answers to the nearest tenth of a year.<BR>\n" );
document.write( "- the original amount to double if interest is compounded annually\\\r<BR>\n" );
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document.write( "A/P=2\r<BR>\n" );
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document.write( "A/P= (1+r)^n\r<BR>\n" );
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document.write( "2= 1.12^n\r<BR>\n" );
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document.write( "log 2= n log 1.12\r<BR>\n" );
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document.write( "log2/log1.12= n\r<BR>\n" );
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document.write( "n= 6.116 years  (approx)\r<BR>\n" );
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document.write( "- the original amount to triple if interest is compounded monthly\r<BR>\n" );
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document.write( "A/P=3<BR>\n" );
document.write( "r=12% = 0.12<BR>\n" );
document.write( "compounded monthly =0.01\r<BR>\n" );
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document.write( "A/P = 1.01^12t\r<BR>\n" );
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document.write( "3= 1.01^12t\r<BR>\n" );
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document.write( "log3= 12t log 1.01\r<BR>\n" );
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document.write( "log 3/12*log 1.01 =t\r<BR>\n" );
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document.write( "t= 9.2 years\r<BR>\n" );
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document.write( " </SPAN>\n" );
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